Convergence/Divergence of an Infinite Series with Natural Logarithms

Question

I've spent a good week and half manipulating and trying different tests to find the convergence or divergence of this series:

$$\sum_{n=0}^\infty \frac{1}{(\ln n)^{\ln n}}$$

I've tried all the Convergence Tests I know, and have tried to exhaust the Ratio Test for many different manipulations of the above, including $\ln n^{-\ln n}$, $\ln n^{\ln(1/n)}$, and, using the relationship $a^x = e^{x\ln a}$, $e^{\ln(1/n)*\ln(\ln(n))}\implies(1/n)^{\ln(\ln n)}$.

Basically, I'm just burnt out trying to solve this, and came for any help someone on here can offer me.

Cheers.

EDIT: the only hint we're given is to note that $n > e^2$.

Answer 1

HINT:

$$(\ln n)^{\ln n} = n ^{\ln \ln n} > n^2$$ for all $n > n_0$. So $$\frac{1}{n ^{\ln \ln n}} < \frac{1}{n^2}$$

Then use comparison test