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

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.

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Hint: Borel-Cantelli lemma shows that $\Bbb{P}(|X_{n}| > n \text{ i.o.}) = 0$. –  sos440 Jul 7 '13 at 13:14
Crossposted: stats.stackexchange.com/q/63561/2970 –  cardinal Jul 7 '13 at 15:24
@cardinal so what? i can not post my question in two different sections? –  johan paul Jul 7 '13 at 15:33
@johan: See this meta.SO answer. This is the quasi official policy on this topic. Cheers. –  cardinal Jul 7 '13 at 15:36
didn't know that i am really sorry, but i think it is not big deal to spent your time! –  johan paul Jul 7 '13 at 15:39

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.
If $|X_n|$ can't surpass $n$ for $n$, then what happens to $\frac{|X_n|}{n}$?. Also remember that the limsup of a sequence is its largest acummulation point. –  Bunder Jul 7 '13 at 13:42
We used the "original" Borel Cantelli which does not need independence of the events. The converse (i.e $\sum P(\ldots) = \infty \Rightarrow P( \ldots i.o) = 1$) requires independence. Check en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma –  Bunder Jul 7 '13 at 15:01