Chose $a_n$ so that $\frac{M_n}{a_n}$ converges to 0 in distribution Let $X_1, ... , X_n$ be i.i.d. random variables with distribution Function $F$ and $M_n = \max\left\{X_1, \dots , X_n\right\}$ . Now i need to find $ a_n \in (0, \infty )$ so that $\frac{M_n}{a_n}$ convergence to 0 in distribution. 
Since the mean of the random variables can be $\infty$ many tools aren't available, i struggle with this issue and don't really have an approach.
Thank you in advance.
 A: For any sequence of random variables $\{M_n\}$ we can find $a_n$'s such that $\frac {M_n} {a_n} \to 0$ on probability (hence in distribution). For this choose $a_n$ such that $P\{|M_n |>\frac {a_n} n\} <\frac 1 n$. For any $\epsilon >0$ we have $P\{\frac {|M_n|} {a_n}>\epsilon\} \leq P\{|M_n| >\frac {a_n} n\} <\frac 1 n$ provided $\frac 1 n <\epsilon$. Hence  $\frac {M_n} {a_n} \to 0$ on probability .
A: You want to show that it converges in distribution to $0$. i.e:
$\underset{n\rightarrow \infty}{\lim} \mathbb{P}\Big(\frac{M_n}{a_n}\leq t \Big) =\begin{cases} 1 & t>0 \\ 0  & t<0 \end{cases}$
You can see that:
$\mathbb{P}\Big(\frac{M_n}{a_n}\leq t \Big) = \mathbb{P}\Big( M_n\leq t\cdot a_n \Big)= \mathbb{P}\Big(\cap \{  X_k \leq t\cdot a_n \} \Big)= \overset{n}{\underset{k=1}{\prod}}\mathbb{P}\Big(   X_k \leq t\cdot a_n  \Big)= $
$= \overset{n}{\underset{k=1}{\prod}} F(t\cdot a_n)= \Big( F(t\cdot a_n) \Big)^n$
Recall that $F(s)\overset{s\rightarrow \infty}{\rightarrow}1$ and $F(s)\overset{s\rightarrow -\infty}{\rightarrow}0$, and thus since $F(s_0)<1$ for some $s_0$, it is enough that $ t\cdot a_n \leq s_0$  for the term to tend to $0$. Which will happen eventually for any $0<a_n\rightarrow \infty$ and $t<0$. For $\Big( F(t\cdot a_n) \Big)^n$ to tend to $1$, it is enough that $F(t\cdot a_n)\geq \big(\frac{n-1}{n}\big)^{\frac{1}{n}}$ by the squeeze theorem. And since $F(s)\overset{s\rightarrow \infty}{\rightarrow}1$ you can indeed find such $a_n$.
I know you've been given a great answer above, but this is a more direct approach and I already started writing from before.
