Let $X_1, X_2,..$ be iid random variables with mean $\mu$ and variance $\sigma^2$. Show $$ \displaystyle \frac{\sum_{i = 1}^n (X_i - \mu)}{\sqrt{\sum_{i = 1}^n (X_i - \overline{X}_n)^2}} \to N(0,1) $$ in distribution
I want to do this using characteristic functions. It looks similar to $$ \displaystyle \frac{\sum_{i = 1}^n (X_i) - n \mu}{\sqrt{n\sigma^2}} \to N(0,1) \,\,(1)$$ Which follows from Central Limit Theorem. Now, $$\displaystyle \frac{\sum_{i = 1}^n (X_i - \mu)}{\sqrt{\sum_{i = 1}^n (X_i - \overline{X}_n)^2}} = \displaystyle \frac{\sum_{i = 1}^n (X_i) - n\mu}{\sqrt{\sum_{i = 1}^n (X_i - \sum_{i=1}^n \frac{X_i}{n })^2}} = \displaystyle \frac{\sum_{i = 1}^n (X_i) - n\mu}{\sqrt{\sum_{i = 1}^n (X_i^2) - n\overline{X}_n^2}}$$
The denominator looks almost like sample variance but missing a factor so I can re-write as $$ \frac{\sum_{i = 1}^n (X_i) - n\mu}{\sqrt{n \bigg( \frac{1}{n}\bigg(\sum_{i = 1}^n (X_i^2) - n\overline{X}_n^2}\bigg)\bigg)} $$
In (1) above you can set $Y_i = \frac{X_i - \mu}{\sigma}$ then use characteristic functions so I would like do something similar here. However, I am not sure what to use as my $Y_i$ here because of the $X_i$'s present in the denominator.
$$\displaystyle \frac{\sum_{i = 1}^n (X_i - \mu)}{\sqrt{\sum_{i = 1}^n (X_i - \overline{X}_n)^2}} = \frac{\sum_{i = 1}^n X_i - n \mu}{\sqrt{n\sigma^2}} \cdot \sqrt\frac{\sigma^2}{\frac{\sum_{i=1}^n X_i^2}{n}-\overline{X}_n^2} $$ The second multiplier converges to $1$ in probability since biased sample variance converges to $\sigma^2$: $$\frac{\sum_{i=1}^n X_i^2}{n}-\overline{X}_n^2 \xrightarrow{p} \mathbb E[X_1^2]-(\mathbb E[X_1])^2=\sigma^2$$ Then you can use Slutsky's theorem.