TITLE: Convergence of Random Variable 
QUESTION [6 upvotes]: Let $X_1, X_2, ...$ be identically distributed nonnegative random variable with $EX_1<\infty$. Prove that $\frac{X_n}{n}\rightarrow 0$ almost surely.
I am trying to use Borel Cantelli Lemma (part one) to prove this.

REPLY [6 votes]: I think Borel-Cantelli applies. Fix $a > 0$. Let $X$ be a random variable that has the same distribution as any of the $X_n$'s. For each $m \geq 1$, define $A_m$ to be the event that $am \leq X \lt a(m+1)$. $\newcommand{\E}{\mathbf{E}}$
$$
\begin{eqnarray*}
\sum\limits_{n=1}^{\infty} \Pr\left[\frac{X_n}{n} \geq a \right] 
&=&
\sum\limits_{n=1}^{\infty} \Pr\left[X \geq an \right] 
\\ &=&
\sum\limits_{n=1}^{\infty} \sum_{m=n}^{\infty} \Pr[am \leq X < a(m+1)] 
\\ &=&
\sum\limits_{n=1}^{\infty} \sum_{m=n}^{\infty} \Pr[A_m] 
\\ &=&
\sum\limits_{m=1}^{\infty} m \Pr[A_m]   \ \ \ \ \ \text{switching the order of summations}
\\ &=&
\frac{1}{a}\sum\limits_{m=1}^{\infty} am \Pr[A_m] 
\\ &\leq&
\frac{1}{a}\sum\limits_{m=1}^{\infty} \E[X \cdot \mathbf{1}_{A_m}] 
\\
&=&
\frac{1}{a} \E \left[X \cdot \sum\limits_{m=1}^{\infty} \mathbf{1}_{A_m} \right] 
\\ &\stackrel{(1)}{\leq}& \frac{\E X}{a} < \infty.
\end{eqnarray*}
$$
The inequality $(1)$ follows from the fact that $A_m$ for different $m$ are disjoint.