Independent random variables, convergence in distribution

Question

Let $(X_n)_{n \in \mathbb{N}} $ be a sequence of independent random variables with pmf of $X_n$: $p_n=p_{-n}=\frac{1}{2n^2}$ and $p_0=1-\frac{1}{n^2}$. I'm wondering whether $ \sum_{k=1}^{n} \frac{X_k}{\sqrt{n}} \rightarrow 0$ in distribution for $n \rightarrow \infty$. I can't utilize the law of large numbers here because of the square root and I'm not quite sure how to find the cumulative distribution function of $\sum_{k=1}^{n} X_k$ since convolution of multiple RV seems to get a bit messy. So, is what I'm trying to show even true and if so what would I need to utilize in order to show it?

Answer 1

$P(X_n \neq 0)=\sum_n \frac 1 {n^{2}} <\infty$. By Borel-Cantelli Lemma $X_n=0$ for all large $n$ with probability $1$. This implies that $\frac 1 {\sqrt n} \sum\limits_{k=1}^{n} X_k \to 0$ almost surely (hence also in distribution).