Sum of series $\sum_{n = 1}^{\infty} \frac{1}{(\log {x})^{\log{n}}}, x \in \mathbb{R}$

Question

I need to find the sum of $\sum_{n = 1}^{\infty} \frac{1}{(\log {x})^{\log{n}}}, x \in \mathbb{R}$. I'm pretty sure this can be solved using a comparison test but I fail to see how this could be done.

Any help is appreciated!

Answer 1

Using that $$ x^{\log n}=e^{\log x \log n}=n^{\log x}$$

we see that the series $\sum_{n=1}^\infty \frac{1}{x^{\log n}}$ converges as long as $\log x>1$

In your case this would be $\log (\log x)>1$.

Answer 2

Obviously, $(\log x)^{\log n}=e^{\log \log x\,\log n}=n^{\log \log x}$, so the series converges for $x>e^e$, and its value is $\zeta(\log \log x)$.