Does convergence in probability of $X-Y$ and convergence in distribution of $X$ imply convergence in distribution of $Y$?

Question

I'm trying to learn about the asymptotic properties of random variables. Suppose that $X$ and $Y$ are random variables such that

$$\sqrt{n}X \xrightarrow{d} N(0,\sigma^2)$$

$$\sqrt{n}(X-Y) \xrightarrow{p} 0$$

Can we conclude that $\sqrt{n}Y \xrightarrow{d} N(0,\sigma^2)$?

Answer 1

If $A_n\to A$ in distribution and $B_n\to 0$ in probability, then $A_n+B_n\to A$ in distribution.

Use this with $A_n=\sqrt nX_n$, $A$ having a centered normal distribution with variance $\sigma^2$ and $B_n=\sqrt n\left(Y_n-X_n\right)$.