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Chapter 4 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 → · Chapter answer key →

Problem 1

Original problem · added worked solution

Use the ratio, root, and comparison test to determine the convergence of the following series.

Original source equation or figure

For integers $n \geq 2$ , $0 < \ln n < n$ , so $\frac{1}{\ln n} > \frac{1}{n}$ .
The harmonic series starting at 2 diverges. Since every term of the displayed positive series is larger, direct comparison proves divergence.

Final answer

Diverges.

Return to problem 1 →

Problem 2

Original problem · added worked solution

Use the ratio, root, and comparison test to determine the convergence of the following series.

Original source equation or figure

For $a_{n} = (\frac{2}{e})^{n}$ , the nth root is $| a_{n} |^{1 / n} = \frac{2}{e}$ .
Since $e > 2$ , the root-test limit is less than 1, so the series converges absolutely.

Final answer

Converges; root-test limit $\frac{2}{e} < 1$ .

Return to problem 2 →

Problem 3

Original problem · added worked solution

Use the ratio, root, and comparison test to determine the convergence of the following series.

Original source equation or figure

Use absolute values: $| a_{n} | = \frac{n^{3}}{2^{n}}$ . Its nth root is $\frac{n^{3 / n}}{2}$ .
Because $\ln (n^{3 / n}) = \frac{3 \ln n}{n} \to 0$ , $n^{3 / n} \to 1$ .
The root-test limit is $\frac{1}{2} < 1$ , proving absolute convergence of the alternating series.

Final answer

Absolutely convergent; root-test limit $\frac{1}{2}$ .

Return to problem 3 →

Problem 4

Original problem · added worked solution

Use the ratio, root, and comparison test to determine the convergence of the following series.

Original source equation or figure

The nth root of $a_{n} = n 3^{- n}$ is $\frac{n^{1 / n}}{3}$ .
Since $n^{1 / n} \to 1$ , the root-test limit is $\frac{1}{3} < 1$ .

Final answer

Converges.

Return to problem 4 →

Problem 5

Original problem · added worked solution

Use the ratio, root, and comparison test to determine the convergence of the following series.

Original source equation or figure

For $n \geq 1$ , $\ln n \geq 0$ and $\ln n \leq n$ , so $0 \leq \frac{\ln n}{n^{3}} \leq \frac{1}{n^{2}}$ .
The comparison p-series has exponent 2 and converges. Direct comparison therefore proves convergence of the displayed series.

Final answer

Converges.

Return to problem 5 →

Problem 6

Original problem · added worked solution

Use the ratio, root, and comparison test to determine the convergence of the following series.

Original source equation or figure

For $a_{n} = \frac{n!}{n^{n}}$ , calculate $\frac{a_{n + 1}}{a_{n}} = \frac{(n + 1)!}{(n + 1)^{n + 1}} \frac{n^{n}}{n!} = (\frac{n}{n + 1})^{n}$ .
Take logarithms: $n \ln (\frac{n}{n + 1}) = - n \ln (1 + \frac{1}{n}) \to - 1$ .
Thus the ratio limit is $e^{- 1} < 1$ , proving convergence.

Final answer

Converges; ratio-test limit $e^{- 1}$ .

Return to problem 6 →

Problem 7

Original problem · added worked solution

Use the ratio, root, and comparison test to determine the convergence of the following series.

Original source equation or figure

For $n \geq 1$ , $n^{4} + 2 \geq n^{4}$ , so $0 \leq \frac{n^{2}}{n^{4} + 2} \leq \frac{1}{n^{2}}$ .
Since $\sum \frac{1}{n^{2}}$ converges, the original series converges by direct comparison.

Final answer

Converges.

Return to problem 7 →

Problem 8

Original problem · added worked solution

Use the ratio, root, and comparison test to determine the convergence of the following series.

Original source equation or figure

For $a_{n} = \frac{2^{n}}{n^{3}}$ , $\frac{a_{n + 1}}{a_{n}} = 2 (\frac{n}{n + 1})^{3}$ .
The limit is 2, greater than 1. The terms eventually grow and cannot approach zero, so the series diverges.

Final answer

Diverges; ratio-test limit $2$ .

Return to problem 8 →

Problem 9

Original problem · added worked solution

Use the ratio, root, and comparison test to determine the convergence of the following series.

Original source equation or figure

For $n \geq 2$ , $\cos (n π) \geq - 1$ , so $n^{2} + \cos (n π) \geq n^{2} - 1 \geq \frac{3}{4} n^{2} > 0$ .
Therefore $0 < \frac{1}{n^{2} + \cos (n π)} \leq \frac{4}{3 n^{2}}$ .
The comparison p-series converges, so the displayed series also converges.

Final answer

Converges.

Return to problem 9 →

Problem 10

Original problem · added worked solution

Use the ratio, root, and comparison test to determine the convergence of the following series.

Original source equation or figure

With $a_{n} = \frac{3^{n}}{n^{4}}$ , calculate $\frac{a_{n + 1}}{a_{n}} = 3 (\frac{n}{n + 1})^{4}$ .
The ratio limit is 3, greater than 1, so the series diverges.

Final answer

Diverges; ratio-test limit $3$ .

Return to problem 10 →

Problem 11

Original problem · added worked solution

Use the ratio, root, and comparison test to determine the convergence of the following series.

Original source equation or figure

For $a_{n} = \frac{n^{3}}{3^{n}}$ , calculate $\frac{a_{n + 1}}{a_{n}} = \frac{1}{3} (\frac{n + 1}{n})^{3}$ .
The limit is $\frac{1}{3} < 1$ , so the series converges.

Final answer

Converges; ratio-test limit $\frac{1}{3}$ .

Return to problem 11 →

Problem 12

Original problem · added worked solution

Use the ratio, root, and comparison test to determine the convergence of the following series.

Original source equation or figure

For $a_{n} = \frac{(n^{2} + 1) 3^{n}}{n!}$ , $\frac{a_{n + 1}}{a_{n}} = \frac{3 ((n + 1)^{2} + 1)}{(n + 1) (n^{2} + 1)}$ .
The numerator has degree 2 and the denominator degree 3, so this ratio tends to 0.
Since $0 < 1$ , the ratio test proves convergence.

Final answer

Converges; ratio-test limit $0$ .

Return to problem 12 →

Problem 13

Original problem · added worked solution

Use the ratio, root, and comparison test to determine the convergence of the following series.

Original source equation or figure

The terms are positive. Compare them with $b_{n} = \frac{1}{n}$ .
The ratio is $\frac{a_{n}}{b_{n}} = \frac{n (n^{2} + 1)}{n^{3} + n + 2} \to 1$ .
A finite positive comparison limit transfers convergence behavior. The harmonic series diverges, so this series diverges.

Final answer

Diverges.

Return to problem 13 →

Problem 14

Original problem · added worked solution

Use the ratio, root, and comparison test to determine the convergence of the following series.

Original source equation or figure

For $a_{n} = \frac{(n!)^{2}}{(2 n)!}$ , cancel factorials: $\frac{a_{n + 1}}{a_{n}} = \frac{(n + 1)^{2}}{(2 n + 2) (2 n + 1)}$ .
The ratio tends to $\frac{1}{4} < 1$ , so the series converges.

Final answer

Converges; ratio-test limit $\frac{1}{4}$ .

Return to problem 14 →

Problem 15

Original problem · added worked solution

Use the ratio, root, and comparison test to determine the convergence of the following series.

Original source equation or figure

The nth root of $a_{n} = (\frac{3 n^{3}}{2 n^{3} + 8 n^{2} + 2 n})^{n / 2}$ is $\sqrt{\frac{3 n^{3}}{2 n^{3} + 8 n^{2} + 2 n}}$ .
Divide inside the radical by $n^{3}$ . The limit is $\sqrt{\frac{3}{2}} > 1$ .
The root test therefore proves divergence.

Final answer

Diverges; root-test limit $\sqrt{\frac{3}{2}}$ .

Return to problem 15 →

Problem 16

Added problem and worked solution

Use the ratio test to determine whether $\sum_{n = 1}^{\infty} \frac{n!}{5^{n}}$ converges.

For $a_{n} = \frac{n!}{5^{n}}$ , cancel common factors to get $\frac{a_{n + 1}}{a_{n}} = \frac{n + 1}{5}$ .
This tends to infinity and is eventually greater than 1, so the terms do not tend to zero and the series diverges.

Final answer

Diverges.

Return to problem 16 →

Problem 17

Added problem and worked solution

Use the ratio test to determine whether $\sum_{n = 1}^{\infty} \frac{5^{n}}{n!}$ converges.

For $a_{n} = \frac{5^{n}}{n!}$ , the ratio is $\frac{a_{n + 1}}{a_{n}} = \frac{5}{n + 1}$ .
The limit is 0, less than 1, so the series converges.

Final answer

Converges.

Return to problem 17 →

Problem 18

Added problem and worked solution

Determine convergence of $\sum_{n = 1}^{\infty} (\frac{n + 1}{3 n + 2})^{n}$ using the root test.

The nth root of the positive term is $\frac{n + 1}{3 n + 2}$ .
Divide by $n$ to find the limit $\frac{1}{3} < 1$ . The root test proves convergence.

Final answer

Converges; root-test limit $\frac{1}{3}$ .

Return to problem 18 →

Problem 19

Added problem and worked solution

Determine convergence of $\sum_{n = 1}^{\infty} (1 + \frac{1}{n})^{n^{2}}$ using the root test.

The nth root of $(1 + \frac{1}{n})^{n^{2}}$ is $(1 + \frac{1}{n})^{n}$ .
Its logarithm is $n \ln (1 + \frac{1}{n}) \to 1$ , so the root-test limit is $e > 1$ .
The series diverges by the root test.

Final answer

Diverges; root-test limit $e$ .

Return to problem 19 →

Problem 20

Added problem and worked solution

Determine convergence of $\sum_{n = 1}^{\infty} (\frac{n}{n + 1})^{n^{2}}$ using the root test.

The nth root of $(\frac{n}{n + 1})^{n^{2}}$ is $(\frac{n}{n + 1})^{n}$ .
Take logarithms: $n \ln (\frac{n}{n + 1}) = - n \ln (1 + \frac{1}{n}) \to - 1$ .
Hence the root-test limit is $e^{- 1} < 1$ , proving convergence.

Final answer

Converges; root-test limit $e^{- 1}$ .

Return to problem 20 →

Problem 21

Added problem and worked solution

Use direct comparison to determine convergence of $\sum_{n = 1}^{\infty} \frac{1}{n^{2} + 7}$ .

For every $n \geq 1$ , $0 < \frac{1}{n^{2} + 7} \leq \frac{1}{n^{2}}$ .
The series with terms $\frac{1}{n^{2}}$ converges because it is a p-series with exponent 2. Direct comparison gives the result.

Final answer

Converges.

Return to problem 21 →

Problem 22

Added problem and worked solution

Use limit comparison to determine convergence of $\sum_{n = 1}^{\infty} \frac{1}{\sqrt{n^{2} + 1}}$ .

Compare with the positive harmonic terms $b_{n} = \frac{1}{n}$ .
The ratio is $\frac{n}{\sqrt{n^{2} + 1}} = \frac{1}{\sqrt{1 + 1 / n^{2}}} \to 1$ .
Since the harmonic series diverges and the ratio limit is finite and positive, the original series diverges.

Final answer

Diverges.

Return to problem 22 →

Problem 23

Added problem and worked solution

Use limit comparison to determine convergence of $\sum_{n = 1}^{\infty} \frac{3 n^{2} + 4}{n^{4} + 2}$ .

The leading powers suggest comparison with $b_{n} = \frac{1}{n^{2}}$ .
Compute $\frac{a_{n}}{b_{n}} = \frac{n^{2} (3 n^{2} + 4)}{n^{4} + 2} \to 3$ .
This positive finite limit transfers convergence from $\sum \frac{1}{n^{2}}$ .

Final answer

Converges.

Return to problem 23 →

Problem 24

Added problem and worked solution

Apply the ratio test to $\sum_{n = 1}^{\infty} \frac{1}{n^{2}}$ . Explain its conclusion, then determine the actual convergence with another valid test.

For $a_{n} = \frac{1}{n^{2}}$ , $\frac{a_{n + 1}}{a_{n}} = (\frac{n}{n + 1})^{2} \to 1$ .
A ratio-test limit of 1 is inconclusive; it does not prove either convergence or divergence.
The p-series test independently proves convergence because the exponent is $2 > 1$ .

Final answer

Ratio test inconclusive; the series converges by the p-series test.

Return to problem 24 →

Problem 25

Added problem and worked solution

Determine convergence of $\sum_{n = 2}^{\infty} \frac{1}{n + \ln n}$ using direct comparison.

For $n \geq 2$ , $\ln n \leq n$ , so $n + \ln n \leq 2 n$ . Both denominators are positive.
Taking reciprocals reverses this inequality: $\frac{1}{n + \ln n} \geq \frac{1}{2 n}$ .
Half of the harmonic series diverges, so direct comparison proves divergence.

Final answer

Diverges.

Return to problem 25 →

Problem 26

Added problem and worked solution

Determine convergence of $\sum_{n = 1}^{\infty} \frac{(n!)^{2}}{(3 n)!}$ with the ratio test.

For $a_{n} = \frac{(n!)^{2}}{(3 n)!}$ , cancel factorials to obtain $\frac{a_{n + 1}}{a_{n}} = \frac{(n + 1)^{2}}{(3 n + 3) (3 n + 2) (3 n + 1)}$ .
The numerator grows quadratically and the denominator cubically, so the ratio tends to 0.
A ratio-test limit less than 1 proves convergence.

Final answer

Converges.

Return to problem 26 →

Problem 27

Added problem and worked solution

Determine convergence of $\sum_{n = 1}^{\infty} \frac{(2 n)!}{(n!)^{2} 5^{n}}$ using the ratio test.

For $a_{n} = \frac{(2 n)!}{(n!)^{2} 5^{n}}$ , form the ratio and cancel: $\frac{a_{n + 1}}{a_{n}} = \frac{(2 n + 2) (2 n + 1)}{5 (n + 1)^{2}}$ .
The leading coefficients give ratio limit $\frac{4}{5} < 1$ .
The ratio test therefore proves convergence.

Final answer

Converges; ratio-test limit $\frac{4}{5}$ .

Return to problem 27 →

Problem 28

Added problem and worked solution

For a real parameter $q \geq 0$ , determine exactly when $\sum_{n = 1}^{\infty} n^{2} q^{n}$ converges. Check the boundary cases.

If $q = 0$ , all terms vanish, so the series converges. For $q > 0$ , the ratio is $\frac{a_{n + 1}}{a_{n}} = q (\frac{n + 1}{n})^{2} \to q$ .
Thus $0 < q < 1$ gives convergence, while $q > 1$ gives divergence by the ratio test.
At $q = 1$ , the ratio test is inconclusive, but the terms equal $n^{2}$ and do not tend to zero. This boundary case diverges.

Final answer

Converges for $0 \leq q < 1$ ; diverges for $q \geq 1$ .

Return to problem 28 →

Problem 29

Added problem and worked solution

For real $p$ , determine convergence of $\sum_{n = 1}^{\infty} \frac{1}{n^{p} (n + 1)}$ using limit comparison.

For real $p$ , all terms are positive. Compare $a_{n} = \frac{1}{n^{p} (n + 1)}$ with $b_{n} = \frac{1}{n^{p + 1}}$ .
Their ratio is $\frac{a_{n}}{b_{n}} = \frac{n}{n + 1} \to 1$ , so they have the same convergence behavior.
A p-series with exponent $p + 1$ converges exactly when $p + 1 > 1$ , or $p > 0$ . At $p = 0$ it is harmonic, and for $p < 0$ it also diverges.

Final answer

Converges exactly for $p > 0$ ; diverges for $p \leq 0$ .

Return to problem 29 →

Problem 30

Added problem and worked solution

Determine convergence of $\sum_{n = 1}^{\infty} (\frac{n}{n^{3} + 1} + (\frac{1}{4})^{n})$ by giving a useful upper bound.

For $n \geq 1$ , $0 \leq \frac{n}{n^{3} + 1} \leq \frac{1}{n^{2}}$ because $n^{3} + 1 \geq n^{3}$ .
Thus the whole positive term is bounded above by $\frac{1}{n^{2}} + (\frac{1}{4})^{n}$ .
The first comparison series is a convergent p-series and the second a convergent geometric series. Their sum converges, so direct comparison proves the original series converges.

Final answer

Converges.

Return to problem 30 →

Chapter answer key and worked steps

Answers and worked steps for the original problems, including verified corrections where needed.

Original worked answer

The series diverges through the comparison test
Because ln n < n for n>2,

The series diverges.
Problem 1 →
Original worked answer

The series converges through the root test.
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The series converges.
Problem 2 →
Original worked answer

The series converges through the root test.
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The series converges.
Problem 3 →
Original worked answer

The series converges through the root test.
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= < 1
The series converges.
Problem 4 →
Original worked answer

The series converges through the comparison test.
Because

The series converges.
Problem 5 →
Original worked answer

The series through the ratio test.
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= < 1
The series converges.
Problem 6 →
Original worked answer

7. The series converges through the comparison test.
Because and ...

The series is convergent.
The series diverges through the ratio test.
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Problem 7 →
Original worked answer

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The series diverges.
Problem 8 →
Original worked answer

The series converges through the comparison test.
Because

The series converges.
Problem 9 →
Original worked answer

The series diverges through the ratio test.
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=
=
The series diverges.
Problem 10 →
Original worked answer

The series converges through the ratio test.
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=
The series converges.
Problem 11 →
Original worked answer

The series converges through the ratio test.
=
=
=
=
= 3
= 0 < 1
The series converges.
Problem 12 →
Original worked answer

The series diverges through the comparison test.
Because and

The series diverges.
Problem 13 →
Original worked answer

The series converges through the ratio test.
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=
=
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= < 1
The series converges.
Problem 14 →
Original worked answer

The series diverges through the root test.
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The series diverges.
Problem 15 →