Converge in Sum vs Convergence Let $X_1, X_2, \ldots$ be a sequence of independent, integrable, random variables on $(\Omega, F, P)$, with $E[ X_k] = 0$ for all $k=1,2,\ldots$ Show that 
$$
\mbox{If } \frac{S_n}{n} \xrightarrow{P} 0 \mbox{ then } \frac{X_n}{n} \xrightarrow{P} 0
$$
I am having difficulties solving this problem, I am not sure how the 0 mean conditions are used here. We know that $E[S_n]=0$ as well but apart from that I dont see how this is used any further.
Let $X_1, X_2, \ldots$ be a sequence of independent, integrable, random variables , with $E X_k = 0$ for all $k=1,2,\ldots$ Show that 
$$
\mbox{If } \frac{S_n}{n} \xrightarrow{\mbox{a.s.}} 0 \mbox{ then }  \sum_{n=1}^\infty P\{|X_n| \ge \epsilon n\} < \infty \quad 
\mbox{for all $\epsilon > 0$.}
$$
For this second one, there is a statement that $\sum_{n=1}^\infty P\{|X_n| \ge \epsilon\} < \infty$ if and only if $X_n\to 0$ (easily proved from Borel Cantelli lemmas). I think that is used here but I am not sure how. And help will be great. Thanks!
 A: First Question
By definition $\frac{S_n}{n} \xrightarrow{P} 0$ if and only if  $P\left(\frac{|S_n|}{n} \geqslant \epsilon\right) \to 0$ as $ n \to \infty$ for all $\epsilon > 0$.
We also have,
$$\tag{*}\frac{X_n}{n} = \frac{S_n}{n} - \left(\frac{n-1}{n} \right)\frac{S_{n-1}}{n-1}$$
Since $(n-1)/n \to 1$ it is easy to show that the second term on the RHS of (*) converges to zero in probability (as well as the first term by hypothesis).
That the LHS of (*) converges to zero in probability follows from the fact that $A_n \xrightarrow{P} 0$ and $B_n \xrightarrow{P} 0$ implies $A_n + B_n \xrightarrow{P} 0$.
To prove this note that 
$$\left\{|A_n| < \epsilon/2 \right\} \cap \left\{|B_n| < \epsilon/2 \right\} \subset \left\{|A_n + B_n| < \epsilon \right\}$$
Taking complements, it follows that
$$\left\{|A_n + B_n| \geqslant \epsilon \right\} \subset \left\{|A_n| \geqslant \epsilon/2 \right\} \cup \left\{|B_n| \geqslant \epsilon/2 \right\}, $$
and
$$P(|A_n + B_n| \geqslant \epsilon) \leqslant P(|A_n| \geqslant \epsilon/2 ) + P(|B_n| \geqslant \epsilon/2) $$
Second Question
Hint: $\frac{S_n}{n} \xrightarrow{\mbox{a.s.}} 0 \implies \frac{X_n}{n} \xrightarrow{\mbox{a.s.}} 0$
