# Convergence of $\sum_{n=2}^{\infty}\left(\frac{\log n}{\log(\log n)}\right)^{-\log n/\log(\log n)}$

Initially, I need to prove, that

$$\forall \lambda > 0 \ \forall\, \xi_i \sim \text{Pois}(\lambda), \xi_i\text{ are independent } \implies P\left(\limsup\limits_{n \rightarrow \infty}\frac{\xi_n \log{\log{n}}}{\log{n}} = 1\right) = 1$$

I used Borel-Cantelli lemma and Taylor series with Lagrange remainder, and reduced problem to the following:

$$\forall a > 0, \ \forall\, \varepsilon > 0 \ \sum_{n=2}^{\infty}{(a\ \log(n)/\log(\log(n)))^{-(1-\varepsilon)\log(n)/\log(\log(n)}}$$ - diverges and

$$\sum_{n=2}^{\infty}{(a\log(n)/\log(\log(n)))^{-(1+\varepsilon)\log(n)/\log(\log(n)}}$$ converges.

WolframAlpha tells me that's true, but I can't prove it...

The term $$\exp\left\{-\frac{\log n}{\log\log n}\log\frac{\log n}{\log\log n}\right\}=\frac{1}{n}\exp\frac{\log n\log\log\log n}{\log\log n}$$ is $$\frac{1}{n}$$ times something (slowly) growing. This implies divergence of the original series. The variant with $$a$$ and $$1\pm\varepsilon$$ is handled similarly.