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

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?

• $\frac{1}{n^{1+\frac{1}{n}}} < \frac{1}{n}$, so you can't compare to the harmonic series. – vhspdfg Dec 27 '15 at 20:02
• $\root n\of n \rightarrow 1$. – David Mitra Dec 27 '15 at 20:04
• The fact that $n^{1+1/n} > n$ does not prove convergence. It just means that your divergence proof does not work. – Hans Engler Dec 27 '15 at 20:04
• It diverges by the limit comparison test since $\sum\frac1n$ diverges and $\sqrt[n]{n}\rightarrow1$. – Gregory Grant Dec 27 '15 at 20:08
• Don't ever write $1/n^{(1+1/n)}=1/n$ on an exam. – DanielWainfleet Dec 28 '15 at 4:01

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.
• 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$. You do not directly compare to the harmonic series but rather a multiple of it. – quid Dec 27 '15 at 20:18 • @IlanAizelmanWS you are referring to the limit comparison test i.e. let$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 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.