Proving a series converges almost surely 
Let $X_{1}, X_{2}, \ldots$ be independent random variables uniformly
  distributed on the interval $[-1, 1]$. Let $a_1, a_2, \ldots$ be a
  sequence of real numbers such that $\sum_{n=1}^{\infty} a_n^2$
  converges. Prove that the series $\sum_{n=1}^{\infty} a_n X_n$
  converges almost surely.

I'm studying for my exam by doing problems in a book but I am stuck on this one. I am pretty new to almost sure convergence and I guess it makes it difficult even moreso for me that each $X_i$ is uniform on $[-1, 1]$ rather than just positive numbers like $[0, 1]$. 
I would really appreciate your assistance in helping me answer this question. I tried on this one for almost five  hours with  no luck. I am familiar with common probability inequalities, like Markov's, Chebyshev's etc.
 A: First, I’ll need a lemma.
Let $U_1, \ldots, U_n$ be independent $L^2$ random variables with mean zero and let $X_k=\sum_{l=1}^k{U_l}$. 
Let $M$ be the maximum of $X_0,\ldots,X_n$ and let $V=\|X_n\|_2$. 
I claim that for $a > 0$, $P(M \geq a) \leq \frac{V}{a}$. 
Proof: let $T \in [0,n+1]$ be the first index such that $X_T \geq a$. Note that $T = k$ and the $U_l$ with $l > k$ are independent. As a consequence, for all $0 \leq k \leq n$, $X_k1(T=k)$ and $X_n1(T=k)$ have the same mean. Summing over $0 \leq k \leq n$, it follows that $X_T1(T \leq n)$ and $X_n1(T \leq n)$ have the same mean. 
Now $X_T1(T \leq n) \geq a1(T=n)=a1(M \geq a)$, while the mean of $X_n1(T \leq n)$ is clearly under $V$. 
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Let’s come back to the original question. The maximal inequality yields $P(\max_{n \geq m}\,|S_n-S_m| > \alpha)^2 \leq \epsilon^{-2} \sum_{k > n}{a_k^2}$. 
It follows that for all $\epsilon > 0$ there exists as some $n$ such that for all $m \geq n$ $|S_m-S_n| \leq \epsilon$ and the rest is standard. 
A: This is an immediate consequence of Kolomogorov's Three Series Theorem (you can take $A=1$ in that theorem): https://en.wikipedia.org/wiki/Kolmogorov%27s_three-series_theorem
