Q:

If $X_j$ is iid Unif$(-1,1)$ and $Y_n=\frac{\sum X_j}{\sum X_j^2+\sum X_j^3}$, show that $\sqrt{n}Y_n\rightarrow N(0,3)$ in distribution.

If $X_j$ is iid Unif$(-1,1)$ and $\displaystyle Y_n=\frac{\sum X_j}{\sum X_j^2+\sum X_j^3}$, show that $\sqrt{n}Y_n\rightarrow N(0,3)$ in distribution.
I know from central limit theorem that $\sqrt{n} \frac{1}{n}\sum X_j \rightarrow N(0,\frac{1}{3})$ but I'm not sure how to show the denominator of $Y_n$ goes to some constant in probability or something else in distribution.

A:

I convert my comments into an answer.
First remark that: $$
\sqrt{n}\cdot Y_n  = \frac{\frac{1}{\sqrt{n}}\sum X_i}{\frac{1}{n}\sum X_i^2 + \frac{1}{n}\sum X_i^3}.
$$
The numerator converges in law to a normal random variable by the Central Limit Theorem.
The denominator converges almost surely (in probability is sufficient) to some constant by the Law of Large Numbers.
Determine the variance of the normal random variable, the limit constant for the denominator, and finally conclude with Slutsky's lemma.

