Show that for $(X_n)$ i.i.d. $\Bbb E [X_1] = 0$: $\frac{\sum_{k=1}^n X_k}{\sqrt{\sum_{k=1}^n X_k^2}} \xrightarrow[]{d} \mathcal N(0,1)$

Let $(X_n)_{n \in \Bbb N}$ be i.i.d. ; $\Bbb E [X_1] = 0$; $0 \lt \mathrm{Var}[X_1]\lt \infty$. Now i want to show that $$Y_n := \frac{\sum_{k=1}^n X_k}{\sqrt{\sum_{k=1}^n X_k^2}} \xrightarrow[]{d} \mathcal N(0,1)$$ I know that convergence in distribution is equivalent to $$P(Y_n \le x)\xrightarrow[]{n\to \infty}P(Z \le x)$$ with $Z$ having a $\mathcal N(0,1)$ distribution.
I don't really know how to start, can someone give me a hint on how to show this? Thanks in advance!

• This follows by mixing 3 theorems: classical Central Limit Theorem, Law of Large Numbers, and the Slutsky's theorem. Jul 18 '17 at 10:47

Let $\sigma^2=Var(X_1)$ ... then just a little hint: \begin{align} \frac{\sum_{k=1}^n X_k}{\sqrt{\sum_{k=1}^n X_k^2}}&=\frac{\sigma\sqrt n}{\sqrt{\sum_{k=1}^n X_k^2}}\frac{\sum_{k=1}^n X_k}{\sigma\sqrt n}\\ &=\frac{\sigma}{\sqrt{\frac1n\sum_{k=1}^n X_k^2}}\frac{\sum_{k=1}^n X_k}{\sigma\sqrt n}\\ \end{align}
Now show using an LLN that $\frac{\sigma}{\sqrt{\frac1n\sum_{k=1}^n X_k^2}}\to1$ and use a classic CLT to establish $\frac{\sum_{k=1}^n X_k}{\sigma\sqrt n}\to \mathcal{N}(0,1)$.
$$\frac{\sum_{k=1}^n X_k}{\sqrt{\sum_{k=1}^n X_k^2}} = \frac{\frac{1}{\sqrt{n}}\sum_{k=1}^n X_k}{\sqrt{\frac{1}{n}\sum_{k=1}^n X_k^2}} = \frac{\sqrt{n}\bar{X}_n}{\sqrt{\frac{1}{n}\sum_{k=1}^n X_k^2}}.$$ Where by WLLN and the continuous mapping theorem $\sqrt{\frac{1}{n}\sum_{k=1}^n X_k^2} \xrightarrow{p} \sqrt{EX^2}=\sqrt{Var(X)}$, and by CLT $$\sqrt{n}\bar{X}_n \xrightarrow{D}N(0,Var(X)),$$ hence by Slutsky $$\frac{\sum_{k=1}^n X_k}{\sqrt{\sum_{k=1}^n X_k^2}} \xrightarrow{D} \frac{N(0, Var(X))}{\sqrt{Var(X)}} = N(0,1).$$