A test for convergence involving logarithms $$\sum_{n=2}^\infty \frac{n^{\log n}}{(\log n)^n}$$
Tried to use the inequality of $\ln x< x-1$ and apply the comparison theorem but no success. 
 A: Let's upper-bound each term with a term whose sum converges. Note that, for $n$ large enough, $\log \log n \geq 1$, therefore:
$$\frac{n^{\log n}}{(\log n)^n} = \frac{e^{\log^2n}}{e^{n\log\log n}}\leq \frac{e^{\log^2n}}{e^{n}}=e^{-\frac n 2}e^{-\frac n 2+\log^2n}$$
and because for $n$ large enough, $\left(-\frac n 2+\log^2n\right)\leq 0$ (left as an exercise),
we must have $$\frac{n^{\log n}}{(\log n)^n} \leq e^{-\frac n 2}$$
Conclusion: The sum converges
A: We consider 
$$\lim_{n\to \infty}\sqrt[n]{\frac{n^{\log n}}{(\log n)^n}}
=\lim_{n\to \infty}\frac{n^{\frac{\log n}{n}}}{\log n}
=\lim_{n\to \infty}\frac{e^{\frac{(\log n)^2}{n}}}{\log n}=0.$$
By Cauchy test, this series is convergent.
A: Too long for a comment.
The ratio test is interesting to work
$$a_n=\frac{n^{\log n}}{(\log n)^n}\implies \log(a_n)=\log ^2(n)-n \log (n)$$ Now, using Taylor for infinitely large values of $n$,
$$\log(a_{n+1})-\log(a_n)=-1-\log \left({n}\right)+\frac{2 \log
   \left({n}\right)-\frac{1}{2}}{n}+O\left(\frac{1}{n^2}\right)$$ and, continuing with Taylor expansions,
$$\frac{a_{n+1} } {a_n }=e^{\log(a_{n+1})-\log(a_n) }=\frac1 {e n}+\frac{2 \log
   \left({n}\right)-\frac{1}{2}}{en^2}+O\left(\frac{1}{n^3}\right)$$
