For $X_{k}\sim N(0,1)$ iid find diverging constants $a_{n}$ s.t. $$a_{n}(max_{1\leq k\leq n}X_{k}-a_{n})\stackrel{d}{\to}L$$
for some limit L.
Or $a_{n}$ s.t. $\frac{n}{a_{n}}e^{-\frac{1}{2}a_{n}^{2}}\to c$ for $c\neq \infty$ or 0.
Attempt
Let $M_{n}:=max_{1\leq k\leq n}X_{k}$ then we have
$$P(M_{n}\leq \frac{x}{a_{n}}+a_{n})=(1-P(X_{k}\geq \frac{x}{a_{n}}+a_{n}))^{n}$$
$$\sim (1- \frac{1}{\frac{x}{a_{n}}+a_{n}}e^{-\frac{1}{2}(\frac{x}{a_{n}}+a_{n})^{2}} )^{n}.$$
So we want $a_{n}$ s.t.
$$\frac{n}{\frac{x}{a_{n}}+a_{n}}e^{-\frac{1}{2}(\frac{x}{a_{n}}+a_{n})^{2}}\to f(x),$$ some function $f(x)$ so that $e^{-f(x)}$ is chf.
One can show by Borel Cantelli that $lim \frac{M_{n}}{\sqrt{2logn}}\stackrel{as}{\to} 1$. So that is a good guess:
$$\frac{n}{\frac{x}{\sqrt{2logn}}+\sqrt{2logn}}e^{-\frac{1}{2}(\frac{x}{\sqrt{2logn}}+\sqrt{2logn})^{2}}=c\frac{1}{\frac{x}{\sqrt{2logn}}+\sqrt{2logn}}\to 0$$
This implies $P(M_{n}\leq \frac{x}{\sqrt{2logn}}+\sqrt{2logn})\to 1,$
which is not a cdf.
For $a_{n}=\sqrt{2loglogn}$, we get $P(M_{n}\leq \frac{x}{\sqrt{2logn}}+\sqrt{2logn})\sim e^{-\infty}=0,$
which is also not a cdf. Now I am trying to work in between them.
Please only hints. Thank you
