Prove $ \frac{X_n}{n} $ converges to zero Let $\{X_n\}$ be a sequence of i.i.d. positive random variables and $\mathbb{E}(X_1)<\infty$.
Can we prove that $$ \frac{X_n}{n} \to 0 ~~\text{a.e.}~? $$
Every $X_n$ is finite almost everywhere, but how to control all $X_n$'s ?
 A: $\frac {X_n} n=\frac {S_n-S_{n-1}} n$ where $S_n =X_1+X_2+...+X_n$. Hence $\frac {X_n} n=\frac {S_n} n - \frac  {n-1} n\frac {S_{n-1}} {n-1}\to EX_1-(1)(EX_1)=0$ a.e. by SLLN.
A: @Kavi Rama Murthy made very simple and elegant proof. I will prove the claim without assumption that $\{X_n\}_{n \in \mathbb N}$ are independent. Note that for any $\varepsilon > 0$ we have:
$$\sum_{n=1}^\infty \mathbb P(\frac{|X_n|}{n} > \varepsilon) = \sum_{n=1}^\infty \mathbb P(\frac{1}{\varepsilon}|X_1| > n) \le \frac{1}{\varepsilon}\mathbb E|X_1|$$
Where first equality is due to the same distribution of sequence $(X_n)$ and the inequality is a consequence of applying expected value to pointwise inequality of the form $\sum_{n=1}^\infty 1_{\{|Z| > n\}} \le |Z|$ (in our case $Z = \frac{1}{\varepsilon}{X_1}$)
Hence by assumption the series converge, so by borel cantelli $\mathbb P(\limsup \{ |X_n| > n\varepsilon\}) = 0$, which means $\mathbb P(\liminf \{|X_n| \le n\varepsilon\})=1$.
Hence almost surely we have:
$$0 \le \liminf \frac{|X_n|}{n} \le \limsup \frac{|X_n|}{n} \le \varepsilon$$
Since $\varepsilon >0$ was arbitrary, we can now take sequence $\varepsilon_m = \frac{1}{m}$ (just to have countable intersection to still have full measure set) to conclude that almost surely: $$ 0 \le \liminf \frac{|X_n|}{n} \le \limsup \frac{|X_n|}{n} \le 0$$ hence $\lim \frac{|X_n|}{n} = 0$ almost surely
