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.


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