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?