Let $\{X_n\}$ be a sequence of independent random variables such that $P(X_n=1)=P(X_n=-1)=\dfrac 12 -\dfrac 1{2n^2}$ and $P(X_n=n)=P(X_n=-n)=\dfrac 1{2n^2}$.
Then does the sequence $Y_n=\dfrac 1{\sqrt n}\sum_{k=1}^n X_k$ converge in distribution ? If it does, what is the limiting distribution ?
My work: $E(X_n)=0$ and $Var(X_n)=E(X_n^2)=2-\dfrac 1{n^2}$. $E(Y_n^2)=2-\dfrac {\sum_{k=1}^n\dfrac 1{k^2}}{n}$ . So at least there is no constant $c$ such that $Y_n \to c$ in $L^2$. I don't know what to do next. Please help
$\{Y_n\}$ converges to $N(0,1)$ in distribution. To prove this let $Z_n=1$ if $X_n\in \{1,n\}$ and $Z_n=-1$ if $X_n\in \{-1,-n\}$. It is easy to check that $Z_n$ takes the values $\pm 1$ with probability $\frac 1 2 $ each. Of course, $Z_n$'s are independent. Hence, by CLT, $\frac 1 {\sqrt n} \sum\limits_{k=1}^{n} Z_k \to N(0,1)$ in distribution. Now consider $\frac 1 {\sqrt n} \sum\limits_{k=1}^{n} (X_k-Z_k)$. Observe that $\sum\limits_{k=1}^{\infty} P\{X_k \neq Z_n\} <\infty$. By Borel Cantelli Lemma it follows that $X_k=Z_k$ eventually with probability $1$, so $\frac 1 {\sqrt n} \sum\limits_{k=1}^{n} (X_k-Z_k) \to 0$ with probability $1$. This completes the proof.