Let $\{X_n\}_{n\ge1}$ be independent $N(0,1)$ random variables. Show that $$\limsup\limits_{n\to\infty} \frac{\left|X_n\right|}{\sqrt{\log(n)}}=\sqrt{2} \qquad \text{a.s.}$$
I aim to prove this using the fact that
$$\limsup\limits_{n\to\infty} X_n = b \quad \iff \quad \text{for all } \varepsilon>0 \ : \ \Biggl\{ \begin{array}{l} \mathbb{P}(X_n \le b+\varepsilon \text{ eventually})=1, \text{ and} \\ \mathbb{P}(X_n > b-\varepsilon\text{ i.o.})=1. \end{array}$$
I show the first of these two conditions as follows:
\begin{align*} &\hspace{-2em}\mathbb{P}\left(\frac{\left|X_n\right|}{\sqrt{\log(n)}} > \sqrt{2} + \varepsilon\right)\\ &= \mathbb{P}\bigl(|X_1|>(\sqrt{2}+\varepsilon)\sqrt{\smash[b]{\log(n)}}\bigr) & \text{as $X_n$'s are identically distributed}\\ &=\int_{(\sqrt{2}+\varepsilon)\sqrt{\smash[b]{\log(n)}}}^{\infty} |x|\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} \, dx\\ &=\int_{(\sqrt{2}+\varepsilon)\sqrt{\smash[b]{\log(n)}}}^{\infty} x\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} \, dx\\ &=\frac{1}{\sqrt{2\pi}}\int_{(\sqrt{2}+\varepsilon)^2\log(n)}^{\infty} e^{-u}du & \text{by making the substitution $u=\frac{x^2}{2}$}\\ &=-\frac{1}{\sqrt{2\pi}}\bigl[e^{-\infty}-e^{-\log(n)(\sqrt{2}+\varepsilon)^2}\bigr]\\ &=\frac{1}{\sqrt{2\pi}}n^{-(\sqrt{2}+\varepsilon)^2} \end{align*}
Thus, we have: \begin{align*} \sum_{n=1}^\infty \mathbb{P}\Biggl(\frac{\left|X_n\right|}{\sqrt{\log(n)}} > \sqrt{2} + \varepsilon\Biggr)&=\sum_{n=1}^{\infty}\frac{1}{\sqrt{2\pi}}n^{-(\sqrt{2} + \varepsilon)^2}\\ &<\infty \qquad \text{ since $\sqrt{2}+\varepsilon>1$} \end{align*}
So, by the Borel–Cantelli Lemmas:
\begin{align} &\mathbb{P}\Biggl(\frac{\left|X_n\right|}{\sqrt{\log(n)}} > \sqrt{2} + \varepsilon\text{ i.o.}\Biggr)=0\\ &\implies \mathbb{P}\Biggl(\frac{\left|X_n\right|}{\sqrt{\log(n)}} \le \sqrt{2} + \varepsilon\text{ eventually}\Biggr)=1 \end{align}
It remains then to show that
$$\mathbb{P}\Biggl(\frac{\left|X_n\right|}{\sqrt{\log(n)}} > \sqrt{2} - \varepsilon \text{ i.o.}\Biggr)=1.$$
To do this, I would like to run a symmetric argument to the one above but here we need the final series to diverge which will happen if and only if $\sqrt{2}-\varepsilon \le1$, which happens if and only if $\varepsilon\ge \sqrt{2}-1=0.414\ldots$ and so $\varepsilon$ is getting away from zero, which we can't have. Unless I am making a stupid mistake or missing something obvious, I don't see a way around this issue. Is this approach doomed to fail or is there a way to fix it up? Or is there just a better approach in general for such a problem. Thanks in advance.