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Exercises: Ratio Test

Theory

Determine whether the following series converge or diverge.

Exercises

Exercise 1
Question
∑n=1∞n22n\sum_{n=1}^{\infty} \frac{n^{\sqrt{2}}}{2^{n}}
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Solution

Let an=n22na_n=\frac{n^{\sqrt{2}}}{2^n}. Then

\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| =\lim_{n\to\infty}\frac{(n+1)^{\sqrt{2}}}{n^{\sqrt{2}}}\cdot\frac{1}{2} =\frac12<1.

Hence the series converges.

Exercise 2
Question
∑n=1∞n!e−n\sum_{n=1}^{\infty} n!e^{-n}
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Solution

Let an=n!e−na_n=n!e^{-n}. Then

lim⁡n→∞∣an+1an∣=lim⁡n→∞(n+1)!e−(n+1)n!e−n=lim⁡n→∞n+1e=∞.\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| =\lim_{n\to\infty}\frac{(n+1)!e^{-(n+1)}}{n!e^{-n}} =\lim_{n\to\infty}\frac{n+1}{e}=\infty.

Thus the series diverges.

Exercise 3
Question
∑n=1∞n1010n\sum_{n=1}^{\infty} \frac{n^{10}}{10^{n}}
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Solution

Let an=n1010na_n=\frac{n^{10}}{10^n}. Then

\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| =\lim_{n\to\infty}\frac{(n+1)^{10}}{n^{10}}\cdot\frac{1}{10} =\frac{1}{10}<1.

Hence the series converges.

Exercise 4
Question
∑n=1∞ln⁡nen\sum_{n=1}^{\infty} \frac{\ln n}{e^{n}}
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Solution

Let an=ln⁡nena_n=\frac{\ln n}{e^n}. Then

\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| =\lim_{n\to\infty}\frac{\ln(n+1)}{\ln n}\cdot\frac{1}{e} =\frac{1}{e}<1.

Thus the series converges.

Exercise 5
Question
∑n=1∞(n+1)(n+2)n!\sum_{n=1}^{\infty} \frac{(n+1)(n+2)}{n!}
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Solution

Let an=(n+1)(n+2)n!a_n=\frac{(n+1)(n+2)}{n!}. Then

lim⁡n→∞∣an+1an∣=lim⁡n→∞(n+2)(n+3)(n+1)!⋅n!(n+1)(n+2)=lim⁡n→∞n+3(n+1)2=0.\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| =\lim_{n\to\infty}\frac{(n+2)(n+3)}{(n+1)!}\cdot\frac{n!}{(n+1)(n+2)} =\lim_{n\to\infty}\frac{n+3}{(n+1)^2}=0.

Hence the series converges.

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

Let an=n!(2n+1)!a_n=\frac{n!}{(2n+1)!}. Then

∣an+1an∣=(n+1)!(2n+3)!⋅(2n+1)!n!=n+1(2n+3)(2n+2)→0.\left|\frac{a_{n+1}}{a_n}\right| =\frac{(n+1)!}{(2n+3)!}\cdot\frac{(2n+1)!}{n!} =\frac{n+1}{(2n+3)(2n+2)}\to0.

Thus the series converges.

Exercise 7
Question
∑n=1∞2nn!n!(2n)!\sum_{n=1}^{\infty} \frac{2^{n}n!n!}{(2n)!}
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Solution

Let an=2nn!n!(2n)!a_n=\frac{2^n n!n!}{(2n)!}. Then

\left|\frac{a_{n+1}}{a_n}\right| =\frac{2(n+1)^2}{(2n+2)(2n+1)} =\frac{n+1}{2n+1}\to\frac12<1.

Hence the series converges.

Exercise 8
Question
∑n=1∞1⋅3⋅…⋅(2n−1)4n2nn!\sum_{n=1}^{\infty} \frac{1\cdot3\cdot\ldots\cdot(2n-1)}{4^{n}2^{n}n!}
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Solution

Let an=1⋅3⋯(2n−1)4n2nn!a_n=\frac{1\cdot3\cdots(2n-1)}{4^n2^n n!}. Then

\left|\frac{a_{n+1}}{a_n}\right| =\frac{2n+1}{8(n+1)}\to\frac14<1.

Thus the series converges.

Exercise 9
Question
∑n=1∞5n+nn!+3\sum_{n=1}^{\infty} \frac{5^{n}+n}{n!+3}
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Solution

Since n!n! dominates 5n5^n,

∣an+1an∣∼5n+1(n+1)!⋅n!5n=5n+1→0.\left|\frac{a_{n+1}}{a_n}\right| \sim\frac{5^{n+1}}{(n+1)!}\cdot\frac{n!}{5^n} =\frac{5}{n+1}\to0.

Hence the series converges.

Exercise 10
Question
∑n=1∞32+n⋅5n\sum_{n=1}^{\infty} \frac{3}{2+n\cdot5^{n}}
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Solution

Let an=32+n5na_n=\frac{3}{2+n5^n}. Then

\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| =\lim_{n\to\infty}\frac{n5^n}{(n+1)5^{n+1}} =\frac15<1.

Thus the series converges.

Exercise 11
Question
∑n=1∞(2n)!n!(2n)n\sum_{n=1}^{\infty} \frac{(2n)!}{n!(2n)^n}
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Solution

Let an=(2n)!n!(2n)na_n=\frac{(2n)!}{n!(2n)^n}. Then

\left|\frac{a_{n+1}}{a_n}\right| =\frac{(2n+2)(2n+1)}{(n+1)(2n+2)}\cdot\frac{n^n}{(n+1)^{n+1}} \to\frac12<1.

Hence the series converges.

Exercise 12
Question
∑n=1∞(2n)!3n\sum_{n=1}^{\infty} \frac{(2n)!}{3^{n}}
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Solution

Let an=(2n)!3na_n=\frac{(2n)!}{3^n}. Then

∣an+1an∣=(2n+2)(2n+1)3→∞.\left|\frac{a_{n+1}}{a_n}\right| =\frac{(2n+2)(2n+1)}{3}\to\infty.

Thus the series diverges.