Convergence of independent random variables with different distributions Let $Y_1,Y_2,...,$ be independent r.v. with distribution $P(Y_k=k)=P(Y_k=−k)=1/2$ for all k. Here, $S_n=Y_1+Y_2+...+Y_n \ \ ∀k>0$.
Does ${S_n/n^{3/2}}$ converge in distribution? If yes, please write the limit distribution.
Note: each Yi has a different distribution
I know that the Law of Large Numbers requires i.i.d. random variables but I don't know how to start given this condition.

A simple calculation shows $E\Big(\frac{S_n}{n\sqrt{n}}\Big)=0$ while $$V\bigg(\frac{S_n}{n\sqrt{n}}\bigg)=\frac{n(n+1)(2n+1)}{6n^3}\longrightarrow\frac{1}{3}$$ as $n \longrightarrow \infty$. The latter calculation suggests $\frac{S_n}{n\sqrt{n}}$ won't converge in distribution to the constant random variable $0$ since its limiting variance is non$-$vanishing. Moreover, since $\frac{S_n}{n\sqrt{n}}$ resembles an average of some sort, we may hypothesize that $\frac{S_n}{n\sqrt{n}}$ converges in distribution to a $N(0,1/3)$ random variable. Let's use moment generating function to see this. The moment generating function of $\frac{S_n}{n\sqrt{n}}$ is $$E\bigg(e^{\frac{tS_n}{n\sqrt{n}}}\bigg)=E\bigg(e^{\frac{tY_1}{n\sqrt{n}}}\bigg)\times \ldots \times E\bigg(e^{\frac{tY_n}{n\sqrt{n}}}\bigg)=\prod_{k=1}^n\cosh\bigg(\frac{kt}{n\sqrt{n}}\bigg)$$ You can see here that the right hand side in the above expression  converges to the function $y=e^{x^2/6}$ which is precisely the moment generating function of a $N(0,1/3)$ random variable.
