Prove almost sure convergence related to i.i.d random variables I have following question and stuck at the 1st one.  


  $\left\{X_n\right\}$ are i.i.d. random variables with non-zero finite mean. Let $S_n=X_1+X_2+\dots+X_n$. Prove
  1. $|X_n|/n \xrightarrow{a.s.}0$;
  2. $(\max_{1\le k \le n}|X_k|)/n \xrightarrow{a.s.} 0$
  3. $(\max_{1\le k \le n}|X_k|)/(1+S_n) \xrightarrow{a.s.} 0$


I tried to use Borel-Cantelli lemma and Markov inequality, but failed.
1st Try: 
It's equivalent to show for any $\epsilon &gt; 0$, we have $P(|X_n|/n &gt; \epsilon \text{ i.o.})=0$

$$\sum_{n=1}^{\infty} P(|X_n|/n &gt; \epsilon) \le \sum_{n=1}^{\infty}E[|X_n|]/n\epsilon=\infty$$
I figured it didn't work probably because i.i.d. condition was not used.
2nd Try: $$\begin{align}P(|X_n|/n &gt; \epsilon \text{ i.o.}) &amp;= \lim_{m \rightarrow \infty}P(\bigcup_{n=m}^\infty |X_n|/n &gt; \epsilon) \\ &amp;= \lim_{m \rightarrow \infty}1-P((\bigcup_{n=m}^\infty |X_n|/n &gt; \epsilon)^c) \\ &amp;=\lim_{m \rightarrow \infty}1-\prod_{n=m}^\infty P(|X_1|/n \le \epsilon)\end{align}$$ Stuck again.
The Markov inequaltiy is too weak in this case. Instead, use the fact that the $X_n$ are all identically distributed and the integral comparison test.

$$\sum \limits_{n = 1}^\infty P(|X_n|/n &gt; \varepsilon) = \sum \limits_{n = 1}^\infty P(|X_1|/\varepsilon &gt; n) \le 1 + \int_0^\infty P(|X_1|/\varepsilon &gt; x) \, dx$$

Now note that for any nonnegative random variable $Y$ the equation
$$\begin{align*}
\int_0^\infty P(Y &gt; x) \, dx &amp;= \int_0^\infty \int_0^\infty I\{y &gt; x\} \, dP^Y(y) \, dx = \int_0^\infty \int_0^\infty I\{y &gt; x\} \, dx \, dP^Y(y)\\
&amp;= \int_0^\infty y \, dP^Y(y) = E[Y]
\end{align*}$$
holds.
I guess following is the answer. 
$$\begin{align}P(|X_n|/n &gt; \epsilon \text{ i.o.}) &amp;= \lim_{m \rightarrow \infty}P(\bigcup_{n=m}^\infty |X_n|/n &gt; \epsilon)\\ &amp;= \lim_{m \rightarrow \infty}P(\bigcup_{n=m}^\infty |X_1|/n &gt; \epsilon)\\ &amp;= \lim_{m \rightarrow \infty}P(|X_1|/m &gt; \epsilon) \\ &amp; \le \lim_{m \rightarrow \infty} E(|X_1|)/n \epsilon \rightarrow 0\end{align}$$