Infinite Series

CHAPTER 37

Infinite Series 37.1

Prove that, if E an converges, then Let

37.2

an=Q.

Then

Show that the harmonic series

diverges.

etc. fore,

37.3

There-

Alternatively, by the integral test,

imply that E an converges?

Does

No. The harmonic series E 1/n (Problem 37.2) is a counterexample.

37.4

Let Sn = a + ar + •• • + ar" ', with r^l. Show that rS=ar + ar2 + - - - + ar" Thus,

37.5

Let

+ ar". S. = a + ar + ar2 + • • • + ar"~\ Hence, (r- 1)5_ = ar" - a = a(r" - I)

a T^ 0. Show that the infinite geometric series

and diverges if

if M < i ,

By Problem 37.4,

since

r"-*0;

JSJ—»+°°, since |r| -*+\,

a¥=0.

If

INFINITE SERIES

37.10

313

Investigate the series Hence, the partial sum

The series converges to 1. (The method used here is called "telescoping.")

37.11

Study the series So

Thus, the series converges to

37.12

Find the sum of the series 4 — 1 + j — & + • • • . and first term a = 4. Hence, it converges to

This is a geometric series with ratio

37.13

Test the convergence of This is a geometric series with ratio

37.14

r = \ > 1. Hence, it is divergent.

Test the convergence of 3+I + I + I + - - - . The series has the general term Hence, by Problem 37.1, the series diverges.

37.15

(starting with n = 0), but lim an = lim

Investigate the series Rewrite the series as and 37.10.

37.16

by Problems 37.11

Test the convergence of Hence, by Problem 37.1, the series diverges.

37.17

Study the series So the partial sum

37.18

Study the series

Thus, The partial sum

314 37.19

CHAPTER 37 Study the series Hence, the partial sum

37.20

Study the series So

37.21

Evaluate So the partial sum or

37.22

is either

In either case, the partial sum approaches 1.

Evaluate The partial sum

37.23

Evaluate so, by Problem 37.1, the series diverges.

37.24

Evaluate The partial sum The series diverges.

37.25

Find the sum

and show that it is correct by exhibiting a formula which, for each

e > 0, specifies

an integer m for which \Sn — S\< e holds for all n > m (where Sn is the nth partial sum) In fact,

is a geometric series with ratio r = 5 and first term a = 1. So the sum assume e > 0. Then, by Problem 37.4, Sn = 1

Now

We want Choose m to be the least positive integer that exceeds 37.26

Determine the value of the infinite decimal 0.666 + • • -. is a geometric series with ratio the sum is

and first term

Hence.

INFINITE SERIES 37.27

315

Evaluate is the harmonic series minus the first 99 terms. However, convergence or divergence is not affected by deletion or addition of any finite number of terms. Problem 37.2), so is the given series.

37.28

Since the harmonic series is divergent (by

Evaluate Since the harmonic series is divergent, so is the given series.

37.29

Evaluate

In [nl(n + 1)] = In n - In (n + 1), and 5,, = (In 1 - In 2) + (In 2 - In 3) + - • • + [In n - In (n + 1)] = -In (« + !)-» -oo. Thus, the given series diverges. 37.30

Evaluate This is a geometric series with ratio

37.31

r=\le1/nsince np £ n. Therefore, by the comparison test and the fact that is divergent, is divergent. 37.37

Determine whether

is convergent.

For n > l , l/n! = l / ( l - 2

n) 20,000, 37.64

1-? +

1/[2(51) - 1] =

required for an approximation of the sum n > 12. Hence, we should use the first 11 terms.

Show that, if £ an converges by the integral test for a function f(x), the error Rn, if we use the first n terms, satisfies If we approximate by the lower rectangles in Fig. 37-1, then If we use the upper rectangles,

Fig. 37-1 37.65

Estimate the error when

is approximated by the first 10 terms.

By Problem 37.64,

In addition, Hence, the error lies between 0.023 and

37.66

How many terms are necessary to approximate By

37.67

correctly to three decimal places?

Problem 37.64, the error we need 100 < n2, n>10. So at least 11 terms are required.

What is the error if we approximate a convergent geometric series

To

get

by the first n terms a+ar+...+

arn-1? The error is ar" + ar"*1 + • • •, a geometric series with sum ar"l(\ - r).

320

37.68

CHAPTER 37 by means of the first 10 terms, what will the

If we approximate the geometric series error be? By Problem 37.67, the error is

37.69

For the convergent p-series

( p > 1), show that the error Rn after »terms is less than

Bv Problem 37.64,

37.70

Study the convergence of Use the ratio test. lutely convergent.

37.71

Therefore, the series is abso-

Determine whether

converges.

Use the ratio test.

Hence, the series converges. 37.72

converges.

Determine whether Use the ratio test. Hence, the series converges.

37.73

convergent?

Is

Yes. Since In we have series is dominated by a convergent geometric series.

37.74

so the given

converges.

Determine whether

Hence, the series is absolutely convergent by comparison with the convergent p-series

37.75

Study the convergence of The series is geometric series

Note that, for the series is absolutely convergent.

So, by comparison with the convergent

INFINITE SERIES 37.76

Study the convergence of This is the series

By the alternating series test, it is convergent. By the limit

comparison test with is divergent, 37.77

321

Since diverges. Hence, the given series is conditionally convergent.

Prove the following special case of the ratio test: A series of positive terms £ an is convergent if Choose r so that There exists an integer k such that, if and, therefore, Hence, Hence, by comparison with the convergent geometric series

then

So, if

the series

is convergent, and, therefore, the given series is convergent (since it is obtained from a convergent sequence by addition of a finite number of terms). 37.78

or convergence.

Test

Use the limit comparison test with the convergent p-series Hence, the given series converges. 37.79

Test

for convergence.

Use the ratio test. In

37.80

since

In2<

Therefore, the series diverges.

Test

for convergence.

Use the ratio test.

Therefore, the

series converges. 37.81

Test

for convergence.

Use the ratio test. series diverges. 37.82

Therefore, the

Determine the nth term of and test for convergence the series The nth term is

Use the limit comparison test with the convergent p-series T, 1 In2. Therefore, the given series converges.

37.83

Determine the nth term of and test for convergence the series The nth term is l/(n + l)(n+ 2 ) - • • (2n). The nth term is less than 1 / 2 - 2 - • - 2 = 1/2". Hence, the series is convergent by comparison with the convergent geometric series

322

37.84

CHAPTER 37 Determine the nth term of and test for convergence the series The nth term is

Use the limit comparison test with the divergent series E 1/n. Hence, the given series diverges.

37.85

Determine the nth term of and test for convergence the series The nth term is n/(n + 1)". Observe that comparison with the convergent p-series £ 1/n2.

37.86

Hence, the series converges by

Determine the nth term of and test for convergence the series The nth term is (n + l)/(n 2 + 1). Use the limit comparison test with the divergent series E 1/n. Therefore, the given series is divergent.

37.87

Determine the nth term of and test for convergence the series The nth term is

The ratio test yields

Hence, the series converges. 37.88

Determine the nth term of and test for convergence the series The nth term is (In + l)/(n 3 + n). Use the limit comparison test with the convergent p-series E 1/n2. Hence, the series is convergent.

37.89

Determine the nth term of and test for convergence the series The nth term is (2n + l)/(n + l)n3.

Use the limit comparison test with the convergent p-series E 1 /n3. Hence, the given series converges.

37.90

Determine the nth term of and test for convergence the series The nth term is n/[(n + I) 2 - n]. Use the limit comparison test with the divergent series E 1/n. Hence, the given series is divergent.

37.91

Prove the root test: A series of positive terms £ an converges if Assume and, therefore,

and diverges if

Choose r so that L < r < 1. Then there exists an integer k such that, if an1. Then use the first n 2 negative terms until the sum becomes 1, then more negative terms until the sum becomes l. For x = ±l, convergence by Problem 37.50. Hence, the power series converges for — 1 s* s 1. By L'HopitaFs rule, 38.25

we have absolute

(compare Problem 38.15).

Find the radius of convergence of the power series Therefore, the series converges for |x|4. Hence, the radius of convergence is 4.

38.26

Prove that, if a power series £ anx" converges for x = b, \b\.

then it converges absolutely for all x such that

|jc| <

Since £ anb" converges, lim |a n i>"|=0. Since a convergent sequence is bounded, there exists an M such that \aab"\ k, \anx"\ < r"\x\" = \rx\". Thus, eventually, E \anx"\ is term by term less than the convergent geometric series L \rx\ and is convergent by the comparison test. Now, on the other, hand, assume |jt| > 1 I L . = L, there exists Then L > 1/1x1. Choose r so that L > r > l / | x | . Then |rx|>l. Since lim and, therefore, |a n |>r". Hence, for n^k, \anx"\> an integer k such that, if n > k, then r \x\ = |rx| > 1. Thus, we cannot have hm anx =0, and, therefore, L anx cannot converge, ( t h e theorem also holds when L = 0: then the series converges for all x.) 38.30

Find the radius of convergence of Therefore, by Problem 38.29, the radius of convergence is l / e

38.31

Find the radius of convergence of the binomial series Use the ratio test.

Hence, the radius of convergence is 1. 38.32

If E a,,x" has a radius of convergence rl and if E bnx" has a radius of convergence /•-, > r , , what is the radius of convergence of the sum E (an + bn)x"! For |x| l ' contradicting

|*|>1. Then,

lim |aj |*|" =0.

But, for infinitely many values of

Denoting the sum of the hypergeometric series (Problem 38.113) by F(a, b;c;x), show that tan lx = By Problem 38.113, (see Problem 38.36).

CHAPTER 39

Taylor and Maclaurin Series 39.1

Find the Maclaurin series of e*. Let

f("\x) = e'

/(*) = **. Then

for all «>0. Hence,

/'"'(O) = 1 for all n & 0 . Therefore, the

Maclaurin series 39.2

Find the Maclaurin series for sin x. Let

/(;c) = sin X.

Then,

/(0) = sinO = 0, /'(O) = cosO= 1, /"(O) = -sin 0 = 0, /"'(O) = -cosO= -1,

and, thereafter, the sequence of values 0,1,0, — 1 keeps repeating. Thus, we obtain

39.3

Find the Taylor series for sin x about ir/4. Let f(x) = sinx. Then /(ir/4) = sin (ir/4) = V2/2, f ' ( i r / 4 ) = cos(irf4) = V2/2, f(ir/4) = -sin (77/4) = and, thereafter, this cycle of four values keeps repeating. Thus, the Taylor series for sin* about

39.4

Calculate the Taylor series for IIx about 1. Let

Then,

and, in general, /(">(1) = (-!)"«!.

So

39.5

Thus, the Taylor series is

Find the Maclaurin series for In (1 - x). Let/(AT) = In (1 - x). Then,/(0) = 0,/'(0) = -1,/"(0) = -1,/"'(0) = -1 -2,/ < 4 ) = -1 • 2 • 3, and, in general / ( ">(0) = -(„ _ i)i Thus, for n > l,/ ( / 0 (0)/n! = -1/n, and the Maclaurin series is

39.6

Find the Taylor series for In x around 2.

Let

f(x) = lnx.

Then,

and, in general, So

Taylor series is

39.7

/(2) = In 2,

and, for

n > 1,

Thus, the

(x - 2)" = In 2 + i(x - 2) - 4(x - 2)2 + MX - 2)3 - • • •.

Compute the first three nonzero terms of the Maclaurin series for ecos". Let f ( x ) = ecos*. Then, f'(x)=-ecos'sinx, /"(*) = e'05* (sin2 *-cos*), /'"(x) = ecos' (sin jc)(3 cos x + 1-sin 2 *), /(4)(;(0) = 9. Hence, the Maclaurin series is x + $x3 + -jex5 + • • -. x

39.11

If f(x) = 2 an(x - a)" for \x - a\ < r, prove that In other words, if f(x) has a power ii -o series expansion about a, that power series must be the Taylor series for/(*) about a. f(a) = aa. It can be shown that the power series converges uniformly on | * - a | < p < r , allowing differentiation term by term: n(n - 1) • • • [/I - (A: - 1)K(* - «)""*-

39.12

M we let x = a, fm(a) = k(k-l)

1 • ak = k\ • ak. Hence,

Find the Maclaurin series for By Problem 38.34, we know that

for

|*| 3 .

The Maclaurin series for f(x) is the polynomial 39.34

Thus,

f(x) = 2 + x + 2x2 + 2x3.

Thus,

Exhibit the nth nonzero term of the Maclaurin series for In (1 + x 2 ). ln(l + x) = x-x2/2 + x3/3 + (-l)"+Wn + - - for |*|