in nite seriesp. sam johnson (nit karnataka) in nite series january 29, 2019 5 / 62. in nite series...

Infinite Series

P. Sam Johnson

January 29, 2019

P. Sam Johnson (NIT Karnataka) Infinite Series January 29, 2019 1 / 62

Overview

An infinite series is the sum of an infinite sequence of numbers

a1 + a2 + a3 + · · ·+ an + · · · .

The aim of this lecture is to understand the meaning of such an infinitesum and to develop methods to calculate it.


Infinite Series

Since there are infinitely many terms to add in an infinite series, we cannotjust keep adding to see what comes out.

Instead we look at what we get by summing the first n terms of thesequence and stopping.

The sum of the first n terms

Sn = a1 + a2 + a3 + · · ·+ an

is an ordinary finite sum and can be calculated by normal addition. It iscalled the nth partial sum.

As n gets larger, we expect the partial sums to get closer and closer to alimiting value in the same sense that the terms of a sequence approach alimit.


Infinite Series

For example, to assign meaning to an expression like

1 +1

2+

1

4+

1

8+

1

16+ · · ·

We add terms one at a time from the beginning and look for a pattern inhow these partial sums grow.


Infinite Series

Indeed there is a pattern. The partial sums form a sequence whose nthterm is

sn = 2− 1

2n−1·

This sequence of partial sums converges to 2 because limn→∞

1

2n= 0.

We say “the sum of the infinite series 1 + 12 + 1

4 + · · ·+ 12n−1 + · · · is 2.”

Is the sum of any finite number of terms in this series equal to 2? No. Canwe actually add an infinite number of terms one by one? No. But we canstill define their sum by defining it to be the limit of the sequence ofpartial sums as n→∞, in this case 2.

Our knowledge of sequences and limits enables us to break away from theconfines of finite sums.


Infinite Series

Definition 1 (Infinite Series, nth Term, Partial Sum).

Given a sequence of numbers {an}, an expression of the form

a1 + a2 + a3 + · · ·+ an + · · ·

is an infinite series. The number an is the nth term of the series. Thesequence {sn} defined by

s1 = a1

s2 = a1 + a2

...

sn = a1 + a2 + · · ·+ an =n∑

k=1

ak

is the sequence of partial sums of the series, the number sn being thenth partial sum.


Infinite Series

Definition 2 (Converges, Sum).

If the sequence of partial sums converges to a limit L, we say that theseries converges and that its sum is L.

In this case, we also write

a1 + a2 + · · ·+ an + · · · =∞∑n=1

an = L.

If the sequence of partial sums of the series does not converge, we say thatthe series diverges.


Notation

When we begin to study a given series

a1 + a2 + · · ·+ an + · · ·

we might not know whether it converges or diverges.

In either case, it is convenient to use sigma notation to write the series as

∞∑n=1

an,∞∑k=1

ak , or∑

an.


Geometric Series

Geometric series are series of the form

a + ar + ar 2 + · · ·+ arn−1 + · · · =∞∑n=1

arn−1

in which a and r are fixed real numbers and a 6= 0. The series can also be

written as∞∑n=0

arn. The ratio r can be positive, as in

1 +1

2+

1

4+ · · ·+

(1

2

)n−1+ · · · ,

or negative, as in

1− 1

3+

1

9− · · ·+

(− 1

3

)n−1+ · · · .


Infinite Series

If r = 1, the nth partial sum of the geometric series is

sn = a + a(1) + a(1)2 + · · ·+ a(1)n−1 = na,

and the series diverges because limn→∞

sn = ±∞ depending on the sign of a.

If r = −1, the series diverges because the nth partial sums alternatebetween a and 0.


Geometric Series

If |r | < 1, the geometric series

a + ar + ar 2 + · · ·+ arn−1 + · · ·

converges to a1−r :

∞∑n=1

arn−1 =a

1− r, |r | < 1.


Geometric Series

If |r | 6= 1, we can determine the convergence or divergence of the series inthe following way.

sn = a + ar + ar 2 + · · ·+ arn−1

rsn = ar + ar 2 + · · ·+ arn−1 + arn

sn − rsn = a− arn

sn(1− r) = a(1− rn)

sn =a(1− rn)

1− r(r 6= 1).

If |r | < 1, then rn → 0 as n→∞ and sn → a/(1− r). If |r | > 1, then|rn| → ∞ and the series diverges.


Infinite Series

We have determined when a geometric series converges or diverges, and towhat value.

Often we can determine that a series converges without knowing the valueto which it converges.

The formulaa

1− r

for the sum of a geometric series applies only when the summation indexbegins with n = 1 in the expression

∞∑n=1

= arn−1

(or with the index n = 0 if we write the series as∞∑n=0

arn).


Infinite Series

Example 3 (Index starts with n = 1).

The geometric series with a = 1/9 and r = 1/3 is

1

9+

1

27+

1

81+ · · · =

∞∑n=1

1

9

(1

3

)n−1

=1/9

1− (1/3)=

1

6.

Example 4 (Index Starts with n = 0).

The series∞∑n=0

(−1)n5

4n= 5− 5

4+

5

16− 5

64+ · · ·

is a geometric series with a = 5 and r = −1/4. It converges to

a

1− r=

5

1 + (1/4)= 4.


Infinite Series

Example 5 (A Bouncing Ball).

You drop a ball from “a” meters above a flat surface. Each time the ballhits the surface after falling a distance h, it rebounds a distance rh, wherer is positive but less than 1. Find the total distance the ball travels up anddown.


A Bouncing Ball

The total distance is

s = a + 2ar + 2ar 2 + 2ar 3 + · · · = a +2ar

1− r= a

1 + r

1− r.

If a = 6 m and r = 2/3, for instance, the distance is

s = 61 + (2/3)

1− (2/3)= 6

(5/3

1/3

)= 30m.


Infinite Series

Example 6 (Repeating Decimals).

Express the repeating decimal

5.232323 · · ·

as the ratio of two integers.

5.232323 · · · = 5 +23

100+

23

(100)2+

23

(100)3+ · · ·

= 5 +23

100

(1

0.99

)=

518

99


Telephone Series

Unfortunately, formulas like the one for the sum of a convergent geometricseries are rare and we usually have to settle for an estimate of a series’sum.

The next example, however, is another case in which we can find the sumexactly.

Example 7 (A Nongeometic but Telescoping Series).

Find the sum of the series

∞∑n=1

1

n(n + 1).


Solution

We look for a pattern in the sequence of partial sums that might lead to aformula for sk . The key observation is the partial fraction decomposition

1

n(n + 1)=

1

n− 1

n + 1,

sok∑

n=1

1

n(n + 1)=

k∑n=1

(1

n− 1

n + 1

)and

sk =

(1

1− 1

2

)+

(1

2− 1

3

)+

(1

3− 1

4

)+ · · ·+

(1

k− 1

k + 1

).


Solution (contd...)

Removing parentheses and canceling adjacent terms of opposite signcollapses the sum to

sk = 1− 1

k + 1.

We now see that sk → 1 as k →∞.

The series converges, and its sum is 1:

∞∑n=1

1

n(n + 1)= 1.


Necessary condition for convergence

Theorem 8.

If∞∑n=1

an converges, then an → 0.

Proof :

Let S represent the series’ sum and sn = a1 + a2 + · · ·+ an the nth partialsum.

When n is large, both sn and sn−1 sre close to S , so their difference, an, isclose to zero. More formally,

an = sn − sn−1 → S − S = 0.


The nth-Term Test for Divergence

The above theorem leads to a test for detecting the kind of divergence,called the nth term test.

Theorem 9 (The nth-Term Test for Divergence).∞∑n=1

an diverges if limn→∞

an fails to exist or is different from zero.

Theorem (8) does not say that∑∞

n=1 an converges if an → 0.

It is possible for a series to diverge when an → 0.


Divergent Series

One reason that a series may fail to converge is that its terms don’tbecome small.

Example 10 (Partial Sums Outgrow Any Number).

(a) The series ∞∑n=1

n2 = 1 + 4 + 9 + · · ·+ n2 + · · ·

diverges because the partial sums grow beyond every number L. Aftern = 1, the partial sum sn = 1 + 4 + 9 + · · ·+ n2 is greater than n2.

(b) The series ∞∑n=1

n + 1

n=

2

1+

3

2+

4

3+ · · ·+ n + 1

n+ · · ·

diverges because the patial sums eventually outgrow every preassignednumber. Each term is greater than 1, so the sum of n terms is greaterthan n.


Infinite Series

Example 11 (Applying the nth-Term Test).

(a)∞∑n=1

n2 diverges because n2 →∞.

(b)∞∑n=1

n + 1

ndiverges because n+1

n → 1.

(c)∞∑n=1

(−1)n+1 diverges becaues limn→∞

(−1)n+1 does not exist.

(d)∞∑n=1

−n

2n + 5diverges because lim

n→∞

−n

2n + 5= −1

26= 0.


Infinite Series

We have seen that if∑

an converge, then an → 0.

The converse need not be true. The following example illustrates this.an → 0 but the series diverges.

Example 12.

The series

1 +1

2+

1

2+

1

4+

1

4+

1

4+

1

4+ · · ·+ 1

2n+

1

2n+ · · ·+ 1

2n+ · · ·

diverges because the terms are grouped into clusters that add to 1, so thepartial sums increase without bound.

However, the terms of the series form a sequence that converges to 0. Weshall see that the harmonic series also behaves in this manner.


Combining Series

Whenever we have two convergent series, we can add them term byterm, subtract them term by term, or multiply them by constants to makenew convergent series.

Theorem 13.

If∑

an = A and∑

bn = B are convergent series, then

1. Sum Rule :∑

(an + bn) =∑

an +∑

bn = A + B

2. Difference Rule :∑

(an − bn) =∑

an −∑

bn = A− B

3. Constant Multiple Rule :∑

kan = k∑

an = kA (any number k)

The three rules for series from the analogous rules for sequences.


Proof (Sum Rule)

To prove the Sum Rule for series, let

An = a1 + a2 + · · ·+ an, Bn = b1 + b2 + · · ·+ bn.

Then the partial sums of∑

(an + bn) are

sn = (a1 + a2) + (a2 + b2) + · · ·+ (an + bn)

= (a1 + · · ·+ an) + (b1 + · · ·+ bn)

= An + Bn.

Since An → A and Bn → B, we have sn → A + B by the Sum Rule forsequences.


Infinite Series

The proof of the Difference Rule is similar.

To prove the Constant Multiple Rule for series, observe that the partialsums of

∑kan form the sequence

sn = ka1 + ka2 + · · ·+ kan = k(a1 + a2 + · · ·+ an) = kAn,

which converges to kA by the Constant Multiple Rule for sequences.


Infinite Series

As corollaries of the theorem (13), we have the following results. We omitproof.

Theorem 14.

(a) Every nonzero constant multiple of a divergent series diverges.

(b) If∑

an converges and∑

bn diverges, then∑

(an + bn) and∑(an − bn) both diverge.

The following example is given to show that∑

(an + bn) can convergewhen

∑an and

∑bn both diverge.

Example 15.∑an = 1 + 1 + 1 + · · · and

∑bn = (−1) + (−1) + (−1) + . . . diverge,

whereas∑

(an + bn) = 0 + 0 + 0 + · · · converges to 0.


Infinite Series

Example 16.

(a) The sum of∞∑n=1

3n−1

6n−1 − 1is 4

5 :

∞∑n=1

3n−1

6n−1 − 1=

∞∑n=1

(1

2n−1− 1

6n−1

)

=∞∑n=1

1

2n−1−∞∑n=1

1

6n−1= 2− 6

5=

4

5

(b) The sum of∞∑n=1

4

2nis 8 :

∞∑n=1

4

2n= 4

∞∑n=0

1

2n= 8


Adding or Deleting Terms

We can add a finite number of terms to a series or delete a finite numberof terms without altering the serie’s convergence or divergence, althoughin the case of convergence this will usually change the sum.

If∞∑n=1

= an converges, then∞∑n=k

an converges for any k > 1 and

∞∑n=1

= a1 + a2 + · · ·+ ak−1 +∞∑n=k

an.

Conversely, if∞∑n=k

an converges for any k > 1, then∞∑n=1

an converges.


Reindexing

As long as we preserve the order of its terms, we can reindex any serieswithout altering its convergence.

To raise the starting value of the index h units, replace the n in theformula for an by n − h.

∞∑n=1

an =∞∑

n=1+h

an−h = a1 + a2 + a3 + · · · .


Reindexing

To lower the starting value of the index h units, replace the n in theformula for an by n + h :

∞∑n=1

an =∞∑

n=1−han+h = a1 + a2 + a3 + · · · .

It works like a horizontal shift. We saw this in starting a geometric serieswith the index n = 0 instead of the index n = 1, but we can use any otherstarting index value as well.

We usually give preference to indexings that lead to simple expressions.


Reindexing

Example 17 (Reindexing a Geometric Series).

We can write the geometric series

∞∑n=1

1

2n−1= 1 +

1

2+

1

4+ · · · .

As∞∑n=1

1

2n,

∞∑n=5

1

2n−5, or even

∞∑n=−4

1

2n+4.

The partial sums remain the same no matter what indexing we choose.


Finding nth Partial Sums

Exercise 18.

Find a formula for the nth partial sum of each series and use it to find theseries sum if the series converges.

1. 2 + 23 + 2

9 + 227 + · · ·+ 2

3n−1 + · · ·2. 9

100 + 91002 + 9

1003 + · · ·+ 9100n + · · ·

3. 1− 2 + 4− 8 + · · ·+ (−1)n−12n−1 + · · ·4. 1

2.3 + 13.4 + 1

4.5 + · · ·+ 1(n+1)(n+2) + · · ·


Solution

1. 3

2. 111

3. diverges

4. 12


Series with Geometric Terms

Write out the first few terms of each series to show how the series starts.Then find the sum of the series.

Exercise 19.

1.∞∑n=0

(−1)n

4n

2.∞∑n=0

(5

2n− 1

3n

)

3.∞∑n=0

(1

2n+

(−1)n

5n

)


Solution

1. 45

2. 172

3. 176


Exercise

Exercise 20.

Use the nth-Term Test for divergence to show that the series is divergent,or state that the test is inconclusive.

1.∞∑n=1

n(n + 1)

(n + 2)(n + 3)

2.∞∑n=1

cos(1

n

)3.

∞∑n=1

en

en + π

4.∞∑n=1

ln1

n

5.∞∑n=1

cos nπ


Solution

1. diverges

2. diverges

3. diverges

4. diverges

5. diverges


Telescoping Series

Exercise 21.

Find a formula for the nth partial sum of the series and use it to determineif the series converges or diverges. If a series converges, find its sum.

1.∞∑n=1

( 3

n2− 3

(n + 1)2

)2.

∞∑n=1

(ln√

n + 1− ln√

n)

3.∞∑n=1

(tan n − tan(n − 1)

)4.

∞∑n=1

(cos−1

( 1

n + 1

)− cos−1

( 1

n + 2

))5.

∞∑n=1

(√n + 4−

√n + 3

)P. Sam Johnson (NIT Karnataka) Infinite Series January 29, 2019 41 / 62

Solution

1. converges to 3

2. diverges

3. diverges

4. converges to π6

5. diverges


Telescoping Series

Exercise 22.

1.∞∑n=1

6

(2n − 1)(2n + 1)

2.∞∑n=1

(1√n− 1√

n + 1

)

3.∞∑n=1

(1

ln(n + 2)− 1

ln(n + 1)

)

4.∞∑n=1

(tan−1(n)− tan−1(n + 1)

)


Solution

1. converges to 3

2. converges to 1

3. converges to − 1ln 2

4. converges to −π4


Convergence or Divergence

Exercise 23.

Which of the following converge, and which diverge? Give reasons for youranswers. If a series converges, find its sum.

1.∞∑n=1

(−1)n+1 3

2n

2.∞∑n=0

cos nπ

5n

3.∞∑n=0

1

xn, |x | > 1

4.∞∑n=0

n!

1000n

5.∞∑n=1

nn

n!

6.∞∑n=1

ln

(n

2n + 1

)

7.∞∑n=0

enπ

πne


Solution

1. converges to 1

2. converges to 56

3. converges to xx−1

4. diverges

5. diverges

6. diverges

7. divergent geometric series


Geometric Series

Exercise 24.

In each of the following geometric series, write out the first few terms ofthe series to find a and r , and find the sum of series. Then express theinequality |r | < 1 in terms of x and find the values of x for which theinequality holds and the series converges.

1.∞∑n=0

(−1)nx2n

2.∞∑n=0

(−1)n

2

(1

3 + sinx

)n


Solution

1. converges to 11+x2 for |x | < 1

2. converges to 3+sin x8+2 sin x for all x


Geometric Series

Exercise 25.

Find the values of x for which the given geometric series converges.Also, find the sum of the series (as a function of x) for those values of x.

1.∞∑n=0

2nxn

2.∞∑n=0

(−1

2

)n

(x − 3)n

3.∞∑n=0

sinnx

4.∞∑n=0

(ln x)n


Solution

1. converges to 11−2x for |x | < 1

2

2. converges to 2x−1 for 1 < x < 5

3. converges to 11−sin x for x 6= (2k + 1)π2 , k is an integer

4. converges to 11−ln x for e−1 < x < e


Repeating Decimals

Exercise 26.

Express each of the following numbers as the ratio of two integers.

1. 0.23 = 0.232323 · · ·2. 0.7 = 0.7777 · · ·3. 3.142857 = 3.142857142857 · · ·


Solution

1. 2399

2. 79

3. 116,40237,037


Theory and Examples

Exercise 27.

1. Write the series∞∑n=1

1

(n + 1)(n + 2)as a sum beginning with

(a) n = −2 (b) n = 0 (c) n = 5

2. Make up an infinite series of nonzero terms whose sum is

(a) 1 (b) −3 (c) 0

Can you make an infinite series of nonzero terms that converges toany number you want? Explain.

3. Show by example that∑

(an/bn) may diverge even though∑

an and∑bn converge and no bn equals 0.


Solution

1. (a)∞∑

n=−2

1

(n + 4)(n + 5)

(b)∞∑n=0

1

(n + 2)(n + 3)(c)

∞∑n=5

1

(n − 3)(n − 2)

2. Students should try.

3. Students should try.

4. diverges.


Theory and Examples

Exercise 28.

1. Find convergent geometric series A =∑

an and B =∑

bn thatillustrate the fact that

∑anbn may converge without being equal to

AB.

2. Show by example that∑

(an/bn) may converge to something otherthan A/B even when A =

∑an,B =

∑bn 6= 0, and no bn equals 0.

3. If∑

an converges and an > 0 for all n, can anything be said about∑(1/an)? Give reasons for your answer.


Theory and Examples

Exercise 29.

1. What happens if you add a finite number of terms to a divergentseries or delete a finite number of terms from a divergent series? Givereasons for your answer.

2. If∑

an converges and∑

bn diverges, can anything be said abouttheir term-by-term sum

∑(an + bn)? Give reasons for your answer.

3. Make up a geometric series∑

arn−1 that converges to the number 5

(a) a = 2 (b) a = 13/2

4. Find the value of b for which 1 + eb + e2b + e3b + · · · = 9.

5. For what values of r does the infinite series

1 + 2r + r 2 + 2r 3 + r 4 + 2r 5 + r 6 + · · ·

converge? Find the sum of the series when it converges.


Theory and Examples

Exercise 30.

1. Show that the error (L− sn) obtained by replacing a convergentgeometric series with one of its partial sums sn is arn/(1− r).

2. A ball is dropped from a height of 4m. Each time it strikes thepavement after falling from a height of h meters it rebounds to aheight of 0.75h meters. Find the total distance the ball travels up anddown.Find the total number of seconds the ball is travelling. (Hint: Theformula s = 4.9t2 gives t =

√s/4.9.)


Theory and Examples

Exercise 31.

The accompanying figure shows the first five a sequence of squares. Theoutermost square has an area of 4m2. Each of the other squares isobtained by joining the midpoints of the sides of the squares before it.Find the sum of the areas of all the squares.


Theory and Examples

Exercise 32.

The accompanying figure shows the first three rows and part of the fourthrow of a sequence of rows of semicircles. There are 2n semicircles in the4th row, each of radius 1/2”. Find the sum of the areas of all thesemicircles.


Theory and Examples

Exercise 33 (Helga von Kochs snowflake curve).

Helga von Kochs snow-flake is a curve of infinite length that encloses aregion of finite area. To see why this is so, suppose the curve is generatedby starting with an equilateral triangle whose sides have length 1.

(a) Find the length Ln of the nth curve Cn and show that limn→∞ Ln =∞.

(b) Find the area An of the region enclosed by Cn and calculatelimn→∞ An.


Theory and Examples

Exercise 34.

The accompanying figure provides an informal proof that

∞∑n=1

(1/n2)

is less than 2. Explain what is going on.


References

1. M.D. Weir, J. Hass and F.R. Giordano, Thomas’ Calculus, 11thEdition, Pearson Publishers.

2. N. Piskunov, Differential and Integral Calculus, Vol I & II (Translatedby George Yankovsky).

3. S.C. Malik and Savitha Arora, Mathematical Analysis, New AgePublishers.

4. R. G. Bartle, D. R. Sherbert, Introduction to Real Analysis, WileyPublishers.


in nite seriesp. sam johnson (nit karnataka) in nite series january 29, 2019 5 / 62. in nite series...

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