I am a beginner in probability theory (also a beginner in analysis) and am currently having trouble with regard to the following problem.
I would like to show:
Let $X_n$ be a sequence of standard normal random variables. Show that for any $\gamma > 1$, $|X_n| \le (2 \gamma \log{n})^{1/2}$ for all but finitely many $n$ almost surely.
My attempt is the following:
Consider $\gamma > 1$ and $\{ x: |X_n| \le (2 \gamma \log{n} )^{1/2} \text{ for all but finitely many $n$} \}$. Then, $\{ x: |X_n| \le (2 \gamma \log{n} )^{1/2} \text{ for all but finitely many $n$} \} = \bigcup_{N=1}^\infty \bigcap_{n=N}^\infty \{ x: |X_n| \le (2 \gamma \log{n} )^{1/2} \}$. We want to show $P(\bigcup_{N=1}^\infty \bigcap_{n=N}^\infty \{ x: |X_n| \le (2 \gamma \log{n} )^{1/2} \}) = 1$; or to show $P(\bigcap_{N=1}^\infty \bigcup_{n=N}^\infty \{ x: |X_n| > (2 \gamma \log{n} )^{1/2} \}) = 0$; I show this in the edit.
By way of contradiction, suppose $P(\bigcap_{N=1}^\infty \bigcup_{n=N}^\infty \{ x: |X_n| > (2 \gamma \log{n} )^{1/2} \}) > 0$. Then, since the infinitely often events are tail events, $P(\bigcap_{N=1}^\infty \bigcup_{n=N}^\infty \{ x: |X_n| > (2 \gamma \log{n} )^{1/2} \}) = 1$ by the Kolomolgorov Zero-One Law
EDIT:
By Borel-Cantelli Lemma, it suffices to show $\sum_{n \ge 1} P(|X_n| > (2 \gamma \log{n})^{1/2}) < \infty$. Since $X_n$ has standard normal distribution, \begin{align*} \sum_{n \ge 1} P(|X_n| > (2 \gamma \log{n})^{1/2}) &\le \sum_{n \ge 1} 2 \dfrac{exp( -2\gamma \log{n/2} ) } { \sqrt{2\gamma} \sqrt{2 \pi \log{n}}} \\ &= \sum_{n\ge1} \dfrac{2}{\sqrt{2\gamma} n^\gamma \sqrt{2\pi \log{n}}} \\ &< \infty \end{align*} if $\sqrt{2 \gamma} > \sqrt{2}$. Since $\gamma > 1$ by hypothesis, $\sum_{n \ge 1} P(|X_n| > (2 \gamma \log{n})^{1/2}) < \infty$ and by Borel-Cantelli Lemma, $P(\bigcap_{N=1}^\infty \bigcup_{M=N}^\infty \{ |X_n| > (2 \gamma \log{n} )^{1/2} \}) = 0$.
From here, I am stuck. Is there another approach? Any insight? Thank you in advance.