Proving this sequence of random variables satisfies $\lim_{n \to +\infty} \frac{\operatorname{Var}(S_n)}{\operatorname{Var}(T_n)} = 1$ I have been trying to solve the following exercise:

Let $T_1, T_2, \cdots$ and $S_1, S_2, \cdots$ be sequences of random variables such that $\frac{T_{n}-E\left(T_{n}\right)}{\sqrt{\operatorname{Var}\left(T_{n}\right)}}$ converges in distribution to $N(0, 1)$ and $$\lim _{n \rightarrow +\infty} \frac{E\left(T_{n}-S_{n}\right)^{2}}{\operatorname{Var}\left(T_{n}\right)}=0$$


Prove that:


a) $$
\lim _{n \rightarrow +\infty} \frac{E\left(T_{n}\right)-E\left(S_{n}\right)}{\sqrt{\operatorname{Var}\left(T_{n}\right)}}=0 \text { and } \lim _{n \rightarrow +\infty} \frac{T_{n}-S_{n}}{\sqrt{\operatorname{Var}\left(T_{n}\right)}}=0 \text{ in probability }
$$


b)$$
\lim _{n \rightarrow +\infty} \frac{\operatorname{Var}\left(S_{n}\right)}{\operatorname{Var}\left(T_{n}\right)}=1 \text { and } \frac{S_{n}-E\left(S_{n}\right)}{\sqrt{\operatorname{Var}\left(S_{n}\right)}} \text{ converges in distribution to $N(0,1)$ }
$$

I managed to solve a) but I've been stuck on b) for a long time now. I tried to see if $\frac{\operatorname{Var}\left(S_{n}\right)}{\operatorname{Var}\left(T_{n}\right)}$ could somehow be expressed as a sum of the limits that appear in a) but I've had no luck with this. Can anyone help? I'd really appreciate it!
 A: Lemma Let $(Y_n,Z_n)_{n\in\mathbb N}$ be sequences of random variables and $(c_n)_{n\in\mathbb N}$ a sequence of real numbers. Suppose that

*

*$\mathbb E[Y_n]=0$ for every $n\in\mathbb N$;


*$\mathbb E[Y_n^2]=1$ for every $n\in\mathbb N$;


*$\lim_{n\to\infty}\mathbb E[Z_n^2]=0$; and


*$\lim_{n\to\infty} c_n=0$.
Define $$X_n\equiv Y_n+Z_n+c_n\quad\text{for every $n\in\mathbb N$}.$$Then, $\lim_{n\to\infty} \mathbb E[X_n^2]=1$.
Proof Clearly,
\begin{align*}
\mathbb E[X_n^2]&=\overbrace{\mathbb E[Y_n^2]}^{=1}+\overbrace{\mathbb E[Z_n^2]}^{\to 0}+\overbrace{c_n^2}^{\to 0}\\
&+2\underbrace{\mathbb E[Y_n Z_n]}_{\spadesuit}+2c_n\underbrace{\mathbb E[Y_n]}_{=0}+2\underbrace{c_n}_{\to0}\underbrace{\mathbb E[Z_n]}_{\heartsuit}
\end{align*}
as $n\to\infty$. As for the terms $\spadesuit$ and $\heartsuit$, the Cauchy–Schwarz inequality implies that
\begin{align*}
|\mathbb E[Y_n Z_n]|&\leq\mathbb E[|Y_n Z_n|]\leq\sqrt{\mathbb E[Y_n^2\vphantom{)^2}]\mathbb E[Z_n^2]}=\sqrt{\mathbb E[Z_n^2\vphantom{)^2}]}\to 0,\text{ and}\\
|\mathbb E[Z_n]|&\leq\mathbb E[|Z_n|]\leq\sqrt{\mathbb E[Z_n^2\vphantom{)^2}]}\to 0
\end{align*}
as $n\to\infty$. This completes the proof of the lemma. $\blacksquare$
Now apply the lemma to your setting:
\begin{align*}
\underbrace{\frac{S_n-\mathbb E[S_n]}{\sqrt{\operatorname{Var}[T_n]}}}_{\equiv X_n}=\underbrace{\frac{T_n-\mathbb E[T_n]}{\sqrt{\operatorname{Var}[T_n]}}}_{\equiv Y_n}+\underbrace{\frac{S_n-T_n}{\sqrt{\operatorname{Var}[T_n]}}}_{\equiv Z_n}+\underbrace{\frac{\mathbb E[T_n]-\mathbb E[S_n]}{\sqrt{\operatorname{Var}[T_n]}}}_{\equiv c_n}.
\end{align*}
You can verify that all the premises of the lemma are satisfied, so it follows that
\begin{align*}
\lim_{n\to\infty}\mathbb E\left[\left(\frac{S_n-\mathbb E[S_n]}{\sqrt{\operatorname{Var}[T_n]}}\right)^2\right]=\lim_{n\to\infty}\frac{\operatorname{Var}[S_n]}{\operatorname{Var}[T_n]}=1.
\end{align*}
