Originally Posted by

**Hweengee** How do I determine the convergence/divergence of the following:

$\displaystyle

\sum _{n=1}^{\infty }\left({\sqrt[n] 3}-1 \right)

$

I can't use limit comparison test with $\displaystyle {\sqrt[n] 3}$ because it gives me 0. Root and ratio tests don't seem appropriate and I'm struggling to find a divergent series which is smaller than this to compare it with. Would it be possible to compare it with $\displaystyle \sum _{n=1}^{\infty }\left(\ln\left(n \right)-1 \right)$?

As for

$\displaystyle \sum _{n=3}^{\infty }\frac{1}{\left( lnln\left(n \right) \right)^{\ln \left( n \right) }}$

I'm more or less completely stuck.

Thanks for the help in advance.