Borell-Cantelli lemma If $\sum_{n} P(|X_{n}|>n)<\infty$, then prove that the $\limsup_{n}$ $|X_{n}|/n \leq 1$ a.s.
My approach
Let $E_{n}=|X_{n}>n|.$
$\sum_{n}P(E_{n})<\infty$ implies $P(E_{n}  \text{ i.o})=0$
which further implies $P(\limsup_{n} E_{n})=0$.
Can anyone throw some light on how to approach further!

Your choice of $E_n$ should be $E_n=(|X_n|>n)$. Since $P(E_{n}\, \text{i.o})=0$, it follows that with probability one $|X_n|/n\leq 1$ eventually whence $\limsup |X_n|/n\leq 1$.
