# expectation value, distribution function and the central limit theorem

The problem goes thus:

$${\{X_n\}}$$ is an $$iid$$ sequence of random variables with mean 0 and variance $$\sigma^2$$. If the third moment is finite, show that $$\lim_{n \to \infty} \mathbb{E} \left(\left(\frac{S_n}{\sigma\sqrt n} \right)^3\right)=0$$ where $$S_n=X_1+X_2+\cdots +X_n$$

Seeing the $$\frac{S_n}{\sigma\sqrt n}$$ immediately reminded me of the central limit theorem which says that:

$$\lim_{n \to \infty} \mathbb{P}\left( \frac{S_n}{\sigma\sqrt n} \leq x \right)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x}e^{\frac{-t^2}{2}}dt$$

But I don't really see a connection between the third moment and and the probability given here (distribution function to be more precise)

I did recall the proof of law of large numbers that went along similar lines by considering fourth order moments but we never had any $$\sigma\sqrt{n}$$ to be seen there and this isn't just a constant that I can tackle easily. It depends on $$n$$.

You don't use CLT to get this result. What is needed is a direct evaluation of the term $$E[S_n^3]$$.
To begin with, note that for $$n \geq 3$$: \begin{align*} S_n^3 = (X_1 + X_2 + \cdots + X_n)^3 = \sum_{i = 1}^nX_i^3 + \sum_{i \neq j} X_i^2X_j + \sum_{i, j, k \text{ distinct}}X_iX_jX_k. \end{align*} There are $$3n(n - 1)$$ terms in the second summation and $$n(n - 1)(n - 2)$$ terms in the third summation, but for the reason you will see shortly, counting them is not really needed.
Now by the i.i.d. and zero-mean assumption, when $$i \neq j$$, $$E[X_i^2X_j] = E[X_i^2]E[X_j] = 0$$; when $$i, j, k$$ are all distinct, $$E[X_iX_jX_k] = E[X_i]E[X_j]E[X_k] = 0$$, whence (say $$E[X_1^3] = \xi$$) \begin{align*} E[S_n^3] = \sum_{i = 1}^nE[X_i^3] = nE[X_1^3] = n\xi. \end{align*} It then follows that \begin{align*} E\left[\left(\frac{S_n}{\sigma\sqrt{n}}\right)^3\right] = \frac{1}{n^{3/2}\sigma^3}E[S_n^3] = \frac{\xi}{\sqrt{n}\sigma^{3}} \to 0 \end{align*} as $$n \to \infty$$.