Convergence of the series $sumlimits_{n=3}^infty (loglog n)^{-loglog n}$

I am trying to test the convergence of this series from exercise 8.15(j) in Mathematical Analysis by Apostol:

$$\sum_{n=3}^\infty \frac{1}{(\log\log n)^{\log\log n}}$$

I tried every kind of test. I know it should be possible to use the comparison test but I have no idea on how to proceed. Could you just give me a hint?

Answer

Note that, for every $n$ large enough, $$(\log\log n)^{\log\log n}\leqslant(\log n)^{\log\log n}=\exp((\log\log n)^2)\leqslant\exp(\log n)=n,$$ provided, for every $k$ large enough, $$\log k\leqslant\sqrt{k},$$ an inequality you can probably show, used for $k=\log n$. Hence, for every $n$ large enough, $$\frac1{(\log\log n)^{\log\log n}}\geqslant\frac1n,$$ and the series...

...diverges.