Consider a sequence of random variables $\{X_1,...,X_n\}$ in which $X_k\sim N(0,\sigma^2),\; \forall k=1,...,n$ and $\operatorname{Corr}(X_k,X_{k+h})=\phi^h$ where $0< \phi<1$ and $h=1,...,n-1$. Suppose:
$$\lim_{n\rightarrow \infty} \operatorname{Var} \left[\frac{1}{n}\sum_{i=1}^n X_i^2 \right]=0.$$
Determine the limit distribution of
$$\sqrt{n}\frac{\bar{X}_n}{S_n}$$
where
$$\bar{X}_n=\sum_{i=1}^n \frac{X_i}{n} \quad \text{and} \quad S_n^2=\sum_{i=1}^n \frac{(X_i-\bar{X}_n)^2}{n-1}.$$
Attempt.
The main difficulty here is that the random variables aren't independent, although they are identically distributed. Note that we can rewrite the quantity of interest:
$$ \sqrt{n}\frac{\bar{X}_n}{S_n} = \frac{\sigma}{S_n} \cdot \frac{\sqrt{n} \bar{X}_n}{\sigma} = \frac{\sigma}{S_n} \cdot \frac{\bar{X}_n}{\sigma/\sqrt{n}} $$
Note that since $ \displaystyle S_n \stackrel{P}{\to} \sigma, \frac{\sigma}{S_n} \stackrel{P}{\to} \frac{\sigma}{\sigma}=1$. What's left to do is determine the distribution of $ \displaystyle \frac{\bar{X}_n}{\sigma/\sqrt{n}} $. Then we can simply use Slutsky's Theorem.
The result will give a normal distribution, but how can I prove that it follows this distribution with its respective parameters.
Assuming joint normality, $\sqrt{n}\overline{X}_n$ is normal with mean 0 and some variance $\sigma^2_n$ that we can calculate. Then $\sqrt{n}\overline{X}_n$ converges in distribution to a mean 0 normal distribution with the limit of $\sigma^2_n$ as variance.