If $P(X_n=\pm n)=\frac{1}{2n^2}$ and $P(X_n=\pm 1)=\frac{1}{2}(1-\frac{1}{n^2})$ then $S_n/\sqrt{n}\xrightarrow{d} N(0,1)$

Question

Let $\{X_n\}_{n\ge 1}$ be a sequence of independent random variables in a probability space such that $$P(X_n=n)=P(X_n=-n)=\frac{1}{2n^2}\ \forall n$$ and $$P(X_n=1)=P(X_n=-1)=\frac{1}{2}\left(1-\frac{1}{n^2}\right)\ \forall n\ge 2$$ I want to prove that $\frac{\sum_{i=1}^n X_i}{\sqrt{n}}$ converges in distribution to a standard normal random variable. I only can think about the approach using characteristic functions (and Lévy's continuity theorem), but $\lim_{n\to \infty} \Pi_{k=1}^n \varphi_{X_k}\left(\frac{t}{\sqrt{n}}\right)$ seems difficult to be computed since $$\varphi_{X_k}(t)=\frac{1}{k^2}\cos(nt)+\left(1-\frac{1}{k^2}\right)\cos(t)\ \forall k\ge 1\ \forall t\in \mathbb{R}$$ I would appreciate some hint. Thanks in advance!

Answer 1

Let $Y_n=1$ if $X_n >0$ and $Y_n=-1$ if $X_n <0$. Since $\sum P(X_n \neq Y_n)=\sum P(X_n =\pm n) <\infty$ it follows by Borel-Cantelli Lemma that $X_n=Y_n$ for all $n$ sufficiently large with probabiltiy $1$. This implies that $\frac 1 {\sqrt n} (\sum\limits_{k=1}^{n}X_k-\sum\limits_{k=1}^{n}Y_k) \to 0$ almost surely. Check that $(Y_n)$ is i.i.d. and apply CLT to this sequence to finish the proof.