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Does $\sum_{n=1}^{\infty} \frac{1}{n\sqrt[n]{n}}$ converges?

I said that:

$ \frac{1}{n\sqrt[n]{n}}$ = $ \frac{1}{n^{1+\frac{1}{n}}}$ = $ \frac{1}{n}$ and this is harmonic series which does not converge.

Question is: Is there another way with equation sum series? I thought about dividing it by $1/n$, what do you guys think?

edit: I just noticed that $n^{1+\frac{1}{n}}$ is bigger than one. Which means the series converging. and I was all wrong. is that right?

edit2: $\sum_{n=1}^{\infty} \frac{1}{n\sqrt[n]{n}}$ = Limit of $\frac{1}{n} + 1$. which means it does not converge because the sum of $\frac{1}{n}$ does not converge?

Ilan Aizelman WS
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2 Answers2

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The criterion for convergence is that you have $n^{-a}$ where $a>1$. This $a$ must be independent of the $n$. This is not the case in your case.

What you need to do here is recall that $\sqrt[n]{n}$ tends to $1$ and thus you can bound your series from below by $1/cn$ for a suitable $c$. The use the fact that the harmonic series diverges.

quid
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  • Exactly. So I just divide with $\frac{1}{n}$ and then I can say that limit of $\sqrt[n]n$ is 1 which means the limit will be 1 + 1 = 2, and the sum of $\frac{1}{n}$ is harmonic series, which means it diverges. something along those lines. because $0 < L < \infty$ and I got $\sum $\frac{1}{n}$ as harmonic series. – Ilan Aizelman WS Dec 27 '15 at 20:11
  • The limit of the quotient is in fact $1$ not $1+1$, but this is not directly relevant. What is key is that there exists some constant $c$ such that $1/cn \le 1/n^{1 +1/n} $ at least for all sufficiently large $n$. You do not directly compare to the harmonic series but rather a multiple of it. – quid Dec 27 '15 at 20:18
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    @IlanAizelmanWS you are referring to the limit comparison test i.e. let $a_n\geq 0$ and $b_n \geq 0$ and consider $\sum _n a_n, \sum_n b_n$ then if $ \lim _n \frac{a_n}{b_n}=c$ where $0<c<\infty$ then either both series are convergent or divergenent. – clark Dec 27 '15 at 20:22
  • @clark yes I do. – Ilan Aizelman WS Dec 27 '15 at 20:31
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First show that $f(x) = x^{-1- \frac{1}{x}}$ is decreasing for sufficiently large $x > 0$. This can be done by taking the derivative. Now your series is $\sum_{n = 1}^{\infty} f(n)$. By Cauchy's condensation test the series converges if and only if $\sum_{n = 1}^\infty 2^nf(2^n)$ converges. You should be able to work out the rest.

Hans Engler
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