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Check if the series $\sum_{n=1}^\infty\frac{1}{n \sqrt[n]{n}}$ converges

I've tried using the Cauchy test and the root test but none of them seems to work with this series and so I decided to apply the comparison test.
I don't know if what I'm doing right now is correct:
It is know that $$\sqrt[n]n \rightarrow1$$ And so if we take sufficiently large n, $n \sqrt[n]{n} < 2n$ Which implies that $$\mbox{For sufficiently large n} \\\frac{1}{n\sqrt[n]n}>\frac 1 {2n}$$ And so now, by the virtue of the comparison test, I can claim that the series in question diverges. Is it a correct way to do this? If so, is it possible to pick any positive integer greater than $2$ instead of $2$ for this method to work?

Matthew Leingang
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Aemilius
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    Perfect.${{{}}}$ –  Nov 20 '17 at 19:13
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    Your reasoning is sound. Remember to always include the summation symbol when discussing a series (I've taken the liberty of putting this in for you). If you write the bald term like "$\frac{1}{n\sqrt[n]{n}}$" by itself, you're talking about a sequence. That might introduce ambiguity because the sequence converges, while the associated series does not. – Matthew Leingang Nov 20 '17 at 19:32
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    Yes, any real number bigger than 1 will work (such as $1.001$, or $42$), since $n^{1/n}$ will be less than that number for all sufficiently large $n$. The number 2 is a convenient choice. – Michael Nov 20 '17 at 19:45

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