TITLE: Does the series $\sum \limits_{n=2}^{\infty }\frac{n^{\log n}}{(\log n)^{n}}$ converge?
QUESTION [8 upvotes]: I'm trying  to find out now whether the series
$\sum_{n=2}^{\infty } a_{n}$ converges or not when
$$a_n = \frac{n^{\log n}}{(\log n)^{n}}$$
Again, I tried d'Alembert $\frac{a_{n+1}}{a_{n}}$,   Cauchy condensation test $\sum \limits_{n=2}^{\infty } 2^{n}a_{2^n}$, and they both didn't work for me.
I can't use Stirling, nor the integral test.
Edit: I'm searching for a solution which uses sequences theorems and doesn't involve functions.  
Thank you

REPLY [2 votes]: I used Cauchy condensation, then comparison test, then root test and it seems to converge:
Condensation test:
$$\sum \limits_{n=2}^{\infty }2^n\frac{2^{n^{n\log 2}}}{(n\log 2)^{2^{n}}}$$
Comparison test:
$$\sum \limits_{n=2}^{\infty }2^n\dfrac{2^{n^{n\log 2}}}{(n\log 2)^{2^{n}}}< \sum \limits_{n=2}^{\infty }2^n\dfrac{n^{n}}{n^{2^{n}}}$$
Root test:
$$\sqrt[n]{\frac{2^nn^{n}}{n^{2^{n}}}} \underbrace{\longrightarrow}_{n \to \infty}0$$
Then the series converges