Prove that $\limsup_{n\to\infty} |X_n|/n \le1 $ almost surely

Question

Suppose $\{X_n\}$ a sequence of random variables. If $\sum_{n=1}^\infty P(|X_n|>n)< \infty$

Prove that $$\limsup_{n\to\infty}\frac{ |X_n|}{n} \le1 $$ almost surely

What i have done so far:

I thought using the Borel-Cantelli lemma could lead me somewhere, but i didn't have any luck.

From Borel-Cantelli lemma we know that if $\sum_{n=1}^\infty P(|X_n|>n)< \infty$ then $P(|X_n|>n)=0$

How could I proceed? I would appreciate any help, advice. Thank you all very much in advance for your time and concern.

Answer 1

By Borel Cantelli lemma we have that $$ P( \liminf_{n \to \infty} \{ |X_n| \leq n \}) = P( \{|X_n| \leq n \text{ eventually } \} )= 1$$ In words this means than almost surely, the sequence $|X_n|$ is below $n$ for all $n$ sufficently large. I think you can take it from here.