Define $M_n = \max\{X_1,X_2,\dots,X_n\}$. I want to prove that $$\limsup \frac{M_n}{n} = \limsup \frac{X_n}{n}$$
Since $\frac{M_n}{n} \geq \frac{X_n}{n}$,
$ \limsup \frac{M_n}{n} \geq \limsup \frac{X_n}{n}$.
However, I am stuck at proving $ \limsup \frac{M_n}{n} \leq \limsup \frac{X_n}{n}$. How can you prove this?
False in general. Example: Take $X_n=-n$.
Assuming that $X_n$ are non-negative we can argue as follows:
$$\frac {M_n} n\leq \frac {M_k} n\vee \frac {\max {X_{k+1},X_{k+2},\cdots,X_n}} n$$ $$ \leq \frac {M_k} n\vee{\max {X_{k+1}/(k+1),X_{k+2}/(k+2),\cdots,X_n/n}}$$ We can choose $k$ so large that he second part is less than $\lim \sup ({X_n} /n)+\epsilon$. The first part tends to $0$ as $ n\to \infty$.