Find $\lim\limits_{n \to \infty}\mathbb P\left(\frac1{\sqrt n}\sum\limits_{i=1}^nX_i\le x\right)$ if $P(X_n=0)=1-\frac1{n^2}$ 
Let $(X_{n})_{n}$ be independent random variables such that $\mathbb P(X_{n}=n)=\mathbb P(X_{n}=-n)=\frac{1}{2n^2}$ and $P(X_{n}=0)=1-\frac{1}{n^2}$.
  Find $\lim_{n \to \infty}\mathbb P(\frac{1}{\sqrt{n}}\sum_{i=1}^{n}X_{i}\leq x)$.

Ideas: 
I think it would be appropriate to use Borel-Cantelli.
Set $A:=\{X \neq 0\}:=\{\omega\in\Omega:X_{n}(\omega)\neq0, \forall n \in \mathbb N\}$ and $A_{n}:=\{X_{n}\neq0\}$
Note $\sum_{n=1}^{\infty}P(A_{n})=\sum_{n=1}^{\infty}\frac{1}{n^2}<\infty$
Therefore $P(\limsup A_{n})=P(A)=0$
This means that $P(A^{c})=1-P(A)=1$
So $A^c \to 0$ a.s. 
This then implies stochastic convergence. I am unsure on my selection of $A$ and $A_{n}$ though. Any corrections, guidance or recommendations?
 A: Your idea of using Borel-Cantelli lemma is basically right. Let $$
A=\{X_n \ne 0 \text{ for infinitely many }n\in\mathbb{N}\}.
$$ Since $\sum_{n=1}^\infty P(X_n \ne 0)=\sum_{n=1}^\infty \frac{1}{n^2}<\infty$, Borel-Cantelli Lemma implies that $P(A) = 0$. So, for almost every $\omega \in\Omega$, there exists $N(\omega)\in\mathbb{N}$ such that$$
n\ge N(\omega) \Rightarrow X_n(\omega) = 0.
$$ This shows
$$
\lim_{n\to\infty}\frac{1}{\sqrt{n}}\sum_{i=1}^nX_i(\omega) = \lim_{n\to\infty}\frac{1}{\sqrt{n}}\sum_{i=1}^{N(\omega)-1}X_i(\omega)=0
$$ for almost every $\omega$, hence giving the result
$$
\lim_{n\to \infty} P\left(\frac{1}{\sqrt{n}}\sum_{i=1}^nX_i\le x\right) =\begin{cases} 1,\quad x>0\\0,\quad x<0\end{cases}.
$$ The case $x=0$ is more subtle. Note that by the symmetry of the distribution, we have
$$
2P\left(\frac{1}{\sqrt{n}}\sum_{i=1}^nX_i\le 0\right)=P\left(\frac{1}{\sqrt{n}}\sum_{i=1}^nX_i\le 0\right)+P\left(\frac{1}{\sqrt{n}}\sum_{i=1}^nX_i\ge 0\right)=1+P\left(\frac{1}{\sqrt{n}}\sum_{i=1}^nX_i=0\right).
$$ Therefore,
$$
\lim_{n\to \infty} P\left(\frac{1}{\sqrt{n}}\sum_{i=1}^nX_i\le 0\right)=\frac{1}{2}+\frac{1}{2}\lim_{n\to \infty} P\left(\sum_{i=1}^nX_i =0\right).
$$ My guess is that $\lim_{n\to\infty} P\left(\sum_{i=1}^nX_i =0\right)=0$, but as of now, I'm not sure.
