$\sqrt{n}Y_n$ converges to a random variable Y in distribution

Solution 1:

Note that \begin{gather} \sqrt{n}Y_n = \frac{\frac{1}{\sqrt{n}}\sum_{i=1}^{n} X_i}{\frac{1}{n}\sum_{i=1}^{n}X_i^2 + \frac{1}{n}\sum_{i=1}^{n}X_i^3}. \end{gather} The numerator converges in distribution to $N(0,1/3)$, and the denominator converges in probability to $1/3+0$.

Solution 2:

I think this problem has to do with the central limit theorem somehow.

Yes, CLT along with Slutsky's Theorem. In fact, as already shown,

$$\sqrt{n}Y_n=A_nB_n$$

where

$$A_n\xrightarrow{\mathcal{P}}3$$

and

$$B_n\xrightarrow{\mathcal{L}}B\sim N\left(0;\frac{1}{3} \right)$$

thus

$$\sqrt{n}Y_n\xrightarrow{\mathcal{L}}3B\sim N(0;3)$$

Thank you very much! Could you explain a bit more about the convergent deduction, if possible?

in the denominator you can apply SLLN thus

$$\frac{1}{n}\Sigma_iX_i^2\xrightarrow{a.s.}E(X^2)=1/3$$

converging almost surely, it converges also in probability.

As the second addend in the denominator is concerned, using the same reasoning, it converges almost surely to zero, thus using Continuous Mapping theorem, your denominator converges almost surely to $1/3$