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?

$$\frac{1}{(\log n)^{\log n}}=e^{-\log n(\log(\log n))}=\frac{1}{n^{\log(\log n)}}\leq\frac{1}{n^2}$$
for large $n$, hence the given series is convergent by comparison.
