Prove that the series $\sum_{n =2}^{\infty} \frac{1}{(\ln n)^{\ln n}}$ converges.

Question

I am trying to study the convergence of the series:

$$\sum_{n =2}^{\infty} \frac{1}{(\ln n)^{\ln n}}$$ First I have failed to use the D'Alembert test $$ \lim_{n\to \infty} \frac{(\ln (n+1))^{\ln (n+1)}}{(\ln n)^{\ln n}} =1.$$ So I can Conclude anything from here. I have also tried the integral Comparaison test but it was not soo clear.

Answer 1

A short answer: $$ (\ln n)^{\ln n} = e^{\ln\ln n\cdot \ln n} = n^{\ln\ln n} $$ and since $\ln\ln n>2$ for $n$ big enough,* you can conclude by comparison with $\sum_n\frac{1}{n^2}$.

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${}^*$ Namely, $n > e^{e^2}$.