Showing the series converges [duplicate]

This is a problem from Apostol's Real Analysis book. $$\text{Find if }\sum_{n=1}^{\infty}\dfrac{1}{n^{1+\frac{1}{n}}}\text{ converges or diverges. }$$ I tried to compare with $\displaystyle \sum_{n=1}^{\infty}\dfrac{1}{n^p}$ for suitable $p$, but $p>1$ fails always. I tried to show $\displaystyle \sum_{k=1}^{\infty}2^ka_{2^k}$iconverges, where $\displaystyle a_n=n^{-\left( 1+\frac{1}{n}\right)}$ but again this got too complicated. Can someone give me a proof? Thanks.

Edit : Sorry, I was carried away, because I was thinking it would converge, but the book asked to check for convergence only. I edited it.

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marked as duplicate by David Mitra, LeGrandDODOM, amWhy, ᴡᴏʀᴅs, Hans EnglerAug 6 '14 at 15:49

Are you sure that sum converges? Intuitively, I feel like it shouldn't. – Assaultous2 Aug 6 '14 at 14:30
Hmm... See this. – David Mitra Aug 6 '14 at 14:31
The series clearly diverges. Note that $n^{1/n}$ has limit $1$. So after a while (well, immediately) the $n$-th term is $\gt \frac{1}{2n}$. – André Nicolas Aug 6 '14 at 14:31
Agreed, it diverges. – Assaultous2 Aug 6 '14 at 14:31
Very good, André Nicolas – rehband Aug 6 '14 at 14:35

Outline: One can prove, say by induction, that $2^n\gt n$ for every positive integer $n$.
It follows that $n^{1/n}\lt 2$.
An alternative (and, conceptually, a powerful) way to think about such problems is to use the limit comparison test. Note that $n^{1+1/n} = n\cdot n^{1/n}$. What is $\lim\limits_{n\to\infty}n^{1/n}$?