# Prove that the series $\sum\limits_{n=0}^{\infty}X_n$ converges almost surely

Let $(X_n)_{n\ge 1}$ be a sequence of real valued random variables defined on some probability space $(\Omega, \mathcal{A},P)$. Assume that there exists a sequence of real numbers $(a_n)_{n\ge 0}$ such that both series $$\sum_{n\ge 0}a_n \quad and \quad\sum_{n\ge 0}P(X_n\neq a_n)$$ converge. Prove that the series $\sum\limits_{n=0}^{\infty}X_n$ converges almost surely.

I don't really know how to solve this problem. The first thing that I don't understand is the sentence

Prove that the series $\sum\limits_{n=0}^{\infty}X_n$ converges almost surely.

What is the series intended to converge to? I've thought it is intended to converge almost surely to $0$, so that the statement to prove would be $$P\left(\lim_{n\to \infty}\sum_{n=0}^{\infty}X_n =0 \right)=1$$

But then i can't really move on. I was thinking to use Strong Law of Large Numbers, but to use it the sequence of random variables must be i.i.d. and in the Problem they are not. Do you have any hint?

By "the series $\sum\limits_n X_n$ converge almost surely", it is meant that there exists $\Omega'$ of probability $1$ for which if $\omega\in\Omega'$, then the series of real numbers $\sum\limits_{n=1}^{+\infty}X_n(\omega)$ is convergent.
Define the event $E_n:=\{X_n\neq a_n\}$. We get by the assumption that $\mathbb P\left(\limsup\limits_nE_n\right)=0$. Can you show that if $\omega$ belongs to the complement of $\limsup\limits_nE_n$, then the series $\sum\limits_{n=1}^{+\infty}X_n(\omega)$ is convergent?
• +1! Thank you for your help! I'm trying to follow your hint. What I've understood so far is that by Borel Cantelly and the assumption that $\sum_n P(X_n \neq a_n) < \infty$ follows that $\mathbb P\left(\limsup\limits_nE_n\right)=0$. From this i can understand that there are only finite many $\omega \in \Omega$ such that $X_n(\omega) \neq a_n$. (is it correct?). This means that there are infinitely many $\omega \in \Omega$ in the complement of $E_n$. Now if we pick $\Omega' := E_n^c$ we have an event of probability $1$ right? I think that this would imply that the series – Bman72 Aug 21 '14 at 15:49
• $\sum_{n} X_n$ converges, but i'm not sure why... – Bman72 Aug 21 '14 at 15:49
• More exactly, there is $\Omega'$ of probability $1$ for which if $\omega\in \Omega'$ then $\omega\notin\limsup_n E_n$. ($E_n$ is not necessarily of probability $1$: its probability goes to $0$ after all). If $\omega\notin \limsup_n E_n$, this means that for some $N$ depending of $\omega$, we have $\omega\notin E_n$ if $n\geqslant N$. If you change finitely many terms of a series, its convergence/divergence won't change. – Davide Giraudo Aug 21 '14 at 15:53
• Sorry, i've meant to pick $\Omega' = E_n^c$. Ok, so summarizing: $P(\limsup E_n=0)$ implies that there exists $N(\omega)$ such that $\forall n > N(\omega)$, $\omega \notin E_n$. Since the event $E_n$ is the event of all $\omega$ such that $X_n(\omega) \neq a_n$ and since we have shown that this event has probability $0$, this means that there exists a set $\Omega'$ such that for all $\omega \in \Omega'$ $X_n(\omega) = a_n$ and since the series $\sum_{n}a_n$ converges we have that $\sum_{n} X_n(\omega)$ converges for all $\omega$ in that set right? – Bman72 Aug 21 '14 at 15:59