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Exercises: Absolute and Conditional Convergence

Theory

Determine whether the following series converge absolutely, converge conditionally, or diverge.

Exercises

Exercise 1
∑n=1∞(−35)n\sum_{n=1}^{\infty} \left(-\frac{3}{5}\right)^{n}
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Solution

This is a geometric series with ratio \left|\frac{-3}{5}\right|<1. Hence the series converges absolutely.

Exercise 2
∑n=1∞(−1)n3n2\sum_{n=1}^{\infty} (-1)^{n} \frac{3}{n^{2}}
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Solution

Consider

∑n=1∞∣3n2∣=3∑n=1∞1n2,\sum_{n=1}^{\infty} \left|\frac{3}{n^2}\right|=3\sum_{n=1}^{\infty}\frac{1}{n^2},

which converges. Therefore the series converges absolutely.

Exercise 3
∑n=1∞(−1)n(n3en)\sum_{n=1}^{\infty} (-1)^{n} \left(\frac{n^{3}}{e^{n}}\right)
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Solution

Consider

∑n=1∞n3en.\sum_{n=1}^{\infty} \frac{n^3}{e^n}.

Since exponential growth dominates polynomial growth, this series converges. Thus the given series converges absolutely.

Exercise 4
∑n=1∞cos⁡nπn\sum_{n=1}^{\infty} \frac{\cos n\pi}{n}
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Solution

Since cos⁡nπ=(−1)n\cos n\pi=(-1)^n, the series becomes

∑n=1∞(−1)nn.\sum_{n=1}^{\infty} \frac{(-1)^n}{n}.

This converges by the Alternating Series Test, but

∑1n\sum \frac{1}{n}

diverges. Hence the series converges conditionally.

Exercise 5
∑n=1∞(−1)n(n+23n−1)n\sum_{n=1}^{\infty} (-1)^{n} \left(\frac{n+2}{3n-1}\right)^{n}
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Solution

Consider

∑n=1∞∣(n+23n−1)n∣.\sum_{n=1}^{\infty} \left|\left(\frac{n+2}{3n-1}\right)^n\right|.

Since

\lim_{n\to\infty} \frac{n+2}{3n-1}=\frac13<1,

the series converges absolutely by the Root Test.

Exercise 6
∑n=1∞(−1)nn+2n(n+3)\sum_{n=1}^{\infty} (-1)^{n} \frac{n+2}{n(n+3)}
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Solution

Consider

∑n=1∞∣n+2n(n+3)∣.\sum_{n=1}^{\infty} \left|\frac{n+2}{n(n+3)}\right|.

Since the terms behave like 1n\frac{1}{n}, the absolute series diverges. However, the alternating series converges by the Alternating Series Test. Thus the series converges conditionally.

Exercise 7
∑n=1∞sin⁡nπ2\sum_{n=1}^{\infty} \sin \frac{n\pi}{2}
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Solution

The terms do not approach zero, since sin⁡nπ2\sin \frac{n\pi}{2} oscillates. Therefore the series diverges.

Exercise 8
∑n=2∞(−3ln⁡n)n\sum_{n=2}^{\infty} \left(-\frac{3}{\ln n}\right)^{n}
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Solution

Using the Root Test,

∣(−3ln⁡n)n∣n=3ln⁡n→0.\sqrt[n]{\left|\left(-\frac{3}{\ln n}\right)^n\right|} =\frac{3}{\ln n}\to0.

Hence the series converges absolutely.

Exercise 9
∑n=1∞(−1)nn2+1n3+2\sum_{n=1}^{\infty} (-1)^{n} \frac{n^{2}+1}{n^{3}+2}
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Solution

The absolute series behaves like

∑1n,\sum \frac{1}{n},

which diverges. Since the terms decrease to zero, the alternating series converges by the Alternating Series Test. Thus it converges conditionally.

Exercise 10
∑n=1∞(−1)n+1tan⁡−1nn2+2\sum_{n=1}^{\infty} (-1)^{n+1} \frac{\tan^{-1} n}{n^{2}+2}
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Solution

Since 0<\tan^{-1}n<\frac{\pi}{2},

∣tan⁡−1nn2+2∣≤Cn2.\left|\frac{\tan^{-1} n}{n^2+2}\right|\le \frac{C}{n^2}.

Because ∑1n2\sum \frac{1}{n^2} converges, the series converges absolutely.

Exercise 11
∑n=1∞(−100)nn!\sum_{n=1}^{\infty} \frac{(-100)^{n}}{n!}
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Solution

Using the Ratio Test,

lim⁡n→∞∣an+1an∣=lim⁡n→∞100n+1=0.\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| =\lim_{n\to\infty}\frac{100}{n+1}=0.

Thus the series converges absolutely.

Exercise 12
∑n=1∞cos⁡nπnn\sum_{n=1}^{\infty} \frac{\cos n\pi}{n\sqrt{n}}
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Solution

Since cos⁡nπ=(−1)n\cos n\pi=(-1)^n, consider

∑1n3/2,\sum \frac{1}{n^{3/2}},

which converges. Therefore the series converges absolutely.

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

Using the Root Test,

\sqrt[n]{\left|\frac{(n+1)^n}{(2n)^n}\right|} =\frac{n+1}{2n}\to\frac12<1.

Hence the series converges absolutely.

Exercise 14
∑n=1∞(−1)n+1(2n)!2nn!n\sum_{n=1}^{\infty} (-1)^{n+1} \frac{(2n)!}{2^{n} n! n}
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Solution

The terms do not approach zero. Therefore the series diverges.

Exercise 15
∑n=1∞(−1)n+1n+1+n\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sqrt{n+1}+\sqrt{n}}
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Solution

Rationalizing gives

∣1n+1+n∣∼12n.\left|\frac{1}{\sqrt{n+1}+\sqrt{n}}\right| \sim \frac{1}{2\sqrt{n}}.

Since ∑1n\sum \frac{1}{\sqrt{n}} diverges but the alternating series converges, the series converges conditionally.

Exercise 16
∑k=1∞kcos⁡kπk2+1\sum_{k=1}^{\infty} \frac{k \cos k\pi}{k^{2}+1}
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Solution

Since cos⁡kπ=(−1)k\cos k\pi=(-1)^k and the terms do not decrease to zero fast enough, the series diverges.

Exercise 17
∑n=3∞(−1)nln⁡nn\sum_{n=3}^{\infty} \frac{(-1)^{n} \ln n}{n}
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Solution

The absolute series diverges, but ln⁡nn\frac{\ln n}{n} decreases to zero. Thus the series converges conditionally.

Exercise 18
∑n=2∞(−1ln⁡n)n\sum_{n=2}^{\infty} \left(-\frac{1}{\ln n}\right)^{n}
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Solution

By the Root Test,

∣(−1ln⁡n)n∣n=1ln⁡n→0.\sqrt[n]{\left|\left(-\frac{1}{\ln n}\right)^n\right|} =\frac{1}{\ln n}\to0.

Hence the series converges absolutely.

Exercise 19
∑n=1∞(−1)n+1(n+1−n)\sum_{n=1}^{\infty} (-1)^{n+1} (\sqrt{n+1}-\sqrt{n})
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Solution

The absolute series diverges, but the alternating series converges. Thus the series converges conditionally.

Exercise 20
∑n=1∞(−1)n+1(n+n−n)\sum_{n=1}^{\infty} (-1)^{n+1} (\sqrt{n+\sqrt{n}}-\sqrt{n})
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Solution

The terms do not approach zero. Therefore the series diverges.