Let $X_1,X_2,\ldots$ be IID with $\mathbb{E}[X_i] = 0$ and $\sigma^2 = \mathrm{Var}(X_i)<\infty$ with $0<\sigma<\infty$ for all $i\in\mathbb{N}$. We know from Central Limit Theorem that $(\sum_{i=1}^n X_i)/(\sigma\sqrt{n}) \to N(0,1)$ in distribution as $n\to\infty$. Is this still true after replacing $\sigma\sqrt{n}$ in the statement with $(\sum_{i=1}^n X_i^2)^{1/2}$? This seems to me that it should be true. However, I haven't been able to prove it.
My Idea was to use Lindeberg-Feller Central Limit Theorem for a triangular array $\{X_{n,k}\}$ defined as $$ X_{n,k} = \frac{X_k}{\left(\sum_{i=1}^nX_i^2\right)^{1/2}} $$ Then, we can see that $\mathbb{E}[X_{n,k}]=0$ and $X_{n,1},\ldots, X_{n,n}$ are independent. Also, we have $\sum_{k=1}^n\mathbb{E}[X_{n,k}^2] = 1$ for all $n\in\mathbb{N}$. To continue, I need to show that $\sum_{k=1}^n\mathbb{E}[X_{n,k}^2\cdot 1_{|X_{n,k}|>\varepsilon}]\to 0$ as $n\to\infty$ for all $\varepsilon>0$ at which I'm stuck. How should I proceed from this?
The strong law of large numbers says that $$\frac{1}{n}\sum_i X_i^2\stackrel{as}{\to} \sigma^2$$ which means $$\frac{\sum_i X_i^2}{n\sigma^2)}\stackrel{as}{\to} 1$$ so $$\frac{\left(\sum_i X_i^2\right)^{1/2}}{\sqrt{n}\sigma}\stackrel{as}{\to} 1$$
The result now follows by Slutsky's Lemma.
In your version $X_{n,1},\dots, X_{n,n}$ aren't independent, because they share the same denominator, so you can't straightforwardly use the Lindeberg CLT.