Example where $n^{-1/2}S_n\Rightarrow N(0,1)$ but its variance does not converge to 1 Let $\{X_n\}$ be a sequence of independent random variables such that $X_n$ takes the values $\pm n$ each with probability $1/2n^2$ and $\pm 1$ each with probability $1/2(1-1/n^2).$ Define $S_n=X_1+\cdots+X_n$ for $n\geq 1.$ Show that $S_n/\sqrt{n}$ converges in distribution to $N(0,1).$ 
This example is interesting because although $n^{-1/2} S_n \Rightarrow N(0,1),$ it holds that $\text{Var}[n^{-1/2} S_n]\to 2$ as $n\to \infty.$
My approach: Write $X_n = Z_n + (n-1)Z_n Y_n$ where $Z_n$ are iid, taking values $\pm 1$ with prob. $1/2$ and $Y_n\sim\text{Ber}(1/2n^2),$ such that $Y_n$'s are independent themselves and also independent of the $Z_n$'s. Then, $$\frac{S_n}{\sqrt{n}} = \frac{1}{\sqrt{n}} \sum_{k=1}^n Z_k + \frac{1}{\sqrt{n}} \sum_{k=1}^n (k-1)Z_kY_k.$$ Now the first part in the RHS converges weakly to $N(0,1).$ Hence it suffices to show that the other part converges to $0$ in probability (or in distribution). How can I show this?

$\sum P(Y_n=1)=\sum \frac 1 {2n^{2}} <\infty$. By Borel - Cantelli Lemma  it follows that $Y_n=0$ for all large $n$ with probability $1$. This implies that the second term tends to $0$ almost surely. 
