Let $X_n$ be independent $N(0, \sigma^2)$-distributed random variables with partial sum $S_n := \sum_{k=1}^n X_k$, $n \geq 1$. Then I read the following results.
$$ \sum_{k = 1}^n \mathbb P (S_n > \epsilon\sqrt{n\log\log n}) \sim \sum_{k = 1}^n \frac{\sigma}{\epsilon\sqrt{\log\log n}} (\log n)^{-\frac{\epsilon^2}{2\sigma^2}} = \infty, $$
for any $\epsilon > 0$. I do not know how these results are derived. To be specific,
- How to get the first approximation, please?
- How to know the second series does not converge, please? I tried to apply ratio test, but I got the limit is 1 which is not helpful. Could anyone help me, please? Thank you!
Update of the first question:
By the answer provided below, one has $$\mathbb P(S_n > \epsilon\sqrt{n\log\log n}) = \mathbb P\left(Z > \frac{\epsilon}{\sigma} \sqrt{\log\log n}\right) \leq \frac{1}{\sqrt{2\pi}}\frac{\sigma}{\epsilon\sqrt{\log\log n}} (\log n)^{-\frac{\epsilon^2}{2\sigma^2}}.$$ Then how to get $$\lim_{n \to \infty} \frac{\mathbb P\left(Z > \frac{\epsilon}{\sigma} \sqrt{\log\log n}\right)}{\frac{\sigma}{\epsilon\sqrt{\log\log n}} (\log n)^{-\frac{\epsilon^2}{2\sigma^2}}} = 1,$$ please? Thanks.
Using independence, we obtain that $$\mu\{S_n>\varepsilon\sqrt{n\log\log n}\}=\mu\{N>\varepsilon\sqrt{\log \log n}/\sigma\},$$ where the distribution of $N$ is standard normal. So we are reduce to estimates on the tail function of a standard normal distribution. We can use the equivalent $$\mu\{|N|>x\sqrt 2\}\overset{x\to +\infty}{\sim}\frac 1{x\sqrt{\pi}}\exp(-x^2).$$
Notice that for each positive $p$ the series $\sum_{n=1}^\infty(\log n)^{-p}$ is divergent. Indeed, $\sum_{n=N+1}^{2N}(\log n)^{-p}\geqslant N(\log(2N))^{-p}$.