Sum of random variables is equal to zero infinitely often $X_1,\dots ,X_n$ 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.
 A: One approach would be using the Hewitt-Savage 0-1 law, which is a generalization of Kolmogorov's 0-1 law. Roughly speaking we introduce a concept called exchangeable $\sigma$-field $\mathcal{E}$, and show that event $\limsup_{n\rightarrow \infty}\{S_n = 0\}$ or $\{S_n = 0, i.o.\}$ belongs to $\mathcal{E}$. The Hewitt-Savage 0-1 law claims that if $X_i$ are i.i.d. and $A\in \mathcal{E}$, then $P(A)  =0$ or $1$. You can see that detailed explanation at Page 153-154 of Probability: Theory and Examples, Ed 4.1 by Rick Durrett. (see it here online).
Roughly speaking, for our case, an event $A$ is in $\mathcal{E}$ if the occurence of $A$ does not change if we exchange the value of finitely many $X_i$'s. Thus it can be seen that the tail $\sigma$-field $\mathcal{T}\subset \mathcal{E}$, indicating that this result is more general than the Kolmogorov's 0-1 law.
Actually this can generalize to any Borel set $B$, because $\{S_n \in B, i.o.\}\in \mathcal{E}$. 
Here $\{S_n = B, i.o.\}\in \mathcal{E}$ because exchanging values of finitely many $X_i$'s does not affect $\{S_n = B, i.o.\}$.
A: 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 $0$.
