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Exercises: Limit Comparison Test

Theory

Determine whether the following series converge or diverge.

Exercises

Exercise 1
∑n=1∞12n+n3\sum_{n=1}^{\infty} \frac{1}{2\sqrt{n}+\sqrt[3]{n}}
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Solution

Compare with 1n\frac{1}{\sqrt n}.

lim⁡n→∞12n+n31n=12\lim_{n\to\infty}\frac{\frac{1}{2\sqrt n+\sqrt[3]{n}}}{\frac{1}{\sqrt n}} =\frac12

Hence the series diverges.

Exercise 2
∑n=3∞1ln⁡(ln⁡n)\sum_{n=3}^{\infty} \frac{1}{\ln(\ln n)}
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Solution

Since ln⁡(ln⁡n)→∞\ln(\ln n)\to\infty slowly, the terms do not decrease fast enough. The series diverges.

Exercise 3
∑n=1∞(ln⁡n)2n3\sum_{n=1}^{\infty} \frac{(\ln n)^2}{n^3}
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Solution

Compare with 1n2\frac{1}{n^2}.

lim⁡n→∞(ln⁡n)2/n31/n2=0\lim_{n\to\infty}\frac{(\ln n)^2/n^3}{1/n^2}=0

Hence the series converges.

Exercise 4
∑n=2∞1n ln⁡n\sum_{n=2}^{\infty} \frac{1}{\sqrt n\,\ln n}
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Solution

Compare with 1n\frac{1}{\sqrt n}.

lim⁡n→∞1/(nln⁡n)1/n=0\lim_{n\to\infty}\frac{1/(\sqrt n\ln n)}{1/\sqrt n}=0

Thus the series diverges.

Exercise 5
∑n=1∞11+ln⁡n\sum_{n=1}^{\infty} \frac{1}{1+\ln n}
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Solution

Compare with 1ln⁡n\frac{1}{\ln n}.

lim⁡n→∞1/(1+ln⁡n)1/ln⁡n=1\lim_{n\to\infty}\frac{1/(1+\ln n)}{1/\ln n}=1

Hence the series diverges.

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

Compare with ln⁡nn\frac{\ln n}{n}.

lim⁡n→∞ln⁡(n+1)/(n+1)ln⁡n/n=1\lim_{n\to\infty}\frac{\ln(n+1)/(n+1)}{\ln n/n}=1

Thus the series diverges.

Exercise 7
∑n=1∞1nn2−1\sum_{n=1}^{\infty} \frac{1}{n\sqrt{n^2-1}}
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Solution

Compare with 1n2\frac{1}{n^2}.

lim⁡n→∞1/(nn2−1)1/n2=1\lim_{n\to\infty}\frac{1/(n\sqrt{n^2-1})}{1/n^2}=1

Hence the series converges.

Exercise 8
∑n=1∞10n+1n(n+1)(n+2)\sum_{n=1}^{\infty} \frac{10n+1}{n(n+1)(n+2)}
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Solution

Compare with 1n2\frac{1}{n^2}.

lim⁡n→∞(10n+1)/(n(n+1)(n+2))1/n2=10\lim_{n\to\infty}\frac{(10n+1)/(n(n+1)(n+2))}{1/n^2}=10

Thus the series converges.

Exercise 9
∑n=1∞tan⁡−1nn1.1\sum_{n=1}^{\infty} \frac{\tan^{-1}n}{n^{1.1}}
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Solution

Since tan⁡−1n→π2\tan^{-1}n\to \frac{\pi}{2}, compare with 1n1.1\frac{1}{n^{1.1}}.

lim⁡n→∞tan⁡−1n/n1.11/n1.1=π2\lim_{n\to\infty}\frac{\tan^{-1}n/n^{1.1}}{1/n^{1.1}}=\frac{\pi}{2}

Hence the series converges.

Exercise 10
∑n=1∞coth⁡nn2\sum_{n=1}^{\infty} \frac{\coth n}{n^2}
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Solution

Since coth⁡n→1\coth n\to1, compare with 1n2\frac{1}{n^2}.

lim⁡n→∞coth⁡n/n21/n2=1\lim_{n\to\infty}\frac{\coth n/n^2}{1/n^2}=1

Thus the series converges.

Exercise 11
∑n=1∞1nnn\sum_{n=1}^{\infty} \frac{1}{n\sqrt[n]{n}}
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Solution

Since nn→1\sqrt[n]{n}\to1, compare with 1n\frac{1}{n}.

lim⁡n→∞1/(nnn)1/n=1\lim_{n\to\infty}\frac{1/(n\sqrt[n]{n})}{1/n}=1

Hence the series diverges.

Exercise 12
∑n=1∞ln⁡nn3\sum_{n=1}^{\infty} \frac{\ln n}{n^3}
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Solution

Compare with 1n2\frac{1}{n^2}.

lim⁡n→∞(ln⁡n)/n31/n2=0\lim_{n\to\infty}\frac{(\ln n)/n^3}{1/n^2}=0

Thus the series converges.

Exercise 13
∑n=1∞n!ln⁡nn(n+2)!\sum_{n=1}^{\infty} \frac{n!\ln n}{n(n+2)!}
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Solution

Note that

n!ln⁡nn(n+2)!=ln⁡nn(n+1)(n+2).\frac{n!\ln n}{n(n+2)!}=\frac{\ln n}{n(n+1)(n+2)}.

Comparing with 1n2\frac{1}{n^2} shows convergence.

Exercise 14
∑n=1∞arctan⁡nn\sum_{n=1}^{\infty} \frac{\arctan n}{n}
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Solution

Since arctan⁡n→π2\arctan n\to \frac{\pi}{2}, compare with 1n\frac{1}{n}.

lim⁡n→∞arctan⁡n/n1/n=π2\lim_{n\to\infty}\frac{\arctan n/n}{1/n}=\frac{\pi}{2}

Hence the series diverges.

Exercise 15
∑n=1∞5n2+2n1+n5\sum_{n=1}^{\infty} \frac{5n^2+2n}{\sqrt{1+n^5}}
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Solution

Compare with 1n\frac{1}{\sqrt n}.

lim⁡n→∞(5n2+2n)/n51/n=5\lim_{n\to\infty}\frac{(5n^2+2n)/\sqrt{n^5}}{1/\sqrt n}=5

Thus the series diverges.

Exercise 16
∑n=1∞23n−1\sum_{n=1}^{\infty} \frac{2}{3^n-1}
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Solution

Compare with 13n\frac{1}{3^n}.

lim⁡n→∞2/(3n−1)1/3n=2\lim_{n\to\infty}\frac{2/(3^n-1)}{1/3^n}=2

Hence the series converges.

Exercise 17
∑n=1∞n2−3nn10−4n23\sum_{n=1}^{\infty} \frac{n^2-3n}{\sqrt[3]{n^{10}-4n^2}}
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Solution

Compare with 1n4/3\frac{1}{n^{4/3}}.

lim⁡n→∞(n2−3n)/n10/31/n4/3=1\lim_{n\to\infty}\frac{(n^2-3n)/n^{10/3}}{1/n^{4/3}}=1

Thus the series converges.

Exercise 18
∑n=1∞11+n\sum_{n=1}^{\infty} \frac{1}{1+\sqrt n}
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Solution

Compare with 1n\frac{1}{\sqrt n}.

lim⁡n→∞1/(1+n)1/n=1\lim_{n\to\infty}\frac{1/(1+\sqrt n)}{1/\sqrt n}=1

Hence the series diverges.

Exercise 19
∑n=1∞nln⁡nn3+1\sum_{n=1}^{\infty} \frac{\sqrt n\ln n}{n^3+1}
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Solution

Compare with ln⁡nn5/2\frac{\ln n}{n^{5/2}}.

lim⁡n→∞nln⁡n/n3ln⁡n/n5/2=1\lim_{n\to\infty}\frac{\sqrt n\ln n/n^3}{\ln n/n^{5/2}}=1

Thus the series converges.

Exercise 20
∑n=1∞n(n+3)(n+1)(n+2)(3n+7)\sum_{n=1}^{\infty} \frac{n(n+3)}{(n+1)(\sqrt n+2)(3n+7)}
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Solution

Compare with 1n\frac{1}{\sqrt n}.

lim⁡n→∞n(n+3)(n+1)(n+2)(3n+7)/1n=13\lim_{n\to\infty}\frac{n(n+3)}{(n+1)(\sqrt n+2)(3n+7)}\Big/\frac{1}{\sqrt n} =\frac13

Hence the series diverges.