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Chapter 3 Solutions

Worked solutions to all thirty problems, with the reason for each step. These explanations were added for this edition; the chapter answer key and original worked examples follow below.

Chapter problems →

Problem 1

Original problem · clarified · added worked solution

Determine the type of each of the following series.

Clarified pattern: $\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \dots$ (terms $\frac{(- 1)^{n}}{n}$ for $n \geq 2$ ).

Clarification for part (a): this web edition explicitly uses 1/2−1/3+1/4−… . The source’s displayed 1/2−1/3+1/3−… does not specify an unambiguous continuation. Parts (b)–(e) retain their source expressions.

Part (a): the clarified term is $\frac{(- 1)^{n}}{n}$ for $n \geq 2$ . Its signs alternate and its magnitudes follow the harmonic denominators $2, 3, 4, \dots$ , so it is an alternating harmonic tail.
Part (b): $1 + \frac{1}{2} + \frac{1}{3} + \dots$ is the harmonic series, with terms $\frac{1}{n}$ .
Part (c): $2 + 4 + 8 + \dots$ is geometric with first term $a = 2$ and ratio $r = 2$ .
Part (d): the listed consecutive ratios are $\frac{2}{3}$ and $\frac{(4 / 3)}{2} = \frac{2}{3}$ , identifying the usual geometric pattern with $a = 3$ , $r = \frac{2}{3}$ .
Part (e): $1 - \frac{1}{2} + \frac{1}{3} - \dots$ is the alternating harmonic series with term $\frac{(- 1)^{n - 1}}{n}$ .

Final answer

(a) Alternating harmonic tail starting with denominator 2; (b) harmonic; (c) geometric, $a = 2$ , $r = 2$ ; (d) geometric, $a = 3$ , $r = \frac{2}{3}$ ; (e) alternating harmonic.

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Problem 2

Original problem · added worked solution

Determine if the following series are convergent or divergent.

Part (a): the usual odd-denominator pattern has term $\frac{2}{2 n - 1}$ . Since $\frac{2}{2 n - 1} > \frac{1}{n}$ and the harmonic series diverges, this positive series diverges.
Part (b): $4 + 2 + 1 + \dots$ is geometric with $a = 4$ and $r = \frac{1}{2}$ . Since $| r | < 1$ , it converges to $\frac{4}{1 - 1 / 2} = 8$ .

Final answer

(a) Diverges; (b) converges to $8$ .

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Problem 3

Original problem · added worked solution

Use partial fraction decomposition to determine if the following series is telescopic.

Factor $n^{3} + n^{2} + n + 1 = (n + 1) (n^{2} + 1)$ . Write $\frac{n^{2} - n}{(n + 1) (n^{2} + 1)} = \frac{A}{n + 1} + \frac{B n + D}{n^{2} + 1}$ .
Multiplying through gives $n^{2} - n = A (n^{2} + 1) + (B n + D) (n + 1)$ . Comparing coefficients yields $A = 1$ , $B = 0$ , $D = - 1$ .
Thus the terms are $\frac{1}{n + 1} - \frac{1}{n^{2} + 1}$ . These are not consecutive shifts of the same reciprocal expression, so they do not cancel in a telescoping pattern.
Indeed, the ratio of the original term to $\frac{1}{n}$ tends to 1, so the nonnegative series diverges by limit comparison with the harmonic series.

Final answer

Not telescoping; $\frac{n^{2} - n}{n^{3} + n^{2} + n + 1} = \frac{1}{n + 1} - \frac{1}{n^{2} + 1}$ . The series diverges.

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Problem 4

Original problem · added worked solution

Determine the value of the following series to the 30th term.

For a geometric series with first term $a$ and ratio $r \neq 1$ , the first $N$ terms sum to $S_{N} = \frac{a (1 - r^{N})}{1 - r}$ .
Part (a): $a = 1$ , $r = \frac{3}{2}$ . At $N = 30$ , $S_{30} = 2 ((\frac{3}{2})^{30} - 1)$ .
Part (b): $a = e$ , $r = \frac{1}{\sqrt{e}} = e^{- 1 / 2}$ . At $N = 30$ , $S_{30} = \frac{e (1 - e^{- 15})}{1 - e^{- 1 / 2}}$ .

Final answer

(a) $2 ((\frac{3}{2})^{30} - 1)$ ; (b) $\frac{e (1 - e^{- 15})}{1 - e^{- 1 / 2}}$ .

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Problem 5

Original problem · added worked solution

Determine the value of the following series to an infinite number of terms.

Original source equation or figure

The first term is $a = \sqrt{5}$ and the constant ratio is $r = \frac{1}{\sqrt{5}}$ .
Since $| r | < 1$ , the sum is $\frac{\sqrt{5}}{1 - 1 / \sqrt{5}} = \frac{5}{\sqrt{5} - 1}$ .
Rationalize the denominator to obtain $\frac{5 (\sqrt{5} + 1)}{4}$ .

Final answer

$\frac{5 (\sqrt{5} + 1)}{4}$ .

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Problem 6

Original problem · added worked solution

Create Fourier Series for the following periodic functions.

Use $f (x) \sim \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} (a_{n} \cos (\frac{n π x}{L}) + b_{n} \sin (\frac{n π x}{L}))$ on $- L < x < L$ , with $a_{n} = \frac{1}{L} \int_{- L}^{L} f (x) \cos (\frac{n π x}{L}) d x$ and the analogous sine formula.
Part (a): $L = 1$ and $f (x) = x$ is odd, so $a_{0} = a_{n} = 0$ . Integration by parts gives $b_{n} = 2 \int_{0}^{1} x \sin (n π x) d x = \frac{2 (- 1)^{n + 1}}{n π}$ .
Thus part (a) has series $\frac{2}{π} \sum_{n = 1}^{\infty} \frac{(- 1)^{n + 1}}{n} \sin (n π x)$ . It equals $x$ for $- 1 < x < 1$ ; at the periodic jumps $x = 1 + 2 k$ it converges to the midpoint $0$ .
Part (b): $L = 2$ and $f (x) = x^{2}$ is even, so $b_{n} = 0$ . Calculate $a_{0} = \frac{1}{2} \int_{- 2}^{2} x^{2} d x = \frac{8}{3}$ . Two integrations by parts give $a_{n} = \int_{0}^{2} x^{2} \cos (\frac{n π x}{2}) d x = \frac{16 (- 1)^{n}}{n^{2} π^{2}}$ .
Therefore part (b) has series $\frac{4}{3} + \frac{16}{π^{2}} \sum_{n = 1}^{\infty} \frac{(- 1)^{n}}{n^{2}} \cos (\frac{n π x}{2})$ . The periodic extension is continuous at $x = \pm 2$ , so the series converges to $4$ there and to the periodic function elsewhere.

Final answer

(a) $\frac{2}{π} \sum_{n = 1}^{\infty} \frac{(- 1)^{n + 1}}{n} \sin (n π x)$ ; (b) $\frac{4}{3} + \frac{16}{π^{2}} \sum_{n = 1}^{\infty} \frac{(- 1)^{n}}{n^{2}} \cos (\frac{n π x}{2})$ , with convergence as stated.

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Problem 7

Added problem and worked solution

Find the sum of $6 - 2 + \frac{2}{3} - \frac{2}{9} + \dots$ .

The first term is 6 and the ratio is $- \frac{1}{3}$ . Its absolute value is less than 1.
Use the infinite geometric formula: $S = \frac{6}{1 - (- 1 / 3)} = \frac{6}{4 / 3} = \frac{9}{2}$ .

Final answer

$\frac{9}{2}$ .

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Problem 8

Added problem and worked solution

For $\sum_{n = 0}^{\infty} 5 (\frac{2}{5})^{n}$ , find the sum of its first four terms and its infinite sum.

Because the starting index is 0, the first term is 5, with ratio $\frac{2}{5}$ .
The first four terms sum to $S_{4} = \frac{5 (1 - (2 / 5)^{4})}{1 - 2 / 5} = \frac{203}{25}$ .
For the infinite sum, $| r | < 1$ and $S = \frac{5}{1 - 2 / 5} = \frac{25}{3}$ .

Final answer

$S_{4} = \frac{203}{25}$ and $S = \frac{25}{3}$ .

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Problem 9

Added problem and worked solution

Evaluate $\sum_{n = 0}^{\infty} (\frac{1}{\sqrt{2}})^{n}$ .

The first term is 1 and the ratio is $\frac{1}{\sqrt{2}}$ , whose absolute value is less than 1.
Compute $S = \frac{1}{1 - 1 / \sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2} - 1}$ .
Multiply by the conjugate to obtain $2 + \sqrt{2}$ .

Final answer

$2 + \sqrt{2}$ .

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Problem 10

Added problem and worked solution

Find the first-N-term sum of $\sum_{n = 1}^{\infty} 3 (- 2)^{n - 1}$ and determine whether the infinite series converges.

The first term is 3 and the ratio is $- 2$ . For $N$ terms, $S_{N} = \frac{3 (1 - (- 2)^{N})}{1 + 2} = 1 - (- 2)^{N}$ .
These partial sums alternate with unbounded magnitudes, so they do not approach a finite value.

Final answer

$S_{N} = 1 - (- 2)^{N}$ ; the infinite series diverges.

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Problem 11

Added problem and worked solution

Evaluate $\sum_{n = 3}^{\infty} \frac{4}{2^{n}}$ .

At $n = 3$ , the first included term is $\frac{4}{8} = \frac{1}{2}$ . Each following term is half the preceding term.
Use $S = \frac{1 / 2}{1 - 1 / 2} = 1$ .

Final answer

$1$ .

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Problem 12

Added problem and worked solution

Use a geometric series to express $0.272727 \dots$ as a reduced fraction.

Write the decimal as the geometric series $\frac{27}{100} + \frac{27}{10000} + \frac{27}{1000000} + \dots$ .
The first term is $\frac{27}{100}$ and the ratio is $\frac{1}{100}$ , so the sum is $\frac{27 / 100}{1 - 1 / 100} = \frac{27}{99}$ .
Reduce the fraction by dividing numerator and denominator by 9.

Final answer

$\frac{3}{11}$ .

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Problem 13

Added problem and worked solution

Evaluate $\sum_{n = 1}^{\infty} \frac{1}{n (n + 2)}$ by telescoping.

Decompose $\frac{1}{n (n + 2)} = \frac{1}{2} (\frac{1}{n} - \frac{1}{n + 2})$ .
Summing the first $N$ terms cancels all but $S_{N} = \frac{1}{2} (1 + \frac{1}{2} - \frac{1}{N + 1} - \frac{1}{N + 2})$ .
Take $N \to \infty$ to get $\frac{3}{4}$ .

Final answer

$\frac{3}{4}$ .

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Problem 14

Added problem and worked solution

Evaluate $\sum_{n = 1}^{\infty} (\frac{1}{n + 2} - \frac{1}{n + 5})$ .

Write out the shifted sums. The negative terms beginning with $\frac{1}{6}$ cancel matching positive terms.
The first-N-term sum is $S_{N} = \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{N + 3} - \frac{1}{N + 4} - \frac{1}{N + 5}$ .
The last three terms tend to zero, so the sum is $\frac{20 + 15 + 12}{60} = \frac{47}{60}$ .

Final answer

$\frac{47}{60}$ .

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Problem 15

Added problem and worked solution

Find the partial sums and determine convergence of $\sum_{n = 1}^{\infty} \ln (\frac{n + 1}{n})$ .

Use $\ln (\frac{n + 1}{n}) = \ln (n + 1) - \ln n$ .
The first $N$ terms telescope to $S_{N} = \ln (N + 1) - \ln 1 = \ln (N + 1)$ .
This tends to $+ \infty$ , so telescoping does not by itself guarantee convergence.

Final answer

$S_{N} = \ln (N + 1)$ ; diverges to $+ \infty$ .

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Problem 16

Added problem and worked solution

Evaluate $\sum_{n = 1}^{\infty} (\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}})$ .

In $S_{N}$ , every negative term $- \frac{1}{\sqrt{n + 1}}$ cancels the next positive term.
Thus $S_{N} = 1 - \frac{1}{\sqrt{N + 1}}$ .
As $N \to \infty$ , the remaining reciprocal tends to zero.

Final answer

$1$ .

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Problem 17

Added problem and worked solution

What is the smallest number of terms needed to approximate $\sum_{n = 1}^{\infty} \frac{1}{n (n + 1)}$ with absolute error strictly less than $0.001$ ?

The decomposition $\frac{1}{n (n + 1)} = \frac{1}{n} - \frac{1}{n + 1}$ gives $S_{N} = 1 - \frac{1}{N + 1}$ and total sum 1.
The exact remainder after $N$ terms is $R_{N} = \frac{1}{N + 1}$ .
Require $\frac{1}{N + 1} < \frac{1}{1000}$ . This means $N + 1 > 1000$ , so the smallest integer $N$ is 1000.

Final answer

Use at least $1000$ terms.

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Problem 18

Added problem and worked solution

Evaluate $\sum_{n = 1}^{\infty} (\frac{2}{n} - \frac{2}{n + 1})$ .

Factor out 2. The remaining sum has partial sum $1 - \frac{1}{N + 1}$ .
Therefore $S_{N} = 2 - \frac{2}{N + 1}$ , which tends to 2.

Final answer

$2$ .

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Problem 19

Added problem and worked solution

Prove that $\sum_{n = 1}^{\infty} \frac{1}{n}$ diverges by grouping terms in blocks whose endpoints are powers of 2.

In the block from $n = 2^{k - 1} + 1$ to $n = 2^{k}$ , there are $2^{k - 1}$ terms and each is at least $\frac{1}{2^{k}}$ .
Each block therefore contributes at least $\frac{1}{2}$ .
After $m$ blocks, $H_{2^{m}} \geq 1 + \frac{m}{2}$ , which tends to infinity. Since all terms are positive, the full partial-sum sequence diverges to infinity.

Final answer

The harmonic series diverges.

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Problem 20

Added problem and worked solution

Determine whether $\sum_{n = 1}^{\infty} \frac{1}{2 n - 1}$ converges. Give a comparison with the harmonic series.

For every $n \geq 1$ , $2 n - 1 < 2 n$ , so $\frac{1}{2 n - 1} > \frac{1}{2 n}$ .
The comparison series $\frac{1}{2} \sum \frac{1}{n}$ diverges. A positive series larger term by term also diverges.

Final answer

Diverges.

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Problem 21

Added problem and worked solution

Show that the alternating harmonic series $1 - \frac{1}{2} + \frac{1}{3} - \dots$ sums to $\ln 2$ by integrating a finite geometric identity.

Use the finite geometric identity $\sum_{k = 0}^{N - 1} (- x)^{k} = \frac{1 - (- x)^{N}}{1 + x}$ for $0 \leq x \leq 1$ .
Integrating term by term over this finite sum gives $\sum_{n = 1}^{N} \frac{(- 1)^{n - 1}}{n} = \int_{0}^{1} \frac{1}{1 + x} d x - \int_{0}^{1} \frac{(- x)^{N}}{1 + x} d x$ .
The second integral has absolute value at most $\int_{0}^{1} x^{N} d x = \frac{1}{N + 1} \to 0$ . The first integral equals $\ln 2$ .

Final answer

$\ln 2$ .

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Problem 22

Added problem and worked solution

Evaluate the paired series $\sum_{n = 1}^{\infty} (\frac{1}{n} - \frac{1}{n + 2})$ and explain why its finite partial sums matter.

Keep the terms paired in each finite partial sum, rather than subtracting two separately divergent infinite sums.
Cancellation gives $S_{N} = 1 + \frac{1}{2} - \frac{1}{N + 1} - \frac{1}{N + 2}$ .
The two reciprocal tail terms vanish, so the paired series has sum $\frac{3}{2}$ .

Final answer

$\frac{3}{2}$ .

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Problem 23

Added problem and worked solution

Find the real Fourier series of the $2 π$ -periodic function $f (x) = 1$ .

Use the real Fourier convention on $- π < x < π$ : $f \sim \frac{a_{0}}{2} + \sum (a_{n} \cos n x + b_{n} \sin n x)$ .
For $f = 1$ , $a_{0} = \frac{1}{π} \int_{- π}^{π} 1 d x = 2$ . Integrals of $\cos n x$ and $\sin n x$ over a whole period are zero for every integer $n \geq 1$ .
Thus all nonconstant coefficients vanish, and the series equals 1 at every point.

Final answer

$a_{0} = 2$ , $a_{n} = b_{n} = 0$ for $n \geq 1$ ; Fourier series $1$ .

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Problem 24

Added problem and worked solution

Find the Fourier series of the $2 π$ -periodic extension of $f (x) = x$ on $- π < x < π$ , including its value at the periodic jumps.

The function is odd, so $a_{0} = a_{n} = 0$ . Its sine coefficient is $b_{n} = \frac{2}{π} \int_{0}^{π} x \sin (n x) d x$ .
Integration by parts gives $\int_{0}^{π} x \sin (n x) d x = - \frac{π (- 1)^{n}}{n}$ , hence $b_{n} = \frac{2 (- 1)^{n + 1}}{n}$ .
At interior points the series equals $x$ . At each periodic jump $x = (2 k + 1) π$ , the two limits are $π$ and $- π$ , so the Fourier series equals their average, zero.

Final answer

$2 \sum_{n = 1}^{\infty} \frac{(- 1)^{n + 1}}{n} \sin (n x)$ ; endpoint value $0$ .

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Problem 25

Added problem and worked solution

Find the Fourier series of the $2 π$ -periodic extension of $f (x) = | x |$ on $- π \leq x \leq π$ .

The function is even, so $b_{n} = 0$ . Compute $a_{0} = \frac{2}{π} \int_{0}^{π} x d x = π$ .
Integration by parts gives $a_{n} = \frac{2}{π} \int_{0}^{π} x \cos (n x) d x = \frac{2 ((- 1)^{n} - 1)}{π n^{2}}$ . Thus even indices vanish and odd indices give $- \frac{4}{π n^{2}}$ .
Substitute these coefficients. The periodic extension is continuous at $\pm π$ , so the series equals $| x |$ there as well as inside the interval.

Final answer

$\frac{π}{2} - \frac{4}{π} \sum_{k = 0}^{\infty} \frac{\cos ((2 k + 1) x)}{(2 k + 1)^{2}}$ .

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Problem 26

Added problem and worked solution

Find the Fourier series of the $2 π$ -periodic function equal to $- 1$ for $- π < x < 0$ and $1$ for $0 < x < π$ . State the sum at each jump.

The function is odd, so only sine coefficients can be nonzero. Use $b_{n} = \frac{2}{π} \int_{0}^{π} \sin (n x) d x$ .
The integral gives $b_{n} = \frac{2 (1 - (- 1)^{n})}{π n}$ : it is zero for even $n$ and $\frac{4}{π n}$ for odd $n$ .
The Fourier series equals 1 or −1 at points of continuity. At $x = k π$ , the two one-sided values are opposite, so the series converges to zero.

Final answer

$\frac{4}{π} \sum_{k = 0}^{\infty} \frac{\sin ((2 k + 1) x)}{2 k + 1}$ ; value $0$ at $x = k π$ .

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Problem 27

Added problem and worked solution

Find the Fourier series of the $2 π$ -periodic extension of $f (x) = x^{2}$ on $- π \leq x \leq π$ , including the endpoint sum.

Evenness gives $b_{n} = 0$ . Also $a_{0} = \frac{2}{π} \int_{0}^{π} x^{2} d x = \frac{2 π^{2}}{3}$ .
Two integrations by parts yield $\int_{0}^{π} x^{2} \cos (n x) d x = \frac{2 π (- 1)^{n}}{n^{2}}$ , so $a_{n} = \frac{4 (- 1)^{n}}{n^{2}}$ .
The periodic extension is continuous at the endpoints, and the series equals $π^{2}$ there.

Final answer

$\frac{π^{2}}{3} + 4 \sum_{n = 1}^{\infty} \frac{(- 1)^{n}}{n^{2}} \cos (n x)$ .

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Problem 28

Added problem and worked solution

List the real Fourier coefficients of $f (x) = 3 \cos (2 x) - 2 \sin (3 x)$ in the convention $f = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} (a_{n} \cos n x + b_{n} \sin n x)$ .

Compare the function with $\frac{a_{0}}{2} + \sum_{n = 1}^{\infty} (a_{n} \cos n x + b_{n} \sin n x)$ .
There is no constant term; the cosine of frequency 2 has coefficient 3 and the sine of frequency 3 has coefficient −2.
All other coefficients vanish by orthogonality of the trigonometric functions over a full period.

Final answer

$a_{0} = 0$ , $a_{2} = 3$ , $b_{3} = - 2$ , and all other $a_{n}, b_{n}$ are zero.

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Problem 29

Added problem and worked solution

Find the Fourier series of the 4-periodic extension of $f (x) = x$ on $- 2 < x < 2$ , and state its sum at the jumps.

The period is 4, so use $L = 2$ and basis functions $\sin (\frac{n π x}{2})$ . Oddness makes all cosine coefficients zero.
Compute $b_{n} = \frac{2}{L} \int_{0}^{L} x \sin (\frac{n π x}{L}) d x$ . Integration by parts with $L = 2$ gives $b_{n} = \frac{4 (- 1)^{n + 1}}{n π}$ .
The series equals $x$ on $- 2 < x < 2$ . At the periodic jumps $x = 2 + 4 k$ , the average of the one-sided limits is zero.

Final answer

$\frac{4}{π} \sum_{n = 1}^{\infty} \frac{(- 1)^{n + 1}}{n} \sin (\frac{n π x}{2})$ , with value $0$ at the jumps.

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Problem 30

Added problem and worked solution

A square wave has Fourier series $\frac{4}{π} \sum_{k = 0}^{\infty} \frac{\sin ((2 k + 1) x)}{2 k + 1}$ . Write its first three nonzero terms and evaluate that approximation at $x = \frac{π}{2}$ .

The first three nonzero terms occur at the odd frequencies 1, 3, and 5, so $F_{3} (x) = \frac{4}{π} (\sin x + \frac{\sin 3 x}{3} + \frac{\sin 5 x}{5})$ .
At $x = \frac{π}{2}$ , the sine values are 1, −1, and 1.
Therefore $F_{3} (\frac{π}{2}) = \frac{4}{π} (1 - \frac{1}{3} + \frac{1}{5}) = \frac{52}{15 π}$ . This is an approximation to the square wave’s value 1 there.

Final answer

$F_{3} (x) = \frac{4}{π} (\sin x + \frac{\sin 3 x}{3} + \frac{\sin 5 x}{5})$ and $F_{3} (\frac{π}{2}) = \frac{52}{15 π}$ .

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