Series question with logarithms I want to know how to check the divergence of following sum: 
$\sum_{k=0}^\infty \frac{1}{\sqrt[n]{\log n}}$
I tried to use this result:  $ \lim_{n \rightarrow \infty} \frac{1}{\sqrt[n]{\log n}}=1 \neq 0 $   but I don't know why this result is true. 
Best regards.

First, I assume that the index of summation is $n$ (not $k$), otherwise the sum diverges trivially.  Otherwise, note that (for $n>1$) we have $n^n>n>\log n$.  Hence $n>\sqrt[n]{n}>\sqrt[n]{\log n}$.  We take reciprocals to get $$\frac{1}{n}<\frac{1}{\sqrt[n]{\log n}}$$
Now you can use the comparison test to complete the problem.
