If $X_{1}, X_{2},..$ are identically distributed then $M_{n}/n \rightarrow 0$ in probability. If $X_{1}, X_{2},..$ are identically distributed then $M_{n}/n \rightarrow_{p} 0$ where $M_{n}=max\{|X_{1}|,...,|X_{n}|\}$ and $E(|X_{1}|)$ is finite. 
$M_{n}/n \rightarrow_{p} 0$ by definition means given any $\epsilon$, $P(M_{n}/n > \epsilon) \rightarrow 0$ as $n \rightarrow \infty$. Can anyone give any hint?

$M_n>n\varepsilon$ if and only if $|X_i|>n\varepsilon$ for some $1\leq i\leq n$. Therefore
$$ \{M_n>n\varepsilon\}=\bigcup_{i=1}^n\{|X_i|>n\varepsilon\}$$
hence
$$ \mathbb{P}(M_n>n\varepsilon)\leq \sum_{i=1}^n\mathbb{P}(|X_i|>n\varepsilon)=n\mathbb{P}(|X_1|>n\varepsilon) $$
Therefore it is enough to show that
$$ \lim_{n\to\infty}n\mathbb{P}(|X_1|>n\varepsilon)=0 $$
for any $\varepsilon>0$, and this follows from the dominated convergence theorem, since $\mathbb{E}[|X_1|]<\infty$.
