Show that a.s. $\limsup_n \frac{X_n}{\sqrt{2\log n}} \le 1$ Let $(X_n)$ be a family of independent gaussians.
I am suppose to show that a.s.
$$\limsup_n \frac{X_n}{\sqrt{2\log n}} = 1.$$
I was able to show using Borel-Cantelli that
$$\limsup_n \frac{X_n}{\sqrt{2\log n}} \ge  1$$
Is it also true that
I am a bit clueless about the other direction
$$\limsup_n \frac{X_n}{\sqrt{2\log n}} \le  1$$
 A: Assume $\{X_n\}$ is a family of i.i.d $\mathcal N(0,1)$ random variables. Fix $\varepsilon > 0$. We want to show $\limsup_n \frac{X_n}{\sqrt{2\ln(n)}} \ge 1- \varepsilon $ To this end, it would be enough to show that series $\sum_{n=1}^\infty \mathbb P(X_n \ge (1-\varepsilon)\sqrt{2\ln(n)}) $ is divergent.
Note that for $t > 0$ we have  $$\mathbb P(X_n \ge t) = \frac{1}{\sqrt{2\pi}}\int_{t}^\infty e^{-\frac{x^2}{2}} \ge \frac{1}{\sqrt{2\pi}}\Big( \frac{1}{t} -\frac{1}{t^3}\Big)e^{-\frac{t^2}{2}}$$
This bound can be obtained by differentiating right hand side and writing it as integral of derivative, then bounding.
Having that, we get $$\mathbb P(X_n \ge (1-\varepsilon)\sqrt{2\ln(n)}) \ge \frac{1}{\sqrt{2\pi}}\Big(\frac{1}{(1-\varepsilon)\sqrt{2\ln(n)}} - \frac{1}{\big((1-\varepsilon)\sqrt{2\ln(n)}\big)^3}\Big) \exp(-(1-\varepsilon)^2 \ln(n)) $$
Last $\exp$ can be rewritten as $\frac{1}{n^{(1-\varepsilon)^2}}$ which forms a divergent series (even with multiplying by either $\frac{1}{\sqrt{\ln(n)}}$ or $(\frac{1}{\sqrt{\ln(n)}})^3$. Hence our series diverges, and that + indendence gives us the result (via Borel Cantelli). Since $\varepsilon > 0$ was arbitrary, we can conclude (almost surely)
$$ \limsup_n \frac{X_n}{\sqrt{2\ln(n)}} \ge 1 $$
As for the second side: Fix $\varepsilon >0$. We want to show $\limsup_n \frac{X_n}{\sqrt{2\ln(n)}} \le 1+\varepsilon $. Note it would be enough to show that $\sum_{n=1}^\infty \mathbb P(X_n > (1+\varepsilon)\sqrt{2\ln(n)})$ is a convergent series. Note that for $t > 0$ we have $$ P(X_n > t) = \frac{1}{\sqrt{2\pi}}\int_t^\infty e^{-\frac{x^2}{2}}dx \le \frac{1}{t\sqrt{2\pi}}e^{-\frac{t^2}{2}} $$
Proof of this is a consequence of pointwise inequality $(x>t)$ gives $e^{-x^2}{2} \le \frac{x}{t}e^{-x^2}{2}$ and integrating.
Having that we see $$ \mathbb P(X_n > (1+\varepsilon)\sqrt{2\ln(n)}) \le \frac{1}{\sqrt{2\pi}(1+\varepsilon)\sqrt{2\ln(n)}}\exp(-(1+\varepsilon)^2\ln(n)) $$
Again, this $\exp$ can be rewritten as $\frac{1}{{n^{\left(1+\varepsilon \right)}}^2}$ which forms a convergent series (with multiplication by $\frac{1}{\sqrt{\ln(n)}}$, too). Hence, By Borel cantelli (almost surely) we get the result, and since $\varepsilon >0$ was arbitrary, we see that $$ \limsup_n \frac{X_n}{\sqrt{2\ln(n)}} \le 1 $$
