How to show that a series of random variables, $\sum\limits_{n\ge1}X_n$, converges almost surely. 
Let $X_n$ be random variables such that for some $a_n\in \mathbb{R}$:
\begin{align}
\sum\limits_{n\ge1}\mathbb{P}(X_n\ne a_n)<\infty \quad \text{and} \quad \sum\limits_{n\ge1}a_n \  \ \text{converges}
\end{align}
Show that $\sum\limits_{n\ge1}X_n$ converges a.s.

I feel like I don't know a whole lot about how to show that a series of random variables converges almost surely, so any help on this one or even general techniques to show almost sure convergence of a series would be greatly appreciated.
 A: Let $E_n=\{X_n\ne a_n\}$ and $E=\cap_{n=1}^\infty \cup_{k=n}^\infty E_k$. Note that if $\omega\notin E$ then we have $X_n(\omega)\ne a_n$ for only finitely many values of $n$, so the sequence $X_n(\omega)$ eventually just becomes the sequence $a_n$, and hence $\sum_{n=1}^\infty X_n(\omega)$ converges. So now we just have to prove that the probability of the event $\{\omega\notin E\}$ is $1$, or equivalently that $\mathbb{P}(E)=0$. But this just follows immediately from the first Borel-Cantelli lemma.
A: From the first statement, it follows that the probability of event $E_n \equiv X_n \neq a_n $ happening infintely often, $ P(E_n i.o.)=0$, i.e. $P(\limsup_n X_n = 0) = 1$, which follows from the convergence of the second sum (summand $\to_n 0$).
A: I did not see the answers before I wrote my comment so I will just post the fleshed out solution here:
\begin{align}
\sum\limits_{n\ge1}\mathbb{P}(X_n\ne a_n)<\infty &\implies \mathbb{P}(X_n\ne a_n \ \ \text{i.o})=0 \quad \text{by Borel-Cantelli}\\
&\implies \mathbb{P}(X_n=a_n \ \ \text{eventually})=1
\end{align}
So $\exists$ a $N>0$ such that $X_n=a_n$ a.s. for all $n\ge N$ and thus,
\begin{align}
\mathbb{P}\big(\sum\limits_{n\ge1}X_n < \infty\big)&=\mathbb{P}\big(\sum\limits_{n\ge N}X_n < \infty\big)\\
&=\mathbb{P}\big(\sum\limits_{n\ge N}a_n < \infty\big)\\
&=1 \quad \text{by assumption}
\end{align}
