Convergence Rate of Sample Variance Question: Given $X_1,X_2,...$ as an sequence of iid distributed random variables with $E(X_i)=0 $ and $V(X_i)=σ^2$ and the fourth order moment $E(X_i^4)<\infty$. Show that:
$\sqrt{n}(S_n^2-\sigma^2)\xrightarrow{d}N(0,E[(X_i^2-\sigma^2)^2])$, where $S_n^2$ is the sample variance.
I am sure that we have to employ the fact that $S^2_n\xrightarrow{p} \sigma^2$ together with the Central Limit Theorem. But I still can not figure out the exact proof of this problem.

$S_n^2 = \frac{1}{n-1} \sum_{i=1}^n (X_i - \bar{X}_n)^2 = \frac{1}{n-1} \sum_{i=1}^n X_i^2 - \frac{n}{n-1}\bar{X}_n^2$.
The second term converges to $0$ almost surely and can be ignored. Apply the central limit theorem to the first term.
