Why the almost sure convergence of the SLLN doesn't hold here

Question

Let $X$ be a random variable that takes value $1$ with probability $\frac12$ and value $-1$ with probability $\frac12$. For every $i>1$, define the random variable $X_i=X$ with $\mathbb E\:X_i=0$ for all $i$. Show that, for each $n$, $$\mathbb P\left(\frac1nS_n=0\right)=0$$

But isn't the stronger law of large numbers says $$\frac1n S_n\rightarrow0\textrm { a.s.}$$ Then isn't $\mathbb P\left(\frac1nS_n=0\right)=1?$ Why the almost sure convergence of the SLLN doesn't hold here?

Answer 1

The $X_i$ are all equal to $X$, so $S_n = n X$. This implies $\frac 1n S_n$ is $1$ with probability $\frac 12$ and $-1$ with probability $\frac 12$, in particular $\frac 1n S_n \ne 0$ for all $n$. The law of large numbers doesn't apply here because the $X_i$ are not independent.

Answer 2

For $n$ odd, $P(\frac{S_n}{n}=0)=0$. For $n$ even, $P(\frac{S_n}{n}=0)=\frac{n!}{((n/2)!)^2 2^n}\approx \sqrt{\frac{2}{\pi n}}$. However the density function for large $n$ clusters around $0$ leading to the law of large numbers result.