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.

Note that $n^{\log(n)}=e^{\log(n)^2}$ and $\log(n)^n=e^{n\log(\log(n))}$. Thus $$a_n=e^{\log(n)^2-n\log(\log(n))},$$ and $a_n\rightarrow0$ iff $\log(n)^2-n\log(\log(n))\rightarrow-\infty$ (which I'm pretty sure it does). If you can find a sequence which bounds $\log(n)^2-n\log(\log(n))$ from above and also goes to $-\infty$ fast enough, you should be able to prove that the sum converges.

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(this is for the previous version with $\log(n^n)$ instead of $\log(n)^n$)

If $a_n\not\rightarrow0$, then the series $\sum a_n$ must diverge. Note that $a_n=f(n)$ where $$f(x)=\frac{x^{\log(x)}}{\log(x^x)}=\frac{x^{\log(x)}}{x\log(x)}=\frac{x^{\log(x)-1}}{\log(x)}.$$ Thus, if we show that $\lim_{x\rightarrow\infty}f(x)\neq0$, then the series must diverge.

We have that $$\lim_{x\rightarrow\infty}\frac{x^{\log(x)-1}}{\log(x)}=\frac{\infty}{\infty}$$ so using L'Hopital this equals $$\lim_{x\rightarrow\infty}\frac{(\log(x)-1)x^{\log(x)-2}\cdot\frac{1}{x}}{\frac{1}{x}}=\lim_{x\rightarrow\infty}(\log(x)-1)x^{\log(x)-2}=\infty$$

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.