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Does the series $\sum\limits_{n=1}^{\infty}\frac{1}{n\sqrt[n]{n}}$ converge?

VERA
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    What do you know about $\lim_{n \to \infty} \sqrt[n]{n}$? Do you know the limit comparison test?. Show some effort or this is going to be closed. – RRL Oct 19 '18 at 05:24
  • While it won't help with most of the details, eventually, for intuition sake, $\sqrt[n]{n}\approx 1$. – David P Oct 19 '18 at 05:27
  • If I know the limit of comparison – VERA Oct 19 '18 at 05:38

2 Answers2

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For $n>0$, $n < 2^n$. So, $\sqrt[n]{n} < 2$, and $\frac{1}{\sqrt[n]{n}} > \frac{1}{2}$.

Thus, $\frac{1}{n \sqrt[n]{n}} > \frac{1}{2n}$, so $$ \sum_{n=1}^{k} \frac{1}{n \sqrt[n]{n}} > \sum_{n=1}^{k} \frac{1}{2n}$$ and since the right side diverges to infinity, so does the left.

msm
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Limit Comparison Test:

Suppose that we have two series $\displaystyle\sum a_n $ and $\displaystyle\sum b_n$ with $a_n\geq0$,$b_n>0 $ $\forall n$. Define, $$ c = \displaystyle\lim_{n\to \infty} \frac{a_n}{b_n} $$ if $c$ is positive (i.e. $c>0$) and is finite (i.e. $c<\infty$) then either both series converge or both series diverge.

Take $a_n=\sum_{n=1}^{\infty} \frac{1}{n}$ and $b_n=\sum_{n=1}^{\infty} \frac{1}{n \sqrt[n]{n}}$

$$\lim_{n\to \infty} \frac{a_n}{b_n}=\frac{n \sqrt[n]{n}}{n}=\lim_{n\to\infty}\sqrt[n]{n}=1$$ Proof: Proof that $\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$

Therefore the series in the question diverges.

tensor product
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  • Note that the LCT also works when $c=0$ (for convergence case) and when $c=\infty $(for the divergent case). – user Oct 19 '18 at 06:50