TITLE: Sum of random variables is equal to zero infinitely often
QUESTION [5 upvotes]: $X_1,\dots ,X_n$ are i.i.d random variables with $P(X_1=1)=p$, $P(X_1=-1)=1-p$, $p\neq\frac{1}{2}$. And $S_n=\sum_{k=1}^nX_k$.
I need to show that $P(\limsup_{n\to\infty}\{S_n=0\})\in\{0,1\}$.
It seems to me that $\limsup_{n\to\infty}\{S_n=0\}$ is not in tail sigma-algebra. Without this, i don't know how to proceed with it.

REPLY [3 votes]: One can do things explicitly here, without $0$-$1$ laws, and show that the event in question has probability $0$.
Observe that $\mathbb{P}(S_n = 0) = 0$ if $n $ is odd, and if $n= 2k$ with $k\in \mathbb{N}$, then
$$
(1) \qquad \mathbb{P}(S_{2k} = 0) =  \left(\begin{array}{c}
2k \\ 
k
\end{array} \right) p^{k}(1-p)^{k}.
$$
Now, using Stirling's formula for factorials, we get
$$
\left(\begin{array}{c}
2k \\ 
k
\end{array} \right) = \frac{(2k)! }{(k!)^2} \leq C \frac{\sqrt{2\pi 2k } (2k)^{2k} e^{-2k}  }{2\pi k \ k^{2k} e^{-2k}   } = C \frac{4^k}{\sqrt{\pi k}},
$$
where $C>0$ is a constant independent of $k$.
From this, getting back to $(1)$ we obtain
$$
\qquad \mathbb{P}(S_{2k } = 0) \leq C \frac{1}{\sqrt{\pi k}} 4^k p^k (1-p)^k, \qquad k = 1,2,...
$$
It is easy to see that $p(1-p) < 1/4$ where $0\leq p \leq 1$ and  $p\neq 1/2$, and hence 
$$
\sum_{k=1}^{\infty}\mathbb{P}(S_k = 0) < \infty. 
$$
This implies, by Borel-Cantelli, that the event $ \limsup \{{S_k = 0} \}$ has probability $0$.