Question: Does the following series coverge or diverge?

$$\sum\_{n=2}^{\infty} \frac{1}{(\log n)^{\log\log n}}$$

My attempts : I know that $\log(\log n) > 2$

So $\displaystyle\sum\_{n=2}^{\infty}\frac{1}{(\log n)^{\log\log n}}\le\sum\_{n=2}^{\infty} \frac{1}{(\log n)^{2}}\le\sum\frac{1}{n^2}$

so the given series converges.

Is it correct?

Answer: 
It's incorrect. We don't have $\log n \ge n$.

Instead, $(\log n)^{\log \log n} = \exp((\log \log n)^2) \le \exp(((\log n)^{1/2})^2) = n$ for $n \ge N\_0$, so $$\sum\_{n=2}^\infty \frac 1 {(\log n)^{\log \log n}} \ge C + \sum\_{n=N\_0}^\infty \frac 1 n = \infty$$

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Implicit result used: $\log x \le x^\varepsilon$ for sufficiently large $x$ (depending on $\varepsilon$) for any $\varepsilon > 0$.
The series is divergent. Note that $a\_n= \frac{1}{(\log n)^{\log\log n}}=\exp(-(\log\log n)^2)$ is positive and decreasing and, by [Cauchy condensation test](https://en.wikipedia.org/wiki/Cauchy_condensation_test), the convergence of the given series is equivalent to the convergence of
$$\sum\_{n=2}^\infty 2^n\exp(-(\log\log 2^n)^2)=\sum\_{n=2}^\infty \exp(n\log 2-(\log n+\log\log 2)^2)$$
which is divergent because $n\log 2-(\log n+\log\log 2)^2\to +\infty$.