Does the series $\sum \limits_{n=2}^{\infty }\frac{n^{\log n}}{(\log n)^{n}}$ converge?

A comparison test will work here; the key is to write both numerator and denominator in terms of exponentials with bases not involving $n$. Note that the numerator is $e^{\log^2 n}$, which is less than $e^{n/2}$ for sufficiently large $n$. The denominator is $e^{n \log \log n}$, which is greater than $e^n$ for sufficiently large $n$; so for all sufficiently large $n$ the terms are less than $e^{-n/2}$ and thus the series converges.

Assuming he means $\log(n^n)$:

Intuitively, $\log n \leq n$ for all $n$ greater some $n_0$. So picking an $n_0$ such that $\log n \leq n$ and $(\log n) - 1 \geq 1$ will yield $\frac{n^{(log n)-1}}{\log n} \geq 1$. So the series diverges.