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Exercises: Alternating Series

Theory

Determine whether the following series converge or diverge.

Exercises

Exercise 1
Question
∑n=1∞(−1)n+12n+1\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2n+1}
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Solution

Let an=12n+1a_n=\frac{1}{2n+1}. Then

a_{n+1}=\frac{1}{2n+3}<\frac{1}{2n+1}=a_n, \quad \text{and} \quad \lim_{n\to\infty} a_n=0.

By the Alternating Series Test, the series converges.

Exercise 2
Question
∑n=1∞(−1)n+1n+13n+1\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n+1}{3n+1}
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Solution

Let an=n+13n+1a_n=\frac{n+1}{3n+1}. Then

lim⁡n→∞an=13≠0.\lim_{n\to\infty} a_n=\frac13\neq0.

Hence the series diverges by the Test for Divergence.

Exercise 3
Question
∑n=1∞(−1)n+1e−n\sum_{n=1}^{\infty} (-1)^{n+1} e^{-n}
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Solution

Let an=e−na_n=e^{-n}. Then

a_{n+1}=e^{-(n+1)}<e^{-n}=a_n, \quad \text{and} \quad \lim_{n\to\infty} a_n=0.

Thus the series converges by the Alternating Series Test.

Exercise 4
Question
∑n=1∞(−1)n+11n2\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n^{2}}
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Solution

Let an=1n2a_n=\frac{1}{n^2}. Then

a_{n+1}=\frac{1}{(n+1)^2}<\frac{1}{n^2}=a_n, \quad \text{and} \quad \lim_{n\to\infty} a_n=0.

Hence the series converges by the Alternating Series Test.

Exercise 5
Question
∑n=1∞(−1)n+1(n10)n\sum_{n=1}^{\infty} (-1)^{n+1} \left(\frac{n}{10}\right)^{n}
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Solution

Let an=(n10)na_n=\left(\frac{n}{10}\right)^n. Then

lim⁡n→∞an=∞.\lim_{n\to\infty} a_n=\infty.

Therefore the series diverges by the Test for Divergence.

Exercise 6
Question
∑n=2∞(−1)n+11ln⁡n\sum_{n=2}^{\infty} (-1)^{n+1} \frac{1}{\ln n}
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Solution

Let an=1ln⁡na_n=\frac{1}{\ln n}. For n≥3n\ge3,

a_{n+1}=\frac{1}{\ln(n+1)}<\frac{1}{\ln n}=a_n, \quad \text{and} \quad \lim_{n\to\infty} a_n=0.

Hence the series converges by the Alternating Series Test.

Exercise 7
Question
∑n=2∞(−1)n+1ln⁡nln⁡n2\sum_{n=2}^{\infty} (-1)^{n+1} \frac{\ln n}{\ln n^{2}}
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Solution

Since ln⁡n2=2ln⁡n\ln n^{2}=2\ln n, we have

an=ln⁡nln⁡n2=12.a_n=\frac{\ln n}{\ln n^{2}}=\frac12.

Thus lim⁡n→∞an=12≠0\lim_{n\to\infty} a_n=\frac12\neq0, and the series diverges.

Exercise 8
Question
∑n=1∞(−1)n+1n+1n+1\sum_{n=1}^{\infty} (-1)^{n+1} \frac{\sqrt{n}+1}{n+1}
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Solution

Let an=n+1n+1a_n=\frac{\sqrt n+1}{n+1}. Then

a_{n+1}=\frac{\sqrt{n+1}+1}{n+2}<\frac{\sqrt n+1}{n+1}=a_n, \quad \text{and} \quad \lim_{n\to\infty} a_n=0.

Hence the series converges by the Alternating Series Test.

Exercise 9
Question
∑n=2∞(−1)nln⁡nn\sum_{n=2}^{\infty} (-1)^{n} \frac{\ln n}{n}
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Solution

Let an=ln⁡nna_n=\frac{\ln n}{n}. For n≥3n\ge3,

a_{n+1}=\frac{\ln(n+1)}{n+1}<\frac{\ln n}{n}=a_n, \quad \text{and} \quad \lim_{n\to\infty} a_n=0.

Thus the series converges by the Alternating Series Test.

Exercise 10
Question
∑n=2∞(−1)n+1e−n\sum_{n=2}^{\infty} (-1)^{n+1} e^{-n}
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Solution

Let an=e−na_n=e^{-n}. Then

a_{n+1}=e^{-(n+1)}<e^{-n}=a_n, \quad \text{and} \quad \lim_{n\to\infty} a_n=0.

Therefore the series converges by the Alternating Series Test.