Determine if $\sum \frac{1}{(\ln{n})^{\ln{n}}}$ converges. I currently trying to determine it by comparison. I've tried comparing it with $\frac{1}{n^2}$, and it seems to work but I'm not sure if I did it right.
I've done it like this:
$(\ln {n})^{\ln{n}}>n^2$
$e^{\ln{\ln{n}}\times\ln{n}}>e^{\ln{n}\times2}$
$\ln{\ln{n}}\times\ln{n}>\ln{n}\times2$
$\ln{\ln{n}}>2$
$n>e^{(e^2)}$
So I assume this is enough? A convergent series is greater than the original one after some constant $m$. Anything else I need?

What about the condensation test?
$$2^na_{2^n}=\frac{2^n}{\left(\log2^n\right)^{\log 2^n}}=\frac{2^n}{n^{n\log 2}\left(\log2\right)^{n\log 2}}$$
And now apply the $\,n$-th root test to the above:
$$\sqrt[n]{\frac{2^n}{n^{n\log 2}\left(\log2\right)^{n\log 2}}}=\frac2{n^{log2}\left(\log2\right)^{\log2}}\xrightarrow[n\to\infty]{}0$$
Thus, $\,\sum 2^na_{2^n}\;$ converges $\,\iff\sum a_n\;$ converges.
