Convergence a.e. of the series $\sum_{i=1}^{n^2} \frac{X_i}{n^2}$ Let $(X_n)_{n\geq 1}$ be independent random variables with expected value $m$ and $\sup_n Var(X_n)\leq K < \infty$, and they are uncorrelated. 
Then 
$1)$ $$\sum_{i=1}^n \frac{X_i}{n} $$ converges to $m$ in $L^2$.
$2)$ $$\sum_{i=1}^{n^2} \frac{X_i}{n^2}$$ converges to $m$ almost everywhere.
I have done point $1$ but I'm stuck with point $2$, how to proceed ?
 A: We only need the fact that $X_n$ are uncorrelated, independence is not necessary.
$E\left(\sum_{i=1}^n \dfrac{X_i}{n} - m\right)^2 = E\left(\sum_{i=1}^n \dfrac{X_i-m}{n}\right)^2 = \sum_{i=1}^n E\left(\dfrac{X_i-m}{n}\right)^2 \leq\sum_{i=1}^n \dfrac{K}{n^2} = \dfrac{K}{n} $
So we get $\sum_{i=1}^n \dfrac{X_i}{n}$ converges to $m$ in $L^2$, and we will use the above control to prove the second conclusion.
For any $\epsilon >0$
\begin{align}
P(\left|\sum_{i=1}^{n^2} \dfrac{X_i}{n^2} - m\right| > \epsilon) &\le \dfrac{E\left(\sum_{i=1}^{n^2} \dfrac{X_i}{n^2} - m\right)^2}{\epsilon^2} \le \frac{K}{n^2\epsilon^2}
\end{align}
Thus we have $\sum_{n=1}^\infty P(\left|\sum_{i=1}^{n^2} \dfrac{X_i}{n^2} - m\right| > \epsilon) \le \sum_{n=1}^\infty \frac{K}{n^2\epsilon^2} < +\infty$, i.e.
$$E(\sum_{n=1}^\infty 1_{\left|\sum_{i=1}^{n^2} \dfrac{X_i}{n^2} - m\right| > \epsilon}) < +\infty$$
So we have $\sum_{n=1}^\infty 1_{\left|\sum_{i=1}^{n^2} \dfrac{X_i}{n^2} - m\right| > \epsilon}$ is finite almost surely, i.e. almost surely there exists $N$ such that for $n > N$, we have $\left|\sum_{i=1}^{n^2} \dfrac{X_i}{n^2} - m\right| < \epsilon$. That is the almost surely convergence as desired.
A: This is known as the Law of Large Numbers. (Well, almost. The LLN states that the sum in part 1 converges almost surely/in probability to the expectation $m$. The sum in part 2 is merely a subsequence of the sum in part 1.) Given that we have bounded variance, it is not hard to show the sum converges in probability; perhaps try this for yourself using Chebyshev's inequality. Showing almost sure convergence is much more difficult and typically uses a martingale approach.
