the importance of sequences and infinite series in calculus stems from newton’s idea of...

113
The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series. For instance, in finding areas, he often integrated a function by first expressing it as a series and then integrating each term of the series. INFINITE SEQUENCES AND SERIES

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Page 1: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

The importance of sequences and infinite series in

calculus stems from Newtonrsquos idea of representing

functions as sums of infinite series

For instance in finding areas he often integrated a function by first expressing it as a series and then integrating each term of the series

INFINITE SEQUENCES AND SERIES

We will pursue his idea in a later section

in order to integrate such functions as e-x2

Recall that we have previously been unable to do this

INFINITE SEQUENCES AND SERIES

Many of the functions that arise in physics and

chemistry such as Bessel functions are defined as

sums of series

It is important to be familiar with the basic concepts of convergence of infinite sequences and series

INFINITE SEQUENCES AND SERIES

Physicists also use series in another way they use

them to approximate functions in certain situations

In studying fields as diverse as optics special relativity and electromagnetism they analyze phenomena by replacing a function with the first few terms in the series that represents it

INFINITE SEQUENCES AND SERIES

111

Sequences

In this section we will learn about

Various concepts related to sequences

INFINITE SEQUENCES AND SERIES

SEQUENCE

A sequence can be thought of as a list of numbers

written in a definite order

a1 a2 a3 a4 hellip an hellip

The number a1 is called the first term a2 is the second term and in general an is the nth term or general term

SEQUENCES

We will deal exclusively with infinite sequences

So each term an will have a successor an+1

SEQUENCES

Notice that for every positive integer n there is a

corresponding number an

So a sequence can be defined as

A function whose domain is the set of positive integers

SEQUENCES

However we usually write an instead of the function

notation f (n) for the value of the function at the

number n

The sequence a1 a2 a3 is also denoted by

1orn n n

a a

SEQUENCES

In the following examples we give three

descriptions of the sequence

1 Using the preceding notation

2 Using the defining formula

3 Writing out the terms of the sequence

Example 1

SEQUENCES

In this and the subsequent examples notice that n does not have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an of the

sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions start

with 3 and increase by 1 whenever we go to the

next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers of 5

Thus an has denominator 5n

The signs of the terms are alternately positive and

negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1) n meant we started

with a negative term

Here we want to start with a positive term

Thus we use (ndash1)nndash1 or (ndash1)n+1 Therefore

Example 2

1 2( 1)

5n

n n

na

SEQUENCES

We now look at some sequences that do not have a

simple defining equation

Example 3

SEQUENCES

The sequence pn where pn is the population of

the world as of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place of

the number e then an is a well-defined sequence

whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined recursively

by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n 3

Each is the sum of the two preceding term terms

The first few terms are

1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century Italian

mathematician Fibonacci solved a problem

concerning the breeding of rabbits

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can be

pictured either by Plotting its terms on a

number line Plotting its graph

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 2: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

We will pursue his idea in a later section

in order to integrate such functions as e-x2

Recall that we have previously been unable to do this

INFINITE SEQUENCES AND SERIES

Many of the functions that arise in physics and

chemistry such as Bessel functions are defined as

sums of series

It is important to be familiar with the basic concepts of convergence of infinite sequences and series

INFINITE SEQUENCES AND SERIES

Physicists also use series in another way they use

them to approximate functions in certain situations

In studying fields as diverse as optics special relativity and electromagnetism they analyze phenomena by replacing a function with the first few terms in the series that represents it

INFINITE SEQUENCES AND SERIES

111

Sequences

In this section we will learn about

Various concepts related to sequences

INFINITE SEQUENCES AND SERIES

SEQUENCE

A sequence can be thought of as a list of numbers

written in a definite order

a1 a2 a3 a4 hellip an hellip

The number a1 is called the first term a2 is the second term and in general an is the nth term or general term

SEQUENCES

We will deal exclusively with infinite sequences

So each term an will have a successor an+1

SEQUENCES

Notice that for every positive integer n there is a

corresponding number an

So a sequence can be defined as

A function whose domain is the set of positive integers

SEQUENCES

However we usually write an instead of the function

notation f (n) for the value of the function at the

number n

The sequence a1 a2 a3 is also denoted by

1orn n n

a a

SEQUENCES

In the following examples we give three

descriptions of the sequence

1 Using the preceding notation

2 Using the defining formula

3 Writing out the terms of the sequence

Example 1

SEQUENCES

In this and the subsequent examples notice that n does not have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an of the

sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions start

with 3 and increase by 1 whenever we go to the

next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers of 5

Thus an has denominator 5n

The signs of the terms are alternately positive and

negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1) n meant we started

with a negative term

Here we want to start with a positive term

Thus we use (ndash1)nndash1 or (ndash1)n+1 Therefore

Example 2

1 2( 1)

5n

n n

na

SEQUENCES

We now look at some sequences that do not have a

simple defining equation

Example 3

SEQUENCES

The sequence pn where pn is the population of

the world as of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place of

the number e then an is a well-defined sequence

whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined recursively

by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n 3

Each is the sum of the two preceding term terms

The first few terms are

1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century Italian

mathematician Fibonacci solved a problem

concerning the breeding of rabbits

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can be

pictured either by Plotting its terms on a

number line Plotting its graph

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 3: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

Many of the functions that arise in physics and

chemistry such as Bessel functions are defined as

sums of series

It is important to be familiar with the basic concepts of convergence of infinite sequences and series

INFINITE SEQUENCES AND SERIES

Physicists also use series in another way they use

them to approximate functions in certain situations

In studying fields as diverse as optics special relativity and electromagnetism they analyze phenomena by replacing a function with the first few terms in the series that represents it

INFINITE SEQUENCES AND SERIES

111

Sequences

In this section we will learn about

Various concepts related to sequences

INFINITE SEQUENCES AND SERIES

SEQUENCE

A sequence can be thought of as a list of numbers

written in a definite order

a1 a2 a3 a4 hellip an hellip

The number a1 is called the first term a2 is the second term and in general an is the nth term or general term

SEQUENCES

We will deal exclusively with infinite sequences

So each term an will have a successor an+1

SEQUENCES

Notice that for every positive integer n there is a

corresponding number an

So a sequence can be defined as

A function whose domain is the set of positive integers

SEQUENCES

However we usually write an instead of the function

notation f (n) for the value of the function at the

number n

The sequence a1 a2 a3 is also denoted by

1orn n n

a a

SEQUENCES

In the following examples we give three

descriptions of the sequence

1 Using the preceding notation

2 Using the defining formula

3 Writing out the terms of the sequence

Example 1

SEQUENCES

In this and the subsequent examples notice that n does not have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an of the

sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions start

with 3 and increase by 1 whenever we go to the

next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers of 5

Thus an has denominator 5n

The signs of the terms are alternately positive and

negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1) n meant we started

with a negative term

Here we want to start with a positive term

Thus we use (ndash1)nndash1 or (ndash1)n+1 Therefore

Example 2

1 2( 1)

5n

n n

na

SEQUENCES

We now look at some sequences that do not have a

simple defining equation

Example 3

SEQUENCES

The sequence pn where pn is the population of

the world as of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place of

the number e then an is a well-defined sequence

whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined recursively

by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n 3

Each is the sum of the two preceding term terms

The first few terms are

1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century Italian

mathematician Fibonacci solved a problem

concerning the breeding of rabbits

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can be

pictured either by Plotting its terms on a

number line Plotting its graph

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 4: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

Physicists also use series in another way they use

them to approximate functions in certain situations

In studying fields as diverse as optics special relativity and electromagnetism they analyze phenomena by replacing a function with the first few terms in the series that represents it

INFINITE SEQUENCES AND SERIES

111

Sequences

In this section we will learn about

Various concepts related to sequences

INFINITE SEQUENCES AND SERIES

SEQUENCE

A sequence can be thought of as a list of numbers

written in a definite order

a1 a2 a3 a4 hellip an hellip

The number a1 is called the first term a2 is the second term and in general an is the nth term or general term

SEQUENCES

We will deal exclusively with infinite sequences

So each term an will have a successor an+1

SEQUENCES

Notice that for every positive integer n there is a

corresponding number an

So a sequence can be defined as

A function whose domain is the set of positive integers

SEQUENCES

However we usually write an instead of the function

notation f (n) for the value of the function at the

number n

The sequence a1 a2 a3 is also denoted by

1orn n n

a a

SEQUENCES

In the following examples we give three

descriptions of the sequence

1 Using the preceding notation

2 Using the defining formula

3 Writing out the terms of the sequence

Example 1

SEQUENCES

In this and the subsequent examples notice that n does not have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an of the

sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions start

with 3 and increase by 1 whenever we go to the

next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers of 5

Thus an has denominator 5n

The signs of the terms are alternately positive and

negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1) n meant we started

with a negative term

Here we want to start with a positive term

Thus we use (ndash1)nndash1 or (ndash1)n+1 Therefore

Example 2

1 2( 1)

5n

n n

na

SEQUENCES

We now look at some sequences that do not have a

simple defining equation

Example 3

SEQUENCES

The sequence pn where pn is the population of

the world as of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place of

the number e then an is a well-defined sequence

whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined recursively

by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n 3

Each is the sum of the two preceding term terms

The first few terms are

1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century Italian

mathematician Fibonacci solved a problem

concerning the breeding of rabbits

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can be

pictured either by Plotting its terms on a

number line Plotting its graph

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 5: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

111

Sequences

In this section we will learn about

Various concepts related to sequences

INFINITE SEQUENCES AND SERIES

SEQUENCE

A sequence can be thought of as a list of numbers

written in a definite order

a1 a2 a3 a4 hellip an hellip

The number a1 is called the first term a2 is the second term and in general an is the nth term or general term

SEQUENCES

We will deal exclusively with infinite sequences

So each term an will have a successor an+1

SEQUENCES

Notice that for every positive integer n there is a

corresponding number an

So a sequence can be defined as

A function whose domain is the set of positive integers

SEQUENCES

However we usually write an instead of the function

notation f (n) for the value of the function at the

number n

The sequence a1 a2 a3 is also denoted by

1orn n n

a a

SEQUENCES

In the following examples we give three

descriptions of the sequence

1 Using the preceding notation

2 Using the defining formula

3 Writing out the terms of the sequence

Example 1

SEQUENCES

In this and the subsequent examples notice that n does not have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an of the

sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions start

with 3 and increase by 1 whenever we go to the

next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers of 5

Thus an has denominator 5n

The signs of the terms are alternately positive and

negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1) n meant we started

with a negative term

Here we want to start with a positive term

Thus we use (ndash1)nndash1 or (ndash1)n+1 Therefore

Example 2

1 2( 1)

5n

n n

na

SEQUENCES

We now look at some sequences that do not have a

simple defining equation

Example 3

SEQUENCES

The sequence pn where pn is the population of

the world as of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place of

the number e then an is a well-defined sequence

whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined recursively

by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n 3

Each is the sum of the two preceding term terms

The first few terms are

1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century Italian

mathematician Fibonacci solved a problem

concerning the breeding of rabbits

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can be

pictured either by Plotting its terms on a

number line Plotting its graph

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 6: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

SEQUENCE

A sequence can be thought of as a list of numbers

written in a definite order

a1 a2 a3 a4 hellip an hellip

The number a1 is called the first term a2 is the second term and in general an is the nth term or general term

SEQUENCES

We will deal exclusively with infinite sequences

So each term an will have a successor an+1

SEQUENCES

Notice that for every positive integer n there is a

corresponding number an

So a sequence can be defined as

A function whose domain is the set of positive integers

SEQUENCES

However we usually write an instead of the function

notation f (n) for the value of the function at the

number n

The sequence a1 a2 a3 is also denoted by

1orn n n

a a

SEQUENCES

In the following examples we give three

descriptions of the sequence

1 Using the preceding notation

2 Using the defining formula

3 Writing out the terms of the sequence

Example 1

SEQUENCES

In this and the subsequent examples notice that n does not have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an of the

sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions start

with 3 and increase by 1 whenever we go to the

next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers of 5

Thus an has denominator 5n

The signs of the terms are alternately positive and

negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1) n meant we started

with a negative term

Here we want to start with a positive term

Thus we use (ndash1)nndash1 or (ndash1)n+1 Therefore

Example 2

1 2( 1)

5n

n n

na

SEQUENCES

We now look at some sequences that do not have a

simple defining equation

Example 3

SEQUENCES

The sequence pn where pn is the population of

the world as of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place of

the number e then an is a well-defined sequence

whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined recursively

by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n 3

Each is the sum of the two preceding term terms

The first few terms are

1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century Italian

mathematician Fibonacci solved a problem

concerning the breeding of rabbits

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can be

pictured either by Plotting its terms on a

number line Plotting its graph

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 7: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

SEQUENCES

We will deal exclusively with infinite sequences

So each term an will have a successor an+1

SEQUENCES

Notice that for every positive integer n there is a

corresponding number an

So a sequence can be defined as

A function whose domain is the set of positive integers

SEQUENCES

However we usually write an instead of the function

notation f (n) for the value of the function at the

number n

The sequence a1 a2 a3 is also denoted by

1orn n n

a a

SEQUENCES

In the following examples we give three

descriptions of the sequence

1 Using the preceding notation

2 Using the defining formula

3 Writing out the terms of the sequence

Example 1

SEQUENCES

In this and the subsequent examples notice that n does not have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an of the

sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions start

with 3 and increase by 1 whenever we go to the

next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers of 5

Thus an has denominator 5n

The signs of the terms are alternately positive and

negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1) n meant we started

with a negative term

Here we want to start with a positive term

Thus we use (ndash1)nndash1 or (ndash1)n+1 Therefore

Example 2

1 2( 1)

5n

n n

na

SEQUENCES

We now look at some sequences that do not have a

simple defining equation

Example 3

SEQUENCES

The sequence pn where pn is the population of

the world as of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place of

the number e then an is a well-defined sequence

whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined recursively

by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n 3

Each is the sum of the two preceding term terms

The first few terms are

1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century Italian

mathematician Fibonacci solved a problem

concerning the breeding of rabbits

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can be

pictured either by Plotting its terms on a

number line Plotting its graph

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 8: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

SEQUENCES

Notice that for every positive integer n there is a

corresponding number an

So a sequence can be defined as

A function whose domain is the set of positive integers

SEQUENCES

However we usually write an instead of the function

notation f (n) for the value of the function at the

number n

The sequence a1 a2 a3 is also denoted by

1orn n n

a a

SEQUENCES

In the following examples we give three

descriptions of the sequence

1 Using the preceding notation

2 Using the defining formula

3 Writing out the terms of the sequence

Example 1

SEQUENCES

In this and the subsequent examples notice that n does not have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an of the

sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions start

with 3 and increase by 1 whenever we go to the

next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers of 5

Thus an has denominator 5n

The signs of the terms are alternately positive and

negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1) n meant we started

with a negative term

Here we want to start with a positive term

Thus we use (ndash1)nndash1 or (ndash1)n+1 Therefore

Example 2

1 2( 1)

5n

n n

na

SEQUENCES

We now look at some sequences that do not have a

simple defining equation

Example 3

SEQUENCES

The sequence pn where pn is the population of

the world as of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place of

the number e then an is a well-defined sequence

whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined recursively

by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n 3

Each is the sum of the two preceding term terms

The first few terms are

1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century Italian

mathematician Fibonacci solved a problem

concerning the breeding of rabbits

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can be

pictured either by Plotting its terms on a

number line Plotting its graph

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 9: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

SEQUENCES

However we usually write an instead of the function

notation f (n) for the value of the function at the

number n

The sequence a1 a2 a3 is also denoted by

1orn n n

a a

SEQUENCES

In the following examples we give three

descriptions of the sequence

1 Using the preceding notation

2 Using the defining formula

3 Writing out the terms of the sequence

Example 1

SEQUENCES

In this and the subsequent examples notice that n does not have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an of the

sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions start

with 3 and increase by 1 whenever we go to the

next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers of 5

Thus an has denominator 5n

The signs of the terms are alternately positive and

negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1) n meant we started

with a negative term

Here we want to start with a positive term

Thus we use (ndash1)nndash1 or (ndash1)n+1 Therefore

Example 2

1 2( 1)

5n

n n

na

SEQUENCES

We now look at some sequences that do not have a

simple defining equation

Example 3

SEQUENCES

The sequence pn where pn is the population of

the world as of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place of

the number e then an is a well-defined sequence

whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined recursively

by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n 3

Each is the sum of the two preceding term terms

The first few terms are

1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century Italian

mathematician Fibonacci solved a problem

concerning the breeding of rabbits

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can be

pictured either by Plotting its terms on a

number line Plotting its graph

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 10: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

SEQUENCES

In the following examples we give three

descriptions of the sequence

1 Using the preceding notation

2 Using the defining formula

3 Writing out the terms of the sequence

Example 1

SEQUENCES

In this and the subsequent examples notice that n does not have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an of the

sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions start

with 3 and increase by 1 whenever we go to the

next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers of 5

Thus an has denominator 5n

The signs of the terms are alternately positive and

negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1) n meant we started

with a negative term

Here we want to start with a positive term

Thus we use (ndash1)nndash1 or (ndash1)n+1 Therefore

Example 2

1 2( 1)

5n

n n

na

SEQUENCES

We now look at some sequences that do not have a

simple defining equation

Example 3

SEQUENCES

The sequence pn where pn is the population of

the world as of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place of

the number e then an is a well-defined sequence

whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined recursively

by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n 3

Each is the sum of the two preceding term terms

The first few terms are

1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century Italian

mathematician Fibonacci solved a problem

concerning the breeding of rabbits

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can be

pictured either by Plotting its terms on a

number line Plotting its graph

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 11: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

SEQUENCES

In this and the subsequent examples notice that n does not have to start at 1

Example 1 a

Preceding

Notation

Defining

Formula

Terms of

Sequence

11 n

n

n

1n

na

n

1 2 3 4

2 3 4 5 1

n

n

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an of the

sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions start

with 3 and increase by 1 whenever we go to the

next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers of 5

Thus an has denominator 5n

The signs of the terms are alternately positive and

negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1) n meant we started

with a negative term

Here we want to start with a positive term

Thus we use (ndash1)nndash1 or (ndash1)n+1 Therefore

Example 2

1 2( 1)

5n

n n

na

SEQUENCES

We now look at some sequences that do not have a

simple defining equation

Example 3

SEQUENCES

The sequence pn where pn is the population of

the world as of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place of

the number e then an is a well-defined sequence

whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined recursively

by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n 3

Each is the sum of the two preceding term terms

The first few terms are

1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century Italian

mathematician Fibonacci solved a problem

concerning the breeding of rabbits

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can be

pictured either by Plotting its terms on a

number line Plotting its graph

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 12: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

SEQUENCES Example 1 b

Preceding

Notation

Defining

Formula

Terms of

Sequence

( 1) ( 1)

3

n

n

n

( 1) ( 1)

3

n

n n

na

2 3 4 5

3 9 27 81

( 1) ( 1)

3

n

n

n

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an of the

sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions start

with 3 and increase by 1 whenever we go to the

next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers of 5

Thus an has denominator 5n

The signs of the terms are alternately positive and

negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1) n meant we started

with a negative term

Here we want to start with a positive term

Thus we use (ndash1)nndash1 or (ndash1)n+1 Therefore

Example 2

1 2( 1)

5n

n n

na

SEQUENCES

We now look at some sequences that do not have a

simple defining equation

Example 3

SEQUENCES

The sequence pn where pn is the population of

the world as of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place of

the number e then an is a well-defined sequence

whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined recursively

by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n 3

Each is the sum of the two preceding term terms

The first few terms are

1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century Italian

mathematician Fibonacci solved a problem

concerning the breeding of rabbits

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can be

pictured either by Plotting its terms on a

number line Plotting its graph

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 13: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

SEQUENCES Example 1 c

Preceding

Notation

Defining

Formula

Terms of

Sequence

3

3n

n

3

( 3)na n

n

01 2 3 3n

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an of the

sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions start

with 3 and increase by 1 whenever we go to the

next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers of 5

Thus an has denominator 5n

The signs of the terms are alternately positive and

negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1) n meant we started

with a negative term

Here we want to start with a positive term

Thus we use (ndash1)nndash1 or (ndash1)n+1 Therefore

Example 2

1 2( 1)

5n

n n

na

SEQUENCES

We now look at some sequences that do not have a

simple defining equation

Example 3

SEQUENCES

The sequence pn where pn is the population of

the world as of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place of

the number e then an is a well-defined sequence

whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined recursively

by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n 3

Each is the sum of the two preceding term terms

The first few terms are

1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century Italian

mathematician Fibonacci solved a problem

concerning the breeding of rabbits

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can be

pictured either by Plotting its terms on a

number line Plotting its graph

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 14: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

SEQUENCES Example 1 d

Preceding

Notation

Defining

Formula

Terms of

Sequence

0

cos6 n

n

cos6

( 0)

n

na

n

3 11 0 cos

2 2 6

n

SEQUENCES

Find a formula for the general term an of the

sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions start

with 3 and increase by 1 whenever we go to the

next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers of 5

Thus an has denominator 5n

The signs of the terms are alternately positive and

negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1) n meant we started

with a negative term

Here we want to start with a positive term

Thus we use (ndash1)nndash1 or (ndash1)n+1 Therefore

Example 2

1 2( 1)

5n

n n

na

SEQUENCES

We now look at some sequences that do not have a

simple defining equation

Example 3

SEQUENCES

The sequence pn where pn is the population of

the world as of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place of

the number e then an is a well-defined sequence

whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined recursively

by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n 3

Each is the sum of the two preceding term terms

The first few terms are

1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century Italian

mathematician Fibonacci solved a problem

concerning the breeding of rabbits

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can be

pictured either by Plotting its terms on a

number line Plotting its graph

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 15: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

SEQUENCES

Find a formula for the general term an of the

sequence

assuming the pattern of the first few terms

continues

3 4 5 6 7

5 25 125 625 3125

Example 2

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions start

with 3 and increase by 1 whenever we go to the

next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers of 5

Thus an has denominator 5n

The signs of the terms are alternately positive and

negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1) n meant we started

with a negative term

Here we want to start with a positive term

Thus we use (ndash1)nndash1 or (ndash1)n+1 Therefore

Example 2

1 2( 1)

5n

n n

na

SEQUENCES

We now look at some sequences that do not have a

simple defining equation

Example 3

SEQUENCES

The sequence pn where pn is the population of

the world as of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place of

the number e then an is a well-defined sequence

whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined recursively

by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n 3

Each is the sum of the two preceding term terms

The first few terms are

1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century Italian

mathematician Fibonacci solved a problem

concerning the breeding of rabbits

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can be

pictured either by Plotting its terms on a

number line Plotting its graph

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 16: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

SEQUENCES

We are given that

1 2 3

4 5

3 4 5

5 25 125

6 7

625 3125

a a a

a a

Example 2

SEQUENCES

Notice that the numerators of these fractions start

with 3 and increase by 1 whenever we go to the

next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers of 5

Thus an has denominator 5n

The signs of the terms are alternately positive and

negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1) n meant we started

with a negative term

Here we want to start with a positive term

Thus we use (ndash1)nndash1 or (ndash1)n+1 Therefore

Example 2

1 2( 1)

5n

n n

na

SEQUENCES

We now look at some sequences that do not have a

simple defining equation

Example 3

SEQUENCES

The sequence pn where pn is the population of

the world as of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place of

the number e then an is a well-defined sequence

whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined recursively

by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n 3

Each is the sum of the two preceding term terms

The first few terms are

1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century Italian

mathematician Fibonacci solved a problem

concerning the breeding of rabbits

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can be

pictured either by Plotting its terms on a

number line Plotting its graph

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 17: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

SEQUENCES

Notice that the numerators of these fractions start

with 3 and increase by 1 whenever we go to the

next term

The second term has numerator 4 and the third term has numerator 5

In general the nth term will have numerator n+2

Example 2

SEQUENCES

The denominators are the powers of 5

Thus an has denominator 5n

The signs of the terms are alternately positive and

negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1) n meant we started

with a negative term

Here we want to start with a positive term

Thus we use (ndash1)nndash1 or (ndash1)n+1 Therefore

Example 2

1 2( 1)

5n

n n

na

SEQUENCES

We now look at some sequences that do not have a

simple defining equation

Example 3

SEQUENCES

The sequence pn where pn is the population of

the world as of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place of

the number e then an is a well-defined sequence

whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined recursively

by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n 3

Each is the sum of the two preceding term terms

The first few terms are

1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century Italian

mathematician Fibonacci solved a problem

concerning the breeding of rabbits

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can be

pictured either by Plotting its terms on a

number line Plotting its graph

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 18: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

SEQUENCES

The denominators are the powers of 5

Thus an has denominator 5n

The signs of the terms are alternately positive and

negative

Hence we need to multiply by a power of ndash1

Example 2

SEQUENCES

In Example 1 b the factor (ndash1) n meant we started

with a negative term

Here we want to start with a positive term

Thus we use (ndash1)nndash1 or (ndash1)n+1 Therefore

Example 2

1 2( 1)

5n

n n

na

SEQUENCES

We now look at some sequences that do not have a

simple defining equation

Example 3

SEQUENCES

The sequence pn where pn is the population of

the world as of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place of

the number e then an is a well-defined sequence

whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined recursively

by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n 3

Each is the sum of the two preceding term terms

The first few terms are

1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century Italian

mathematician Fibonacci solved a problem

concerning the breeding of rabbits

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can be

pictured either by Plotting its terms on a

number line Plotting its graph

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 19: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

SEQUENCES

In Example 1 b the factor (ndash1) n meant we started

with a negative term

Here we want to start with a positive term

Thus we use (ndash1)nndash1 or (ndash1)n+1 Therefore

Example 2

1 2( 1)

5n

n n

na

SEQUENCES

We now look at some sequences that do not have a

simple defining equation

Example 3

SEQUENCES

The sequence pn where pn is the population of

the world as of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place of

the number e then an is a well-defined sequence

whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined recursively

by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n 3

Each is the sum of the two preceding term terms

The first few terms are

1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century Italian

mathematician Fibonacci solved a problem

concerning the breeding of rabbits

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can be

pictured either by Plotting its terms on a

number line Plotting its graph

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 20: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

SEQUENCES

We now look at some sequences that do not have a

simple defining equation

Example 3

SEQUENCES

The sequence pn where pn is the population of

the world as of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place of

the number e then an is a well-defined sequence

whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined recursively

by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n 3

Each is the sum of the two preceding term terms

The first few terms are

1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century Italian

mathematician Fibonacci solved a problem

concerning the breeding of rabbits

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can be

pictured either by Plotting its terms on a

number line Plotting its graph

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 21: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

SEQUENCES

The sequence pn where pn is the population of

the world as of January 1 in the year n

Example 3 a

SEQUENCES

If we let an be the digit in the nth decimal place of

the number e then an is a well-defined sequence

whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined recursively

by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n 3

Each is the sum of the two preceding term terms

The first few terms are

1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century Italian

mathematician Fibonacci solved a problem

concerning the breeding of rabbits

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can be

pictured either by Plotting its terms on a

number line Plotting its graph

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 22: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

SEQUENCES

If we let an be the digit in the nth decimal place of

the number e then an is a well-defined sequence

whose first few terms are

7 1 8 2 8 1 8 2 8 4 5 hellip

Example 3 b

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined recursively

by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n 3

Each is the sum of the two preceding term terms

The first few terms are

1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century Italian

mathematician Fibonacci solved a problem

concerning the breeding of rabbits

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can be

pictured either by Plotting its terms on a

number line Plotting its graph

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 23: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

FIBONACCI SEQUENCE

The Fibonacci sequence fn is defined recursively

by the conditions

f1 = 1 f2 = 1 fn = fnndash1 + fnndash2 n 3

Each is the sum of the two preceding term terms

The first few terms are

1 1 2 3 5 8 13 21 hellip

Example 3 c

FIBONACCI SEQUENCE

This sequence arose when the 13th-century Italian

mathematician Fibonacci solved a problem

concerning the breeding of rabbits

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can be

pictured either by Plotting its terms on a

number line Plotting its graph

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 24: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

FIBONACCI SEQUENCE

This sequence arose when the 13th-century Italian

mathematician Fibonacci solved a problem

concerning the breeding of rabbits

Example 3

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can be

pictured either by Plotting its terms on a

number line Plotting its graph

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 25: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

SEQUENCES

A sequence such as

that in Example 1 a

[an = n(n + 1)] can be

pictured either by Plotting its terms on a

number line Plotting its graph

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 26: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

SEQUENCES

Note that since

a sequence is a

function whose

domain is the set

of positive integers

its graph consists of isolated points

(1 a1) (2 a2) (3 a3) hellip (n an) hellip

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 27: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

SEQUENCES

From either figure

it appears that

the terms of

the sequence

an = n(n + 1) are

approaching 1 as n

becomes large

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 28: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

SEQUENCES

In fact the difference

can be made as small as we like by taking n

sufficiently large

We indicate this by writing

11

1 1

n

n n

lim 11n

n

n

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 29: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

SEQUENCES

In general the notation

means that the terms of the sequence an get

arbitrarily close to L as n becomes large

This of course needs to be defined more precisely

lim nn

a L

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 30: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

SEQUENCES

Notice that the following definition of the limit of

a sequence is very similar to the definition of a

limit of a function at infinity given in Section 14

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 31: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMIT OF A SEQUENCE

A sequence an has the limit L and we write

if we can make the terms an as close to L as we

like by taking n sufficiently large

If exists the sequence converges (or is convergent)

Otherwise it diverges (or is divergent)

lim or asn nna L a L n

Definition 1

lim nn

a

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 32: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMIT OF A SEQUENCE

Here Definition 1 is illustrated by showing the

graphs of two sequences that have the limit L

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 33: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMIT OF A SEQUENCE

A more precise version of Definition 1 is as

follows

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 34: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMIT OF A SEQUENCE

A sequence an has the limit L if for every ε gt 0

there is a corresponding integer N such that

if n gt N then |an ndash L| lt ε

In this case we write

lim or asn nna L a L n

Definition 2

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 35: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMIT OF A SEQUENCE

Definition 2 is illustrated by the figure in which

the terms a1 a2 a3 are plotted on a number

line

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 36: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMIT OF A SEQUENCE

No matter how small an interval (L ndash ε L + ε) is

chosen there exists an N such that all terms of the

sequence from aN+1 onward must lie in that interval

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 37: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMIT OF A SEQUENCE

Another illustration of Definition 2 is given here

The points on the graph of an must lie between the horizontal lines y = L + ε and y = L ndash ε if n gt N

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 38: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMIT OF A SEQUENCE

This picture must be valid no matter how small ε is

chosen

Usually however a smaller ε requires a larger N

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 39: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

If you compare Definition 2 with Definition 7 in

Section 14 you will see that the only difference

between and is that n is

required to be an integer

Thus we have the following theorem

lim nna L

lim ( )

xf x L

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 40: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

If and f (n) = an when n is an integer

then

lim ( )x

f x L

lim nna L

Theorem 3

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 41: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

Theorem 3 is illustrated here

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 42: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

In particular since we know that

when r gt 0 we have

lim(1 ) 0r

xx

Equation 4

1lim 0 if 0

rnr

n

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 43: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

If an becomes large as n becomes large we use the

notation

The following precise definition is similar to that given for functions

lim nn

a

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 44: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMIT OF A SEQUENCE

means that for every positive number

M there is an integer N such that

if n gt N then an gt M

Definition 5

lim nn

a

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 45: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

If then the sequence an is divergent

but in a special way

We say that an diverges to

lim nn

a

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 46: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

The Limit Laws given in Section 13 also hold for

the limits of sequences and their proofs are similar

Suppose an and bn are convergent sequences

and c is a constant

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 47: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMIT LAWS FOR SEQUENCES

lim( ) lim lim

lim( ) lim lim

lim lim lim

n n n nn n n

n n n nn n n

n nn n n

a b a b

a b a b

ca c a c c

Then

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 48: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMIT LAWS FOR SEQUENCES

Alsolim( ) lim lim

limlim if lim 0

lim

lim lim if 0 and 0

n n n nn n n

nn n

nn n

n nn

PPn n n

n n

a b a b

aab

b b

a a p a

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 49: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

The Squeeze Theorem can also be adapted for

sequences as follows

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 50: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

SQUEEZE THEOREM FOR SEQUENCES

If an bn cn for n n0 and

thenlim nn

b L

lim limn nn n

a c L

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 51: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

Another useful fact about limits of sequences is

given by the following theorem

The proof is left as an exercise

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 52: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

Theorem 6

lim 0nna

lim 0n

na

If then

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 53: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

Find

The method is similar to the one we used in Section 26

We divide the numerator and denominator by the highest power of n and then use the Limit Laws

Example 4

lim1n

n

n

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 54: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

Thus

Here we used Equation 4 with r = 1

lim11lim lim

1 11 1 lim1 lim

11

1 0

n

n n

n n

n

nn n

Example 4

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 55: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

Calculate

Notice that both the numerator and denominator approach infinity as n rarr infin

Example 5

lnlimn

n

n

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 56: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

Here we cannot apply lrsquoHospitalrsquos Rule directly

It applies not to sequences but to functions of a real variable

Example 5

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 57: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

However we can apply lrsquoHospitalrsquos Rule to the

related function f (x) = (ln x)x and obtain

Example 5

ln 1lim lim 0

1x x

x x

x

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 58: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

Therefore by Theorem 3 we have

Example 5

lnlim 0n

n

n

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 59: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

Determine whether the sequence an = (ndash1)n is

convergent or divergent

If we write out the terms of the sequence we

obtain

ndash1 1 ndash1 1 ndash1 1 ndash1 hellip

Example 6

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 60: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

The graph of the sequence is shown

The terms oscillate between 1 and ndash1 infinitely often Thus an does not approach any number

Example 6

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 61: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

Thus does not exist

That is the sequence (ndash1)n is divergent

Example 6

lim( 1)n

n

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 62: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

Evaluate if it exists

Thus by Theorem 6

Example 7

( 1)lim

n

n n

( 1) 1lim lim 0

n

n nn n

( 1)lim 0

n

n n

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 63: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

The following theorem says that if we apply a

continuous function to the terms of a convergent

sequence the result is also convergent

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 64: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

If and the function f is continuous at L

then

The proof is left as an exercise

Theorem 7

lim nn

a L

lim ( ) ( )nn

f a f L

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 65: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

Find

The sine function is continuous at 0 Thus Theorem 7 enables us to write

Example 8

lim sin( )n

n

lim sin( ) sin lim( )

sin 0 0n n

n n

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 66: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

Discuss the convergence of the sequence an =nnn

where n = 1 2 3 hellip N

Both the numerator and denominator approach

infinity as n rarr infin

However here we have no corresponding function

for use with lrsquoHospitalrsquos Rule

x is not defined when x is not an integer

Example 9

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 67: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

Let us write out a few terms to get a feeling for

what happens to an as n gets large

Therefore

Example 9

1 2 3

1 2 1 2 31

2 2 3 3 3a a a

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 68: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

From these expressions and the graph here it

appears that the terms are decreasing and perhaps

approach 0

Example 9

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 69: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

To confirm this observe from Equation 8 that

Notice that the expression in parentheses is at most 1 because the numerator is less than (or equal to) the denominator

Example 9

1 2 3n

na

n n n n

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 70: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

Thus

We know that 1n rarr 0 as n rarr infin

Therefore an rarr 0 as n rarr infin by the Squeeze Theorem

Example 9

10 na n

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 71: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

For what values of r is the sequence r n

convergent

From the graphs of the exponential functions we know that for a gt 1 and for 0 lt a lt 1

Example 10

lim x

xa

lim 0x

xa

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 72: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

Thus putting a = r and using Theorem 3 we have

It is obvious that

Example 10

if 1lim

0 if 0 1n

n

rr

r

lim1 1 and lim 0 0n n

n n

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 73: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

If ndash1 lt r lt 0 then 0 lt |r | lt 1

Thus

Therefore by Theorem 6

Example 10

lim | | lim | | 0n n

n nr r

lim 0n

nr

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 74: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

If r ndash1 then r n diverges as in Example 6

Example 10

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 75: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

The figure shows the graphs for various values of r

Example 10

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 76: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

The case r = ndash1 was shown earlier

Example 10

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 77: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

The results of Example 10 are summarized for

future use as follows

The sequence r n is convergent if ndash1 lt r 1 and

divergent for all other values of r

Equation 9

0 if 1 1lim

1 if 1n

n

rr

r

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 78: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

LIMITS OF SEQUENCES

A sequence an is called

Increasing if an lt an+1 for all n 1 that is

a1 lt a2 lt a3 lt lt an lt

Decreasing if an gt an+1 for all n 1

Monotonic if it is either increasing or decreasing

Definition 10

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 79: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

DECREASING SEQUENCES

The sequence is decreasing because

and so an gt an+1 for all n 1

Example 11

3

5n 3 3 3

5 ( 1) 5 6n n n

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 80: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

DECREASING SEQUENCES

Show that the sequence

is decreasing

Example 12

2 1n

na

n

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 81: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

DECREASING SEQUENCES

We must show that an+1 lt an that is

2 2

1

( 1) 1 1

n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 82: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

DECREASING SEQUENCES

This inequality is equivalent to the one we get by

cross-multiplication

2 22 2

3 2 3 2

2

1( 1)( 1) [( 1) 1]

( 1) 1 1

1 2 2

1

n nn n n n

n n

n n n n n n

n n

E g 12mdashSolution 1

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 83: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

DECREASING SEQUENCES

Since n 1 we know that the inequality n2 + n gt 1

is true

Therefore an+1 lt an

Hence an is decreasing

E g 12mdashSolution 1

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 84: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

DECREASING SEQUENCES

Consider the function 2( )

1

xf x

x

2 2

2 2

22

2 2

1 2( )

( 1)

10 whenever 1

( 1)

x xf x

x

xx

x

E g 12mdashSolution 2

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 85: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

DECREASING SEQUENCES

Thus f is decreasing on (1 infin)

Hence f (n) gt f (n + 1)

Therefore an is decreasing

E g 12mdashSolution 2

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 86: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

BOUNDED SEQUENCES

A sequence an is bounded

Above if there is a number M such that an M for all n 1

Below if there is a number m such that m an for all n 1

If it is bounded above and below

Definition 11

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 87: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

BOUNDED SEQUENCES

For instance

The sequence an = n is bounded below (an gt 0) but not above

The sequence an = n(n+1) is bounded because 0 lt an lt 1 for all n

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 88: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

BOUNDED SEQUENCES

We know that not every bounded sequence is

convergent

For instance the sequence an = (ndash1)n satisfies ndash1 an 1 but is divergent from Example 6

Similarly not every monotonic sequence is

convergent (an = n rarr infin)

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 89: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

BOUNDED SEQUENCES

However if a sequence is both bounded and

monotonic then it must be convergent

This fact is proved as Theorem 12

However intuitively you can understand why it is true by looking at the following figure

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 90: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

BOUNDED SEQUENCES

If an is increasing and an M for all n then the

terms are forced to crowd together and approach

some number L

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 91: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

BOUNDED SEQUENCES

The proof of Theorem 12 is based on the

Completeness Axiom for the set of real numbers

This states that if S is a nonempty set of real numbers that has an upper bound M (x M for all x in S) then S has a least upper bound b

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 92: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

BOUNDED SEQUENCES

This means

b is an upper bound for S

However if M is any other upper bound then b M

The Completeness Axiom is an expression of the

fact that there is no gap or hole in the real number

line

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 93: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

Every bounded monotonic sequence is convergent

Theorem 12MONOTONIC SEQ THEOREM

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 94: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

MONOTONIC SEQ THEOREM

Suppose an is an increasing sequence

Since an is bounded the set S = an|n 1 has an upper bound

By the Completeness Axiom it has a least upper bound L

Theorem 12mdashProof

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 95: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

Given ε gt 0 L ndash ε is not an upper bound

for S (since L is the least upper bound)

Therefore aN gt L ndash ε for some integer N

MONOTONIC SEQ THEOREM Theorem 12mdashProof

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 96: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

However the sequence is increasing

So an aN for every n gt N

Thus if n gt N we have an gt L ndash ε

Since an le L thus 0 le L ndash an lt ε

MONOTONIC SEQ THEOREM Theorem 12mdashProof

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 97: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

MONOTONIC SEQ THEOREM

Thus

|L ndash an| lt ε whenever n gt N

Therefore

Theorem 12mdashProof

lim nn

a L

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 98: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

MONOTONIC SEQ THEOREM

A similar proof (using the greatest lower bound)

works if an is decreasing

Theorem 12mdashProof

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 99: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

MONOTONIC SEQ THEOREM

The proof of Theorem 12 shows that a sequence

that is increasing and bounded above is

convergent

Likewise a decreasing sequence that is bounded below is convergent

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 100: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

MONOTONIC SEQ THEOREM

This fact is used many times in dealing with

infinite series

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 101: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

MONOTONIC SEQ THEOREM

Investigate the sequence an defined by the

recurrence relation

Example 13

11 1 22 ( 6) for 123n na a a n

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 102: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

MONOTONIC SEQ THEOREM

We begin by computing the first several terms

These initial terms suggest the sequence is increasing and the terms are approaching 6

Example 13

1 11 2 32 2

14 5 62

7 8 9

2 (2 6) 4 (4 6) 5

(5 6) 55 575 5875

59375 596875 5984375

a a a

a a a

a a a

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 103: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

MONOTONIC SEQ THEOREM

To confirm that the sequence is increasing we use

mathematical induction to show

that an+1 gt an for all n 1

Mathematical induction is often used in dealing with recursive sequences

Example 13

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 104: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

MONOTONIC SEQ THEOREM

That is true for n = 1 because a2 = 4 gt a1

Example 13

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 105: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

MONOTONIC SEQ THEOREM

If we assume that it is true for n = k

we have

Hence

and

Thus

Example 13

1k ka a

1 6 6k ka a 1 1

12 2( 6) ( 6)k ka a

2 1k ka a

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 106: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

MONOTONIC SEQ THEOREM

We have deduced that an+1 gt an is true for n = k + 1

Therefore the inequality is true for all n by

induction

Example 13

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 107: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

MONOTONIC SEQ THEOREM

Next we verify that an is bounded by showing

that an lt 6 for all n

Since the sequence is increasing we already know that it has a lower bound an a1 = 2 for all n

Example 13

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 108: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

MONOTONIC SEQ THEOREM

We know that a1 lt 6

So the assertion is true for n = 1

Example 13

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 109: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

MONOTONIC SEQ THEOREM

Suppose it is true for n = k

Then

Thus

and

Hence

6ka

6 12ka

Example 13

1 12 2( 6) (12) 6ka

1 6ka

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 110: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

MONOTONIC SEQ THEOREM

This shows by mathematical induction that an lt 6

for all n

Example 13

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 111: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

MONOTONIC SEQ THEOREM

Since the sequence an is increasing and bounded

Theorem 12 guarantees that it has a limit

However the theorem does not tell us what the value of the limit is

Example 13

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 112: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

MONOTONIC SEQ THEOREM

Nevertheless now that we know exists

we can use the recurrence relation to write

lim nn

L a

11 2

12

12

lim lim ( 6)

(lim 6)

( 6)

n nn n

nn

a a

a

L

Example 13

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13

Page 113: The importance of sequences and infinite series in calculus stems from Newton’s idea of representing functions as sums of infinite series.  For instance,

MONOTONIC SEQ THEOREM

Since an rarr L it follows that an+1 rarr L too

(as n rarr n + 1 rarr too)

Thus we have

Solving this equation for L we get L = 6 as predicted

12 ( 6)L L

Example 13