TITLE: Prove that $\sum_{n=2}^\infty (\ln n)^{- \ln n}$ converges
QUESTION [6 upvotes]: As the title suggests, I'd like to prove that the sum
$$
\sum_{n=2}^\infty (\ln n)^{- \ln n}
$$
is finite.  The root and ratio test both fail here, but WA suggests that there is a comparison that can be used to show convergence.
The only thought I have is that it may help to write the terms as $e^{-\ln(n)\ln(\ln(n))}$, but this has not led me to any particular insight.  Any ideas are appreciated.

REPLY [2 votes]: By Cauchy condensation test we have
$$\sum 2^na_{2n}=\sum \frac{2^n}{(\log 2^n)^{\log 2^n}}=\sum \frac{2^n}{n^{(n\log 2)}\cdot{(\log2)}^{(n\log 2)}}$$
which converges for example by limit comparison test with $\sum \frac1{n^2}$.