Estimates for Law of Large number In many literature it has been state that the necessary condition that $\frac{S_n}{n}$ converges almost surely is that $\frac{X_n}{n}$ converges to 0 almost surely. (where $S_n=\sum_{k=1}^{n}X_k$) However, I fail to see why.
It is easy to see that $\frac{X_n}{n}=\frac{S_n}{n}-\frac{n-1}{n}\frac{S_{n-1}}{n-1}$. So if n tends to infinity, $\frac{S_n}{n}$ and $\frac{S_{n-1}}{n-1}$ converges to the same quantity. But what happens with the factor $\frac{n-1}{n}$. Is there a nice way to prove $\frac{X_n}{n}$ tends to zero?
Another question I have is, $\frac{X_n}{n}$ tends to zero is a necessary condition for $\frac{S_n}{n}$ convergence. Is it also sufficient? If yes, how can I see that? If not, what would be the sufficient condition?
 A: Your first question has nothing to do with probability: if $s_n = a_1 + \cdots + a_n$ and $s_n / n$ converges to $\ell$, then
$$ \frac{a_n}{n} = \frac{s_n}{n} - \frac{n-1}{n} \cdot \frac{s_{n-1}}{n-1} \xrightarrow[\quad n\to\infty \quad]{} \ell - (1 \cdot \ell) = 0. $$
This is because $\frac{n-1}{n}$ converges to $1$ and product of convergent sequences converges to the product of the respective limits.
For your second question, the answer is YES, assuming that $(X_n)$ is a sequence of i.i.d. random variables. In this case, we have the following observation:

Proposition. The followings are equivalent:
  
  
*
  
*$X_1$ is integrable.
  
*$X_n/n$ converges to $0$ almost surely.
  

The proof is standard in view of the Borel-Cantelli lemmas. Indeed,
\begin{align*}
X_n/n \to 0 \quad\text{a.s.}
&\quad \Leftrightarrow \quad \forall \epsilon > 0 \ : \ \mathbb{P}(|X_n| \geq \epsilon n \text{ i.o.}) = 0 \\
&\quad \Leftrightarrow \quad \forall \epsilon > 0 \ : \ \sum_{n=1}^{\infty} \mathbb{P}(|X_n| \geq \epsilon n) < \infty \\
&\quad \Leftrightarrow \quad \mathbb{E}|X_1| < \infty.
\end{align*}
