Let $X_1, X_2, \cdots$ be i.i.d random variables with mean $0$ and finite variable $\sigma^2$. Use the central limit theorem and Slutsky's Theorem to show that $\frac{\sum_{m=1}^{n} X_m}{\sqrt{\sum_{m=1}^{n} X_m^2}} \overset{d}{\rightarrow} N(0,1)$.
What I've done is as follows. We know that $\sqrt{n}\left(\frac{1}{n}\sum_{m=1}^{n} X_m - 0\right) \overset{d}{\rightarrow} N(0, \sigma^2)$ by the CLT. For the denominator, we know that $E(X_m^2) = \sigma^2$. I am trying to apply CLT to the denominator too, but we don't know $V(X_m^2)$ and I'm stuck.
Any help would be appreciated.
Note that $$ \frac{1}{n}\sum X_m^2 \xrightarrow{p} \sigma^2, $$ hence, by the continuous mapping theorem you have $$ \frac{1}{\sqrt{n}}\sqrt{\sum X_m^2} \xrightarrow{p} \sigma. $$ Now, as $$ \sqrt{n}\frac{\sum X_m}{n} \xrightarrow{D}\mathcal{N}(0,\sigma^2), $$ Using Slutsky and again the continuous mapping $$ \frac{\sum X_m /\sqrt{n}}{\sqrt{\sum X_m^2}/\sqrt{n}} =\frac{\sum X_m }{\sqrt{\sum X_m^2}} \xrightarrow{D}\frac{1}{\sigma}\mathcal{N}(0,\sigma^2)=\mathcal{N}(0,1) $$