For what values of y does this series converge?

Question

$\sum_{n=2}^{\inf} \frac{|y|^{1/n}}{nlog^2n}$

What could I use if I wanted to do a comparison test?

Note that no matter what $y$ is, for $n$ large enough, $|y|^{1/n}<2$ — saulspatz, Apr 24 '20 at 21:29

herb steinberg · Accepted Answer · 2020-04-25T10:30:08.100

1

For $y\lt 1$, $y^{1/n}\lt 1$. For $y\ge 1$, $y^{1/n}\le y$. In either case the expression is $\le k\sum_{n=2}^\infty \frac{1}{nlog^2n}$ which converges by integral test.

$\sum_{n=3}^\infty \frac{1}{nlog^2n}\lt\int_2^\infty \frac{dx}{xlog^2(x)}$ Let $u=log(x)$, the integral becomes $\int_{log(2)}^\infty \frac{du}{u^2}=\frac{1}{log(2)}$

edited Apr 25 '20 at 10:30

answered Apr 24 '20 at 21:56

herb steinberg

12,765

This series $\sum_{n \geq 2} \frac{1}{n \log^2 n}$ converges though. – Raoul Apr 24 '20 at 22:05
1

@ Raoul You are right. I corrected my answer. – herb steinberg Apr 25 '20 at 10:32

score 0 · Answer 2 · answered Apr 24 '20 at 21:57

For sufficiently large $n,$ we have that $|y|^{1/n}\approx 1,$ so that the series is dominated by $$\sum_{n>N} \frac{1+\epsilon}{n\log^2n}$$ for some large $N$ and small $\epsilon.$ Then use the integral test to obtain $$\int_N^{+\infty}\frac{k}{n\log^2n}\mathrm dn=k\int_N^{+\infty}\log^{-2}n\mathrm d(\log n),$$ from where you should be able to take it.

For what values of y does this series converge?

2 Answers2