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Does the series $\sum_{n=1}^{\infty}a_n=\frac{1}{n^\frac{1}{n}n}$ converge? I know that I can apply comparison test because this series contains nonnegative members.

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    $\begingroup$ The notation is pretty bad... what do you mean when you say this?$$\sum_{n=1}^\infty a_n=\frac1{n^{1/n}n}$$Frankly, it doesn't make sense to me. $\endgroup$ – Simply Beautiful Art Mar 5 '17 at 22:43
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    $\begingroup$ @SimplyBeautifulArt I agree. I suspect what is meant $$\sum\limits_{n=1}^\infty a_n$$ with $$a_n = \frac{1}{n^{1/n}n}. $$ $\endgroup$ – Eff Mar 5 '17 at 22:44
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You can still use the comparison test as follows: Using the AM-GM inequality: $\sqrt[n]{n} = \sqrt[n]{1\cdot 1\cdot 1\cdots n} \le \dfrac{1+1+\cdots + 1+ n}{n}= \dfrac{1\cdot (n-1)+n}{n}= \dfrac{2n-1}{n}\implies \dfrac{1}{n\sqrt[n]{n}}\ge \dfrac{1}{2n-1}> \dfrac{1}{2n} $. Since the series $\displaystyle \sum_{n=1}^\infty \dfrac{1}{2n}$ diverges, so does $\displaystyle \sum_{n=1}^\infty \dfrac{1}{n\sqrt[n]{n}}$.

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Use the fact that $$\lim \limits_{n \to \infty}\sqrt[n]{n} = 1$$ Therefore, in the limit, your term will behave as $a_n\sim\frac1n$. The sum $\sum_{n}a_n$ will behave as harmonic series and will diverge.

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  • $\begingroup$ @YuriyS, thank you for the correction. Answer is edited now $\endgroup$ – Andrei Kulunchakov Mar 5 '17 at 22:53
  • $\begingroup$ The rigourous phrasing would be to say $n^{\frac1n}\sim_\infty 1$, hence $\frac1{n^{\frac1n}n}\sim_\infty \dfrac1n$, and use the criterion test by equivalence of the general terms. $\endgroup$ – Bernard Mar 5 '17 at 22:58

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