Convergence of $\frac{1}{(\ln n)^{\ln n}}$

Question

Could I have a hint for testing the convergence of the following series please?

$$\sum_{n=2}^\infty\frac{1}{(\ln n)^{\ln n}}$$

I am very appreciative for your help.

Answer 1

Hint: $$\frac{1}{(\ln n)^{\ln n}} = e^{-\ln n \ln\ln n} = \frac{1}{n^{\ln \ln n}}$$ and $\ln\ln n> 2$ for $n$ sufficiently big.

Answer 2

Hint: We have $$\log^{\log\left(n\right)}\left(n\right)>n^{\alpha}\Leftrightarrow\log\left(\log\left(n\right)\right)>\alpha $$ and if $n\geq16 $ holds $$\log\left(\log\left(n\right)\right)>1. $$

Answer 3

Since $(\ln n)^{\ln n}$ is increasing, Cauchy's condensation test may be applied directly: $$2^na_{2^n}=\left(\frac{2}{n^{\ln 2}(\ln 2)^{\ln 2}}\right)^n$$ which rapidly becomes smaller than $2^{-n}$.