this is an example of an infinite series. 1 1 start with a square one unit by one unit: this series...

This is an example of an This is an example of an infinite seriesinfinite series..

1

1

Start with a square one Start with a square one unit by one unit:unit by one unit:

1

21

21

4

1

4

1

81

8

1

161

16

1

32 1

64

1

32

1

64 1

This series This series convergesconverges (approaches a limiting value.)(approaches a limiting value.)

Many series do not converge:Many series do not converge:1 1 1 1 1

1 2 3 4 5

9.1 Power Series

In an infinite series:In an infinite series: 1 2 31

n kk

a a a a a

aa11, a, a22,…,… are are termsterms of the series. of the series. aann is the is the nnthth term term..

Partial sums:Partial sums:1 1S a

2 1 2S a a

3 1 2 3S a a a

1

n

n kk

S a

nnthth partial sum partial sum

If If SSnn has a limit as , then the series converges, has a limit as , then the series converges,

otherwise it otherwise it divergesdiverges..

n

9.1 Power Series

Geometric Series:

In a In a geometric seriesgeometric series, each term is found by multiplying , each term is found by multiplying

the preceding term by the same number, the preceding term by the same number, rr..

2 3 1 1

1

n n

n

a ar ar ar ar ar

This converges to if , and diverges if .This converges to if , and diverges if .1

a

r1r 1r

1 1r is the is the interval of convergenceinterval of convergence..

9.1 Power Series

3 3 3 3

10 100 1000 10000

.3 .03 .003 .0003 .333... 1

3

310

11

10

aa

rr

3109

10

3

9

1

3

9.1 Power Series

1 1 11

2 4 8

11

12

11

12

13

2

2

3

aa

rr

9.1 Power Series

A power series is in this form:

c x c c x c x c x c xnn

nn

n

0 1 22

33

0or

c x a c c x a c x a c x a c x ann

nn

n

( ) ( ) ( ) ( ) ( )

0 1 22

33

0

The coefficients c0, c1, c2… are constants.

The center “a” is also a constant.

(The first series would be centered at the origin if you graphed it. The second series would be shifted left or right. “a” is the new center.)

9.1 Power Series

The partial sum of a geometric series is:The partial sum of a geometric series is: 1

1

n

n

a rS

r

If thenIf then1r 1

lim1

n

n

a r

r

1

a

r

0

If and we let , then:If and we let , then:1x r x

2 31 x x x 1

1 x

9.1 Power Series

1 1 x1

1 xx

x

2x x

2x

2x2 3x x

3x

3x We could generate this same series for

with polynomial long division:

1

1 x

9.1 Power Series

Once we have a series that we know, we can find a new Once we have a series that we know, we can find a new series by doing the same thing to the left and right hand series by doing the same thing to the left and right hand sides of the equation.sides of the equation.

This is a geometric series where This is a geometric series where r=-xr=-x..

1

x

xTo find a series forTo find a series for multiply both sides by multiply both sides by xx..

2 311

1x x x

x

2 3 4

1

xx x x x

x

1

1 x

9.1 Power Series

Given:Given: 2 311

1x x x

x

find:find: 2

1

1 x

1

1

d

dx x 11

dx

dx 2

1 1x

2

1

1 x

So:So:

2 32

11

1

dx x x

dxx

2 31 2 3 4x x x We differentiated term by term.We differentiated term by term.

9.1 Power Series

Given:Given: 2 311

1x x x

x

find:find: ln 1 x

1ln 1

1dx x c

x

2 311

1t t t

t

hmm?hmm?

9.1 Power Series

2 311

1t t t

t

2 3

0 0

11

1

x xdt t t t dt

t

2 3 4

00

1 1 1ln 1

2 3 4

xx

t t t t t

2 3 41 1 1ln 1 ln 1 0

2 3 4x x x x x

2 3 41 1 1ln 1

2 3 4x x x x x 1 1x

9.1 Power Series

2 3 41 1 1ln 1

2 3 4x x x x x 1 1x

The previous examples of infinite series approximated The previous examples of infinite series approximated simple functions such as or .simple functions such as or .1

3

1

1 x

This series would allow us to calculate a transcendental This series would allow us to calculate a transcendental function to as much accuracy as we like using only function to as much accuracy as we like using only pencil and paper!pencil and paper!

9.1 Power Series

Brook Taylor1685 - 1731

Brook Taylor was an accomplished musician and painter. He did research in a variety of areas, but is most famous for his development of ideas regarding infinite series.

9.2 Taylor Series

Suppose we wanted to find a fourth degree polynomial of the form:

2 3 40 1 2 3 4P x a a x a x a x a x

ln 1f x x at 0x that approximates the behavior of

If we make , and the first, second, third and fourth derivatives the same, then we would have a pretty good approximation.

0 0P f

9.2 Taylor Series

ln 1f x x

0 ln 1 0f

2 3 40 1 2 3 4P x a a x a x a x a x

00P a 0 0a

1

1f x

x

10 1

1f

2 31 2 3 42 3 4P x a a x a x a x

10P a 1 1a

2

1

1f x

x

10 1

1f

22 3 42 6 12P x a a x a x

20 2P a 2

1

2a

9.2 Taylor Series

3

12

1f x

x

0 2f

3 46 24P x a a x

30 6P a 3

2

6a

4

4

16

1f x

x

4 0 6f

4424P x a

440 24P a 4

6

24a

2

1

1f x

x

10 1

1f

22 3 42 6 12P x a a x a x

20 2P a 2

1

2a

9.2 Taylor Series

2 3 41 2 60 1

2 6 24P x x x x x

2 3 4

02 3 4

x x xP x x

ln 1f x x

P x

f x

If we plot both functions, we see that near zero the functions match very well!

9.2 Taylor Series

This pattern occurs no matter what the original function was!

Our polynomial: 2 3 41 2 60 1

2 6 24x x x x

has the form: 42 3 40 0 0

0 02 6 24

f f ff f x x x x

or: 42 3 40 0 0 0 0

0! 1! 2! 3! 4!

f f f f fx x x x

9.2 Taylor Series

Maclaurin Series:

(generated by f at )0x

2 30 00 0

2! 3!

f fP x f f x x x

9.2 Taylor Series

If we want to center the series (and it’s graph) at some point other than zero, we get the Taylor Series:

Taylor Series:

(generated by f at )x a

2 3

2! 3!

f a f aP x f a f a x a x a x a

9.2 Taylor Series

2 3 4 5 61 0 1 0 1

1 0 2! 3! 4! 5! 6!

x x x x xP x x

cosf x x 0 1f

sinf x x 0 0f

cosf x x 0 1f

sinf x x 0 0f

4 cosf x x 4 0 1f

2 4 6 8 10

1 2! 4! 6! 8! 10!

x x x x xP x

9.2 Taylor Series

cosy x 2 4 6 8 10

1 2! 4! 6! 8! 10!

x x x x xP x

The more terms we add, the better our approximation.

9.2 Taylor Series

cos 2y x

Rather than start from scratch, we can use the function that we already know:

2 4 6 8 102 2 2 2 2

1 2! 4! 6! 8! 10!

x x x x xP x

9.2 Taylor Series

example: cos at 2

y x x

2 3

0 10 1

2 2! 2 3! 2P x x x x

cosf x x 02

f

sinf x x 12

f

cosf x x 02

f

sinf x x 12

f

4 cosf x x 4 02

f

3 5

2 2

2 3! 5!

x xP x x

9.2 Taylor Series

There are some Maclaurin series that occur often enough that they should be memorized. They are on pg 477

9.2 Taylor Series

)1|(|......11

1

0

2

xxxxxx n

nn

)1|(|)1(...)(...11

1

0

2

xxxxxx n

nnn

)(!

...!

...!2

10

2

xrealalln

x

n

xxxe

n

nnx

9.2 Taylor Series

0

121253

)!12()1(

)!12()1(...

!5!3sin

n

nn

nn

n

x

n

xxxxx

0

2242

)!2()1(

)!2()1(...

!4!21cos

n

nn

nn

n

x

n

xxxx

3 5 7

1tan 3 5 7

x x xx x

0

)12(

)12()1(

n

nn

n

x

When referring to Taylor polynomials, we can talk about number of terms, order or degree.

2 4

cos 12! 4!

x xx This is a polynomial in 3 terms.

9.2 Taylor Series

It is a 4th order Taylor polynomial, because it was found using the 4th derivative.

It is also a 4th degree polynomial, because x is raised to the 4th power.

The 3rd order polynomial for is , but it is degree 2.

cos x2

12!

x

The x3 term drops out when using the third derivative.

This is also the 2nd order polynomial.

2 4

cos 12! 4!

x xx

9.2 Taylor Series

2 30 00 0

2! 3!

f fP x f f x x x

11

11

xx

11 x

21 x

32 1 x

46 1 x

524 1 x

nf x

List the function and itsderivatives.

9.2 Taylor Series

Evaluate column onefor x = 0.

2 30 00 0

2! 3!

f fP x f f x x x

111

1x

x

11 x

21 x

32 1 x

46 1 x

524 1 x

1

1

2

6 3!

24 4!

0nf nf x

2 3 42 31

1 2! 3!

! 4!1

!1

4x x x x

x

2 3 411

1x x x x

x

This is a geometric series witha = 1 and r = x.

9.2 Taylor Series

1 1 x1

1 xx

x

2x x

2x

2x2 3x x

3x

3x We could generate this same series for

with polynomial long division:

1

1 x

9.2 Taylor Series

2 30 00 0

2! 3!

f fP x f f x x x

11

11

xx

11 x

21 x

32 1 x

46 1 x

524 1 x

1

1

2

6 3!

24 4!

0nf nf x 2 3 421

1 2! 3! 4!

3! 4!1 1x x x x

x

2 3 411

1x x x x

x

This is a geometric series witha = 1 and r = -x.

9.2 Taylor Series

We wouldn’t expect to use the previous two series to evaluate the functions, since we can evaluate the functions directly.

We will find other uses for these series, as well.

They do help to explain where the formula for the sum of an infinite geometric comes from.

A more impressive use of Taylor series is to evaluate transcendental functions.

9.2 Taylor Series

2 30 00 0

2! 3!

f fP x f f x x x

cos x

cos x

sin x

cos x

sin x

cos x

1

0

1

0

1

0nf nf x 2 3 4cos

2! 3!

1 0

4!

11 0x x x x x

2 4 6

cos 1 2! 4! 6!

x x xx

Both sides are even functions.

Cos (0) = 1 for both sides.

9.2 Taylor Series

2 30 00 0

2! 3!

f fP x f f x x x

sin x

sin x

cos x

sin x

cos x

sin x

01

0

10

0nf nf x

2 3 4sin2

0 1 00 1

! 3! 4!x x x x x

3 5 7

sin 3! 5! 7!

x x xx x

Both sides are odd functions.

Sin (0) = 0 for both sides.

9.2 Taylor Series

2 3 411

1x x x x

x

2

1

1 x

If we start with this function:

and substitute for , we get:2x x 2 4 6 8

2

11

1x x x x

x

This is a geometric series with a = 1 and r = -x2.

9.2 Taylor Series

2 4 6 82

11

1x x x x

x

If we integrate both sides:

2 4 6 82

1 1

1dx x x x x dx

x

3 5 7

1tan 3 5 7

x x xx x

This looks the same as the series for sin (x), but without the factorials.

9.2 Taylor Series

2 30 00 0

2! 3!

f fP x f f x x x

ln 1 x

ln 1 x

11 x

21 x

32 1 x

46 1 x

0

1

1

2

6 3!

0nf nf x

2 3 4ln 12

1

! 3! 4

20 1

!

3!x x x x x

2 3 4

ln 12 3 4

x x xx x

9.2 Taylor Series

Taylor series are used to estimate the value of functions (at least theoretically - now days we can usually use the calculator or computer to calculate directly.)

An estimate is only useful if we have an idea of how accurate the estimate is.

When we use part of a Taylor series to estimate the value of a function, the end of the series that we do not use is called the remainder. If we know the size of the remainder, then we know how close our estimate is.

9.3 Taylor’s Theorem

For a geometric series, this is easy:

Use to approximate over .2 4 61 x x x 2

1

1 x 1,1

Since the truncated part of the series is: ,8 10 12 x x x

the truncation error is , which is .8 10 12 x x x 8

21

x

x

When you “truncate” a number, you drop off the end.

Of course this is also trivial, because we have a formula that allows us to calculate the sum of a geometric series directly.

9.3 Taylor’s Theorem

Taylor’s Theorem with Remainder

If f has derivatives of all orders in an open interval I containing a, then for each positive integer n and for each x

in I:

2

2! !

nn

n

f fa af f f Rx a a x a x a x a xn

9.3 Taylor’s Theorem

convergesseriesTaylorthethen,0)(If xRn

Lagrange Form of the Remainder

11

1 !

nn

n

f cR x x a

n

Remainder after partial sum Sn

where c is between

a and x.

9.3 Taylor’s Theorem

This is also called the remainder of order n or the error term.

Note that this looks just like the next term in the series, but

“a” has been replaced by the number “c” in . 1nf c

This seems kind of vague, since we don’t know the value of c,

but we can sometimes find a maximum value for . 1nf c

9.3 Taylor’s Theorem

We will call this the Remainder Estimation Theorem.

Lagrange Form of the Remainder

11

1 !

nn

n

f cR x x a

n

Remainder Estimation TheoremNote that this is not the formula that is in our book. It is from another textbook.

If M is the maximum value of on the interval

between a and x, then:

1nf x

1

1 !n

n

MR x x a

n

9.3 Taylor’s Theorem

Prove that , which is the Taylor

series for sin x, converges for all real x.

2 1

0 1!2 1

kk

k

x

k

9.3 Taylor’s Theorem

Since the maximum value of sin x or any of it’s

derivatives is 1, for all real x, M = 1.

110

!1n

nR x xn

1

!1

nx

n

1

lim 0!1

n

n

x

n

so the series converges.

Remainder Estimation Theorem

1

1 !n

n

MR x x a

n

9.3 Taylor’s Theorem

Find the Lagrange Error Bound when is used

to approximate and .

2

2

xx

ln 1 x 0.1x

ln 1f x x

11f x x

21f x x

32 1f x x

22

0 00

1 2!

f ff f x x Rx x

Remainder after 2nd order term

2

202

xf x Rx x

On the interval , decreases, so

its maximum value occurs at the left end-point.

.1,.1 3

2

1 x

3

2

1 .1M

3

2

.9 2.74348422497

9.3 Taylor’s Theorem

Find the Lagrange Error Bound when is used

to approximate and .

2

2

xx

ln 1 x 0.1x

On the interval , decreases, so

its maximum value occurs at the left end-point.

.1,.1 3

2

1 x

3

2

1 .1M

3

2

.9 2.74348422497

Remainder Estimation Theorem

1

1 !n

n

MR x x a

n

32.7435 .1

3!nR x

0.000457nR x

Lagrange Error Bound

x ln 1 x2

2

xx error

.1 .0953102 .095 .000310

.1 .1053605 .105 .000361

Error is less than error bound.

9.3 Taylor’s Theorem

We have saved the best for last!

9.3 Taylor Series

2 30 00 0

2! 3!

f fP x f f x x x

xe

xexexexexe

1

1

1

1

1

0nf nf x

2 3 41 1 11 1

2! 3! 4!xe x x x x

2 3 4

12! 3! 4!

x x x xe x

9.3 Taylor Series

2 3 4

12! 3! 4!

x x x xe x

An amazing use for infinite series:

Substitute xi for x.

2 3 4 5 6

1 2! 3! 4! 5! 6!

xi xi xi xi xi xie xi

2 2 3 3 4 4 5 5 6 6

1 2! 3! 4! 5! 6!

xi x i x i x i x i x ie xi

9.3 Taylor Series

2 2 3 3 4 4 5 5 6 6

1 2! 3! 4! 5! 6!

xi x i x i x i x i x ie xi

2 3 4 5 6

1 2! 3! 4! 5! 6!

xi x x i x x i xe xi

2 4 6 3 5

1 2! 4! 6! 3! 5!

xi x x x x xe i x

Factor out the i terms.

9.3 Taylor Series

2 4 6 3 5

1 2! 4! 6! 3! 5!

xi x x x x xe i x

This is the series for cosine.

This is the series for sine.

cos sinxie ix x Let x

cos sinie i

1 0ie i

1 0ie This amazing identity contains the five most famous numbers in mathematics, and shows that they are interrelated.

9.3 Taylor Series

Convergence

The series that are of the most interest to us are those that converge.

Today we will consider the question:

“Does this series converge, and if so, for what values of x does it converge?”

9.4 Radius of Convergence

The first requirement of convergence is that the terms must approach zero.

nth term test for divergence

1 nna

diverges if fails to exist or is not zero.lim nn

a

Note that this can prove that a series diverges, but can not prove that a series converges.

9.4 Radius of Convergence

0! n

nn x

If then grows without

bound.

1x ! nn x

If then0 1x 1

!lim ! limn

nn n

x

nn x

As , eventually is larger than , therefore

the numerator grows faster than the denominator.

n n1

x The series diverges. (except when x=0)

9.4 Radius of Convergence

1 The series converges over some finite interval:(the interval of convergence).

The series may or may not converge at the endpoints of the interval.

There is a positive number R such that the series diverges for but converges for .x a R x a R

2 The series converges for every x. ( )R

3 The series converges for at and diverges everywhere else. ( )0R

x aThe number R is the radius of convergence.

9.4 Radius of Convergence

This series converges.

So this series must also converge.

Direct Comparison Test

For non-negative series:

If every term of a series is less than the corresponding term of a convergent series, then both series converge.

If every term of a series is greater than the corresponding term of a divergent series, then both series diverge.

So this series must also diverge.

This series diverges.

9.4 Radius of Convergence

Ex. 3: Prove that converges for all real x.

2

20 !

n

n

x

n

There are no negative terms:

2 2

2 !!

nnx xnn

2

0 !

n

n

xn

is the Taylor series for , which converges.

2xelarger denominator

The original series converges.

The direct comparison test only works when the terms are non-negative.

9.4 Radius of Convergence

Absolute Convergence

If converges, then we say converges absolutely.na naThe term “converges absolutely” means that the series formed by taking the absolute value of each term converges. Sometimes in the English language we use the word “absolutely” to mean “really” or “actually”. This is not the case here!

If converges, then converges.na na

If the series formed by taking the absolute value of each term converges, then the original series must also converge.

“If a series converges absolutely, then it converges.”

9.4 Radius of Convergence

0

sin

!

n

n

x

n

We test for absolute convergence: sin 1

! !

nx

n n

Since ,2 3

1 2! 3! !

nx x x x

e xn

9.4 Radius of Convergence

Since ,2 3

1 2! 3! !

nx x x x

e xn

0

1

!n n

converges to

1e e

0

sin

!

n

n

x

n

converges by the direct comparison test.

Since converges absolutely, it converges.

0

sin

!

n

n

x

n

9.4 Radius of Convergence

Ratio Technique

We have learned that the partial sum of a geometric series is given by:

1

1

1

n

n

rS t

r

where r = common ratio between terms

When , the series converges.1r

9.4 Radius of Convergence

Geometric series have a constant ratio between terms. Other series have ratios that are not constant. If the absolute value of the limit of the ratio between consecutive terms is less than one, then the series will converge.

9.4 Radius of Convergence

For , if then:1

nn

t

1lim n

nn

tL

t

if the series converges.1L

if the series diverges.1L

if the series may or may not converge.1L

9.4 Radius of Convergence

2 3 4

ln 12 3 4

x x xx x

Ex:If we replace x with x-1, we get:

2 3 41 1 1ln 1 1 1 1

2 3 4x x x x x 1

1

11 1

n n

n

xn

2 1

1

1 1lim

1 1 1

n n

n nn

x nL

n x

1

1

1nn

n n

aa

a a

9.4 Radius of Convergence

1 1lim

1 1

n

nn

x x n

n x

1lim

1n

x n

n

1x

If the limit of the ratio between consecutive terms is less than one, then the series will converge.

9.4 Radius of Convergence

1 1x 1 1 1x 0 2x

The interval of convergence is (0,2).

The radius of convergence is 1.

If the limit of the ratio between consecutive terms is less than one, then the series will converge.

9.4 Radius of Convergence

1

53

n

nn

nx

2 31 2 35 5 5

3 9 27x x x

1

1

1 5 3lim

3 5

n n

nnn

n xL

n x

1 5 5 3lim

3 3 5

n n

nnn

n x xL

n x

1 5

lim3n

n xL

n

9.4 Radius of Convergence

1

53

n

nn

nx

1 5

lim3n

n xL

n

15 lim

3n

nL x

n

15

3L x

15 1

3x

5 3x

3 5 3x

2 8x

The interval of convergence is (2,8).

The radius of convergence is .8 2

32

9.4 Radius of Convergence

41

!3

n

n

nx

n

2 3 41 2 33 3 3 3

8 27 32x x x x

1 4

4

1 ! 3lim

1 ! 3

n

nn

n x nL

n n x

9.4 Radius of Convergence

41

!3

n

n

nx

n

4

4

! 1 3 3lim

1 ! 3

n

nn

n n x x nL

n n x

4

3 lim 11n

nL x n

n

1

9.4 Radius of Convergence

41

!3

n

n

nx

n

4

3 lim 11n

nL x n

n

1

L for all .3x Radius of convergence = 0.

At , the series is , which converges to zero.3x 0 0 0

Note: If r is infinite, then the series converges for all values of x.

9.4 Radius of Convergence

Another series for which it is easy to find the sum is the telescoping series.

1

1

1n n n

1

1 1

1n n n

1 1 1 11 1

2 3 3 42

3

11

4S

11

1nSn

lim 1nnS

Telescoping Series

11

n nn

b b

converges to 1b

9.4 Radius of Convergence

by partialfractions

This section in the book presents several other tests or techniques to test for convergence, and discusses some specific convergent and divergent series.

9.5 Testing Convergence

The series converges if .1L

The series diverges if .1L

The test is inconclusive if .1L

Nth Root Test:

If is a series with positive terms andna lim nnn

a L

then:

Note that the rules are the same as for the Ratio Test.

9.5 Testing Convergence

2

1 2nn

n

2

2n

n

n 2

2

n n

2lim n

nn

2

lim n

nn

?

9.5 Testing Convergence

lim n

nn

1

lim n

nn

1

lim ln nn ne

1lim lnn

nne

lnlimn

n

ne

1

lim1n

n

e

0e

1Indeterminate, so we use L’Hôpital’s Rule

9.5 Testing Convergence

example:2

1 2nn

n

2

2n

n

n 2

2

n n

2

lim2

n

n

n

2lim n

nn

2

lim n

nn

21 1

1

2 it converges

?

9.5 Testing Convergence

another example:2

1

2n

n n

2

2n

n

n 2

2n n

2

2lim

nn n

2

1 it diverges2

9.5 Testing Convergence

Remember:

The series converges if .1L

The series diverges if .1L

The test is inconclusive if .1L

The Ratio Test:

If is a series with positive terms andna 1lim n

nn

aL

a

then:

9.5 Testing Convergence

This leads to:

The Integral Test

If is a positive sequence and where

is a continuous, positive decreasing function, then:

na na f n f n

and both converge or both diverge.nn N

a

Nf dxx

9.5 Testing Convergence

Example 1: Does converge?1

1

n n n

1

1 dx

x x

3

2

1lim

b

bx dx

1

21

lim 2

b

b x

22lim

b b

2

Since the integral converges, the series must converge.

(but not necessarily to 2.)

9.5 Testing Convergence

p-series Test

1

1 1 1 1

1 2 3p p p pn n

converges if , diverges if .1p 1p

If this test seems backward after the ratio and nth root

tests, remember that larger values of p would make the

denominators increase faster and the terms decrease

faster.

9.5 Testing Convergence

the harmonic series:

1

1 1 1 1 1

1 2 3 4n n

diverges.

(It is a p-series with p=1.)

It diverges very slowly, but it diverges.

Because the p-series is so easy to evaluate, we use it to compare to other series.

9.5 Testing Convergence

Limit Comparison Test

If and for all (N a positive integer)0na 0nb n N

If , then both and

converge or both diverge.

lim 0n

nn

ac c

b na nb

If , then converges if converges.lim 0n

nn

a

b na nb

If , then diverges if diverges.lim n

nn

a

b na nb

21

3 5 7 9 2 1

4 9 16 25 1n

n

n

When n is large, the function behaves like:

2

2 2n

n n

2 1

n n lim n

nn

a

b

2

2 1

1lim1n

n

n

n

2

2 1lim1n

nn

n

2

2

2lim

2 1n

n n

n n

2 Since diverges, the

series diverges.

1

nharmonic series

9.5 Testing Convergence

1

1 1 1 1 1

1 3 7 15 2 1nn

When n is large, the

function behaves like:

1

2n

lim n

nn

a

b

12 1lim

12

n

n

n

2

lim2 1

n

nn

1

Since converges, the series converges.1

2ngeometric series

9.5 Testing Convergence

9.5 Testing Convergence

...)1( 3211

1

uuuunn

n

1. each un is positive;2. un > un+1 for all n > N for some integer N (decreasing);3. lim n→un+ 0

Theorem 12 The Alternating Series Test

The series

converges if all three of the following conditions are satisfied:

Alternating Series

example: 1

1

1 1 1 1 1 1 11

1 2 3 4 5 6n

n n

This series converges (by the Alternating Series Test.)

If the absolute values of the terms

approach zero, then an alternating

series will always converge!

Alternating Series Test

This series is convergent, but not absolutely convergent.

Therefore we say that it is conditionally convergent.

9.5 Testing Convergence

Since each term of a convergent alternating series moves the partial sum a little closer to the limit:

Alternating Series Estimation Theorem

For a convergent alternating series, the truncation error is less than the first missing term, and is the same sign as that term.

This is a good tool to remember, because it is easier than the LaGrange Error Bound.

9.5 Testing Convergence

series diverges

Converges to a/(1-r)if |r|<1. Diverges if

|r|>1

Converges if p>1Diverges if p<1

non-negativeterms and/orabsoluteconvergence

Does Σ |an| converge?Apply Integral Test, RatioTest or nth-root Test

Original SeriesConverges

AlternatingSeries Test

Is Σan = u1-u2+u3-…an alternating series

Is there an integer Nsuch that uN>uN-1…?

Converges if un 0Diverges if un 0

nth-Term Test Is lim an=0 no

no

yes or maybe

no

no or maybe

GeometricSeries Test Is Σan = a+ar+ar2+ … ?

yes

yes

yes

yes

yes

p-Series Test Is series form

1

1

npn