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
Best Answer
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.