I want to solve the following task:
Let $p \in (0,1)$ and $X_n$ be a sequence of i.i.d. random variables with $$\mathbb{P}(X_n = 1)=1-\mathbb{P}(X_n=-1)=p$$ for all $n$. Let $$S_0:=0 \text{ and } S_n:=\sum_\limits{i=1}^n X_i$$ and $A=\limsup \{S_n = 0\}$.
Show $$ p \ne \frac{1}{2} \Rightarrow \mathbb{P}(A) = 0.$$
I'm sure that I need the first Borel Cantelli Lemma here. So I have to show $\sum_\limits{n \in \mathbb{N}} \mathbb{P}(S_n=0) < \infty$ and probably this does only hold if $p \ne \frac{1}{2}$ but how do I calculate $\mathbb{P}(S_n=0)$? I know that $S_n$ is only equal to $0$ if I have as many ones as I have minus ones but this approach didn't help me.
Any ideas?
When $n=2k+1$ is odd, $$\{S_{2k+1}=0\}=\emptyset$$ because the sum will be odd.
When $n=2k$ is even, $$\mathbb{P}(S_{2k}=0)=\frac{1}{2^{2k}}\binom{2k}{k}p^k(1-p)^k$$ because exactly $k$ of $X_i$ have to be $1$, and the other $k$ have to be $-1$.
Using the Stirling's approximation, we get
$$\mathbb{P}(S_{2k}=0)\approx \frac{1}{2^{2k}}\frac{4^{2k}} {\sqrt{2\pi k}}(p(1-p))^k = \frac{2^{2k}}{\sqrt{2\pi k}}(p(1-p))^k = \frac{(4p(1-p))^k}{\sqrt{2\pi k}} $$
and when $p\not = \frac{1}{2}$, we have $$4p(1-p)<1$$
so the series converges
$$\sum \mathbb{P}(S_{2k}=0) < \infty$$
by the Root test.
By Borel-Cantelli lemma,
$$ \mathbb{P}(A) = \mathbb{P}(\lim\sup\{S_n=0\}) = 0$$