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

(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
$$\lim_{x\rightarrow\infty}\frac{(\log(x)-1)x^{\log(x)-2}\cdot\frac{1}{x}}{\frac{1}{x}}=\lim_{x\rightarrow\infty}(\log(x)-1)x^{\log(x)-2}=\infty$$
A: 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.
A: 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
A: 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.
A: root test works fine:
$$\limsup_{n\to\infty} \sqrt[n] {\frac{n^{\log n}}{(\log n)^n}}=\limsup_{n\to\infty} \frac{\sqrt[n] {n^{\log n}}}{\sqrt[n] {(\log n)^n}}=\limsup_{n\to\infty}\frac{1}{\log n}$$
and for all $n>$(the base of the logarithm):  $$\frac{1}{\log n}<1$$ therefor this series converges by the root test (and we are able to use this test because $a_n \geq0$) .
