Convergence of the series $\sum\limits_{n=3}^\infty (\log\log n)^{-\log\log n}$ 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?
 A: To avoid relying on precise estimates, one can apply Cauchy's condensation test whenever one has to check the convergence of a series which contains several (iterated) logarithms.  Usually, this works quite well:
Applying Cauchy's condensation test, we find that convergence of the series
$$\sum_{n \geq 3} \frac{1}{(\log \log n)^{\log \log n}} \tag{1}$$
is equivalent to the convergence of
$$ \sum_{n \geq 3} \frac{2^n}{(\log n)^{\log n}} \tag{2}$$
It is not difficult to see that $$\frac{2^n}{(\log n)^{\log n}}$$
is an increasing (strictly positive) sequence for sufficiently large $n$, e.g. by checking that
$$\frac{d}{dx} \left(\frac{2^x}{(\log x)^{\log x}}\right) > 0.$$
This shows that $(2)$ does not converge; hence, $(1)$ does not converge.
A: 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.

