Show that if $\sum\limits_nnE|X_{n}|^2$ converges for uncorrelated, mean zero $X_n$s, then $\sum\limits_{i=1}^{n}X_i$ converges almost surely Let ($X_n$) be a sequence of uncorrelated random variables of mean zero such that 
$$\sum_{n=1}^{\infty}nE|X_{n}|^2 < \infty $$
Show that $S_n = \sum_{i=1}^{n}X_i$ converges almost surely.
This is the second problem of S.-T. Yau College Student Mathematics Contests 2014.
I tried to solve the problem by using a similar inequality of Kolmogorov, but how to use the coefficient of $n$ before $E|X_n|^2$ makes me confused.
 A: Since the random variables are uncorrelated and have mean $0$ it holds that
$$\mathbb{E}(X_i X_j) = 0$$
for all $i \neq j$. This implies
$$\mathbb{E}((S_n-S_m)^2) = \sum_{i=m+1}^n \sum_{j=m+1}^n \mathbb{E}(X_i X_j) = \sum_{i=m+1}^n \mathbb{E}(X_i^2)$$
for all $n \geq m$, and therefore $(S_n)_{n \in \mathbb{N}}$ is an $L^2$-Cauchy-sequence, hence convergent, i.e. $S_n \to X$ in $L^2$ for some random variable $X \in L^2$. Using that
$$\mathbb{E}((S_n-X)^2) = \sum_{i=n+1}^{\infty} \mathbb{E}(X_i^2)$$
we find by Markov's inequality for any $\epsilon>0$
$$\begin{align*} \sum_{n \geq 1} \mathbb{P}(|S_n-X| \geq \epsilon) &\leq \frac{1}{\epsilon^2} \sum_{n \geq 1} \mathbb{E}((S_n-X)^2) \\ &= \frac{1}{\epsilon^2} \sum_{n \geq 1} \sum_{i=n+1}^{\infty} \mathbb{E}(X_i^2) \\ &= \frac{1}{\epsilon^2} \sum_{i \geq 1} i \mathbb{E}(X_i^2) < \infty. \end{align*}$$
Applying the Borel-Cantelli lemma we conclude that $S_n \to X$ almost surely.
Remark: Let $(Y_n)_{n \in \mathbb{N}}$ be a sequence of random variables such that $Y_n \to Y$ in probability (which is, in particular, satisfied in $Y_n \to Y$ in $L^2$). It is well known that this does, in general, not imply $Y_n \to Y$ almost surely. However, if the probabilities $\mathbb{P}(|Y_n-Y|>\epsilon)$ are decaying sufficiently fast as $n \to \infty$ in the sense that
$$\sum_{n \geq 1} \mathbb{P}(|Y_n-Y| >\epsilon ) <\infty$$
for all $\epsilon>0$, then $Y_n \to Y$ almost surely. That's exactly what we have used in the above proof.
