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.
 A: The first Borel-Cantelli lemma yields
$$\mathbb P\left(\limsup_{n\to\infty}|X_n|>n\right)=0. $$
As for each $n$ $$\{|X_n|>n\}\subset \bigcup_{k=n}^\infty \{|X_k|>k\}, $$
it follows that
\begin{align}
0&=\mathbb P\left(\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty \{|X_k|>k\}\right) \\&=\mathbb P\left(\lim_{n\to\infty}\bigcup_{k=n}^\infty \{|X_k|>k\}  \right)\\
&=\lim_{n\to\infty}\mathbb P\left(\bigcup_{k=n}^\infty \{|X_k|>k\} \right)\\
&\geqslant\lim_{n\to\infty}\mathbb P(|X_n|>n)
\end{align}
and hence $\lim_{n\to\infty}\mathbb P(|X_n|>n)=0$. Further, $$\bigcap_{k=n}^\infty \{|X_k|>k\}\subset\{|X_n|>n\} $$
so that
\begin{align}
\mathbb P\left(\limsup_{n\to\infty} |X_n|\leqslant n\right) &= \mathbb P\left(\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty\{|X_k|\leqslant k\}\right)\\
&=1 - \mathbb P\left(\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty \{|X_k|>k\} \right)\\
&=1 - \mathbb P\left(\lim_{n\to\infty} \bigcap_{k=n}^\infty\{|X_k|>k\} \right)\\
&=1 - \lim_{n\to\infty}\mathbb P\left(\bigcap_{k=n}^\infty\{|X_k|>k\}\right)\\
&\geqslant1 - \lim_{n\to\infty}\mathbb P(|X_n|>n)\\
&= 1,
\end{align}
which implies that
$$\mathbb P\left(\limsup_{n\to\infty} \frac{|X_n|}n\leqslant 1 \right)=1. $$
A: 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.
A: Important inequalities
Williams - Probability with Martingales



Deduced similarly:
(iii) If $\liminf x_n > z$, then
$ \ \ \ \ \ \ \ (x_n > z)$ eventually (that is, for infinitely many n)
(iv) If $\liminf x_n < z $, then
$ \ \ \ \ \ \ \ (x_n < z)$ infinitely often (that is, for infinitely many n)

By BCL1, we have $P(\limsup (|X_n| > n)) = 0$
$$\to P(\liminf (|X_n| \le n)) = 1$$
$$\to P(\liminf (|X_n|/n \le 1)) = 1$$
$$\to P([\limsup |X_n|/n] \le 1)) = 1$$
The last step follows by the contrapositive of 2 (see above)
