Let ${X_n}$ be a sequence of independent random variables N[0,1].
Show that: $$ \mathcal{P}(\underbrace{lim sup}_{n \longmapsto \infty} \frac{|X_n|}{\sqrt{log \ n}} = \sqrt{2}) = 1 $$
I've been asked to compute it by using the Mill's ratio, so using that: 1- $\Phi(x)$ ~ $x^{-1} \phi(x)$. So i was thinking about computing: $$ \mathcal{P}\left(|X_n| \geq \sqrt{log \ n} \sqrt{2}\right) $$
but I don't know how to proceed.