TITLE: Prove that the maximum of $n$ independent standard normal random variables, is asymptotically equivalent to $\sqrt{2\log n}$ almost surely.
QUESTION [16 upvotes]: Lets $(X_n)_{n\in\mathbb{N}}$ be an iid sequence of standard normal random variables. Define $$M_n=\max_{1\leq i\leq n} X_i.$$
Prove that $$\lim_{n\rightarrow\infty} \frac{M_n}{\sqrt{2\log n}}=1\quad\text{a.s.}$$
I used the fact that $$\left(\frac{1}{x}-\frac{1}{x^3}\right)e^{-\frac{x^2}{2}}\leq\mathbb P(X_n>x)\leq \frac{1}{x}e^{-\frac{x^2}{2}},$$
and the Borel Cantelli lemmas to prove that 
 $$\limsup_{n\rightarrow\infty} \frac{X_n}{\sqrt{2\log n}}=1\quad\text{a.s.}$$
I used Davide Giraudo's comment to show $$\limsup_n \frac{M_n}{\sqrt{2\log n}}=1\quad \text{a.s.}$$
I have no idea how to compute the $\liminf$. Borel-Cantelli give us tools to compute the $\limsup$ of sets, I am unsure of how to argue almost sure convergence. Any help would be appreciated.

REPLY [2 votes]: Not an answer, but a related comment that is too long for a comment box ...
This question made me curious to compare:

the pdf of the sample maximum (given a sample of size $n$ drawn on a N(0,1) parent), say $f(x;n):$

$$f(x) = \frac{2^{\frac{1}{2}-n} n e^{-\frac{x^2}{2}} \left(1+\text{erf}\left(\frac{x}{\sqrt{2}}\right)\right)^{n-1}}{\sqrt{\pi }}$$
where erf(.) denotes the error function, to

the asymptote proposed by the question $\sqrt{2 \log n}$

... when $n$ is very large (say $n$ = 1 million).
The diagram compares:

The distribution does not, in fact, converge to the asymptote in any meaningful manner, even when $n$ is very large, such as $n$ = 1 million or 100 million or even a billion.