Prove that $\limsup_{n\to\infty} |X_n|/n \le1 $ almost surely Suppose $\{X_n\}$ a sequence of random variables.
If $\sum_{n=1}^\infty P(|X_n|&gt;n)&lt; \infty$

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 $\sum_{n=1}^\infty P(|X_n|&gt;n)&lt; \infty$ then $P(|X_n|&gt;n)=0$

How could I proceed?
I would appreciate any help, advice. Thank you all very much in advance for your time and concern.
The first Borel-Cantelli lemma yields
$$\mathbb P\left(\limsup_{n\to\infty}|X_n|&gt;n\right)=0. $$
As for each $n$ $$\{|X_n|&gt;n\}\subset \bigcup_{k=n}^\infty \{|X_k|&gt;k\}, $$
it follows that
\begin{align}
0&amp;=\mathbb P\left(\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty \{|X_k|&gt;k\}\right) \\&amp;=\mathbb P\left(\lim_{n\to\infty}\bigcup_{k=n}^\infty \{|X_k|&gt;k\}  \right)\\
&amp;=\lim_{n\to\infty}\mathbb P\left(\bigcup_{k=n}^\infty \{|X_k|&gt;k\} \right)\\
&amp;\geqslant\lim_{n\to\infty}\mathbb P(|X_n|&gt;n)
\end{align}
and hence $\lim_{n\to\infty}\mathbb P(|X_n|&gt;n)=0$. Further, $$\bigcap_{k=n}^\infty \{|X_k|&gt;k\}\subset\{|X_n|&gt;n\} $$
so that
\begin{align}
\mathbb P\left(\limsup_{n\to\infty} |X_n|\leqslant n\right) &amp;= \mathbb P\left(\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty\{|X_k|\leqslant k\}\right)\\
&amp;=1 - \mathbb P\left(\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty \{|X_k|&gt;k\} \right)\\
&amp;=1 - \mathbb P\left(\lim_{n\to\infty} \bigcap_{k=n}^\infty\{|X_k|&gt;k\} \right)\\
&amp;=1 - \lim_{n\to\infty}\mathbb P\left(\bigcap_{k=n}^\infty\{|X_k|&gt;k\}\right)\\
&amp;\geqslant1 - \lim_{n\to\infty}\mathbb P(|X_n|&gt;n)\\
&amp;= 1,
\end{align}
which implies that
$$\mathbb P\left(\limsup_{n\to\infty} \frac{|X_n|}n\leqslant 1 \right)=1. $$
Important inequalities

Williams - Probability with Martingales







Deduced similarly:

(iii) If $\liminf x_n &gt; z$, then

$ \ \ \ \ \ \ \ (x_n &gt; z)$ eventually (that is, for infinitely many n)

(iv) If $\liminf x_n &lt; z $, then

$ \ \ \ \ \ \ \ (x_n &lt; z)$ infinitely often (that is, for infinitely many n)



By BCL1, we have $P(\limsup (|X_n| &gt; n)) = 0$

$$\to P(\liminf (|X_n| \le n)) = 1$$

$$\to P(\liminf (|X_n|/n \le 1)) = 1$$

$$\to P([\limsup |X_n|/n] \le 1)) = 1$$

The last step follows by the contrapositive of 2 (see above)