TITLE: For what values of $x$ does the series $\sum_{n=1}^\infty \frac{1}{(\ln x)^{\ln n}}$ converge?
QUESTION [5 upvotes]: I have to study the values of $x$ for which $$\sum_{n=1}^\infty \frac{1}{(\ln x)^{\ln n}}$$ converges. 
First we say that we must have $x>0$. Then, I have started by rewriting the series as 
$$\sum_{n=1}^\infty \frac{1}{e^{\ln\ln x\ln n}} = \sum_{n=1}^\infty \frac{1}{n^{\ln\ln x}}.$$
It converges if $\ln\ln x >1$. So it converges for $x>e^e$. Is my reasoning correct?

REPLY [3 votes]: It is true that $(\ln x)^{\ln n} = n^{\ln \ln x}$, hence it should hold when $\ln \ln x > 1$, which happens, as you correctly wrote, when $x>e^e$.