TITLE: Convergence of $\sum\limits_{n=3}^\infty \frac1{(\log(\log n))^{\log n}}$
QUESTION [5 upvotes]: I can prove the sum $\sum\limits_{n=3}^\infty \dfrac1{(\log(\log n))^{\log n}}$ converges this way:
I assume that $\exists n_0 \in \mathbb{N}$ such that $\forall n, n \ge n_0$ we have that $$\frac1{(\log(\log n))^{\log n}} \le \frac1{n^2}.$$
That is $$\log(\log n))^{\log n} \ge n^2.$$
Solving this, I get that $n \ge e^{e^{e^2}}$. From here it follows that the given sum converges, as the sum of the first $n_0 - 1$ terms is finite and the sum converges for $n \ge n_0$ by the comparison test.
This feels extremely awkward.
Is this a correct approach? If there is a better way, could you please show me?
Thank you.

REPLY [7 votes]: Yes your proof is valid and here is another idea that isn't very different to your work $(\log(\log(n))^{\log n}=n^{\log(\log(\log n))}$ and clearly there's $n_0$ such that $\log(\log(\log n))\geq2$

REPLY [4 votes]: Use the fact that $a^b = e^{b \log a}$ twice to write $(\log \log n)^{\log n} = e^{\log n \log \log \log n} = n^{\log \log \log n}$. This will be greater than $n^2$ once $\log \log \log n \ge 2$, or once $n \ge e^{e^{e^2}}$ as you observed.

REPLY [3 votes]: Cauchy Condensation says that the given series converges if 
$$
\begin{align}
\sum_{n=2}^\infty\frac{2^n}{\log(n\log(2))^{n\log(2)}}
&=\sum_{n=2}^\infty\left(\frac{e}{\log(n\log(2))}\right)^{n\log(2)}\tag{1}
\end{align}
$$
converges. For $n\ge\frac{e^{2e}}{\log(2)}$, $\frac{e}{\log(n\log(2))}\le\frac12$, and so $(1)$ can be compared to
$$
\sum_{n=2}^\infty\frac1{2^{n\log(2)}}\tag{2}
$$
which is a geometric series.