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Exercises: n-th Root Test

Theory

Determine whether the following series converge or diverge.

Exercises

Exercise 1
Question
∑n=1∞(n3n+1)n\sum_{n=1}^{\infty} \left(\frac{n}{3n+1}\right)^n
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Solution

Let an=(n3n+1)na_n=\left(\frac{n}{3n+1}\right)^n. Then

\sqrt[n]{a_n}=\frac{n}{3n+1}\to\frac13<1.

Hence the series converges.

Exercise 2
Question
∑n=1∞2+(−1)n(1.25)n\sum_{n=1}^{\infty} \frac{2+(-1)^n}{(1.25)^n}
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Solution

Let an=2+(−1)n(1.25)na_n=\frac{2+(-1)^n}{(1.25)^n}. Then

\sqrt[n]{|a_n|} =\frac{\sqrt[n]{|2+(-1)^n|}}{1.25}\to\frac1{1.25}<1.

Thus the series converges.

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

Let an=(1−3n)na_n=\left(1-\frac{3}{n}\right)^n. Then

ann=1−3n→1.\sqrt[n]{a_n}=1-\frac{3}{n}\to1.

Since an→e−3≠0a_n\to e^{-3}\neq0, the series diverges.

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

Let an=n(ln⁡n)na_n=\frac{n}{(\ln n)^n}. Then

ann=n1/nln⁡n→0.\sqrt[n]{a_n} =\frac{n^{1/n}}{\ln n}\to0.

Hence the series converges.

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

Let an=(n!)nn2na_n=\frac{(n!)^n}{n^{2n}}. Then

ann=n!n2.\sqrt[n]{a_n}=\frac{n!}{n^2}.

Since n!n2→∞\frac{n!}{n^2}\to\infty, the series diverges.

Exercise 6
Question
∑n=1∞nn2n2\sum_{n=1}^{\infty} \frac{n^n}{2^{n^2}}
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Solution

Let an=nn2n2a_n=\frac{n^n}{2^{n^2}}. Then

ann=n2n→0.\sqrt[n]{a_n}=\frac{n}{2^n}\to0.

Thus the series converges.

Exercise 7
Question
∑n=1∞(1−e−n)n\sum_{n=1}^{\infty} (1-e^{-n})^n
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Solution

Let an=(1−e−n)na_n=(1-e^{-n})^n. Then

ann=1−e−n→1.\sqrt[n]{a_n}=1-e^{-n}\to1.

Since an→1a_n\to1, the series diverges.

Exercise 8
Question
∑n=1∞(n+1n+2)n2\sum_{n=1}^{\infty} \left(\frac{n+1}{n+2}\right)^{n^2}
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Solution

Let an=(n+1n+2)n2a_n=\left(\frac{n+1}{n+2}\right)^{n^2}. Then

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

Hence the series converges.

Exercise 9
Question
∑n=1∞(1ln⁡(n+2))n\sum_{n=1}^{\infty} \left(\frac{1}{\ln(n+2)}\right)^n
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Solution

Let an=(1ln⁡(n+2))na_n=\left(\frac{1}{\ln(n+2)}\right)^n. Then

ann=1ln⁡(n+2)→0.\sqrt[n]{a_n}=\frac{1}{\ln(n+2)}\to0.

Thus the series converges.

Exercise 10
Question
∑n=1∞(5n+23n−1)n\sum_{n=1}^{\infty} \left(\frac{5n+2}{3n-1}\right)^n
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Solution

Let an=(5n+23n−1)na_n=\left(\frac{5n+2}{3n-1}\right)^n. Then

\sqrt[n]{a_n}=\frac{5n+2}{3n-1}\to\frac53>1.

Hence the series diverges.