Compute the variance of expectation of $n^{1/2}X_{k}/(\sum_{m=1}^{n}X_{m}^{2})^{1/2}$ when $\mathbb{E}X_{1}=0$ and $Var(X_{1})=1$ My question is as follows:

Let $X_{k}$ are i.i.d random variables with $\mathbb{E}X_{1}=0$ and $Var(X_{1})=1$, and set $$Y_{n,k}:=\dfrac{X_{k}}{\sqrt{n^{-1}\sum_{i=1}^{n}X_{i}^{2}}}.$$
How could I show $n^{-1/2}\sum_{k=1}^{n}Y_{n,k}\longrightarrow_{d}\mathcal{N}(0,1)$ in distribution?

I tried to argue that since $X_{k}$ i.i.d, then for each $n$ fixed, $Y_{n,k}$ are i.i.d for all $1\leq k\leq n$. 
Then, I'd like to apply the CLT for i.i.d random variables, which needs:
$(1)$ $\mathbb{E}(Y_{n,k})=0$
$(2)$ $Var(Y_{n,k})=\sigma^{2}\in (0,\infty)$
$(3)$ $(Y_{n,1}+Y_{n,2}+\cdots+Y_{n,n})/n\longrightarrow 0.$
But I am not able to show all of these three. How could I compute expectation and variance in this case? the denominator has random variable, right?
Please help! Thank you!
Edit 1:
Okay I think I figure it out, following "James Yang"'s suggestion. 
Below is the proof:
Observe that $$n^{-1/2}\sum_{k=1}^{n}Y_{n,k}=\dfrac{n^{-1/2}\sum_{k=1}^{n}X_{k}}{\sqrt{n^{-1}\sum_{m=1}^{n}X_{m}^{2}}}.$$
Write $S_{n}:=\sum_{k=1}^{n}X_{k}$, then $\mathbb{E}S_{n}=0$, and $Var(S_{n})=n<\infty$. Then, by Markov, for each $\epsilon>0$, we have $$\mathbb{P}\Big(|\dfrac{S_{n}}{n}-0|\geq\epsilon\Big)=\mathbb{P}\Big(|S_{n}|\geq n\epsilon\Big)\leq\dfrac{1}{n^{2}\epsilon^{2}}Var(S_{n})=\dfrac{1}{\epsilon^{2}}{\dfrac{1}{n}}\longrightarrow 0\ \text{as}\ n\longrightarrow\infty,$$ and thus $S_{n}/n\longrightarrow_{p} 0$. 
Therefore, by CLT for i.i.d random variables, we know that $$n^{-1/2}\sum_{k=1}^{n}X_{n}\Rightarrow\mathcal{N}\Big(0,Var(X_{1})\Big).$$
Now, by SLLN, we know that $n^{-1}\sum_{m=1}^{n}X_{m}^{2}\longrightarrow Var(X_{1})$ a.s. Then since the map $x\mapsto\sqrt{x}$ is continuous for all $x\geq 0$, and $n^{-1}\sum_{m=1}^{n}X_{m}^{2}\geq 0$, by the continuous mapping theorem, we know that $$\sqrt{n^{-1}\sum_{m=1}^{n}X_{m}^{2}}\longrightarrow\sqrt{Var(X_{1})}\ \text{a.s.},$$ and since almost sure convergence implies convergence in probability, we know that $$\sqrt{n^{-1}\sum_{m=1}^{n}X_{m}^{2}}\longrightarrow_{p}\sqrt{Var(X_{1})}.$$
It then follows from Slutsky's theorem that $$\dfrac{n^{-1/2}\sum_{k=1}^{n}X_{k}}{\sqrt{n^{-1}\sum_{m=1}^{n}X_{m}^{2}}}\Rightarrow \dfrac{\mathcal{N}(0,Var(X_{1})}{\sqrt{Var(X_{1})}}=\mathcal{N}(0,1),\ \text{as desired.}$$
I haven't been used to combining SLLN with CLT. Really appreciate Yang's suggestion :)
As usually, I will leave the post open and if there is no more discussion, I will answer the post myself, or I will persuade Yang to post an answer. 
 A: Okay I think I figure it out, following "James Yang"'s suggestion. 
Observe that $$n^{-1/2}\sum_{k=1}^{n}Y_{n,k}=\dfrac{n^{-1/2}\sum_{k=1}^{n}X_{k}}{\sqrt{n^{-1}\sum_{m=1}^{n}X_{m}^{2}}}.$$
Write $S_{n}:=\sum_{k=1}^{n}X_{k}$, then $\mathbb{E}S_{n}=0$, and $Var(S_{n})=n<\infty$. Then, by Markov, for each $\epsilon>0$, we have $$\mathbb{P}\Big(|\dfrac{S_{n}}{n}-0|\geq\epsilon\Big)=\mathbb{P}\Big(|S_{n}|\geq n\epsilon\Big)\leq\dfrac{1}{n^{2}\epsilon^{2}}Var(S_{n})=\dfrac{1}{\epsilon^{2}}{\dfrac{1}{n}}\longrightarrow 0\ \text{as}\ n\longrightarrow\infty,$$ and thus $S_{n}/n\longrightarrow_{p} 0$. 
Therefore, by CLT for i.i.d random variables, we know that $$n^{-1/2}\sum_{k=1}^{n}X_{n}\Rightarrow\mathcal{N}\Big(0,Var(X_{1})\Big).$$
Now, by SLLN, we know that $n^{-1}\sum_{m=1}^{n}X_{m}^{2}\longrightarrow Var(X_{1})$ a.s. Then since the map $x\mapsto\sqrt{x}$ is continuous for all $x\geq 0$, and $n^{-1}\sum_{m=1}^{n}X_{m}^{2}\geq 0$, by the continuous mapping theorem, we know that $$\sqrt{n^{-1}\sum_{m=1}^{n}X_{m}^{2}}\longrightarrow\sqrt{Var(X_{1})}\ \text{a.s.},$$ and since almost sure convergence implies convergence in probability, we know that $$\sqrt{n^{-1}\sum_{m=1}^{n}X_{m}^{2}}\longrightarrow_{p}\sqrt{Var(X_{1})}.$$
It then follows from Slutsky's theorem that $$\dfrac{n^{-1/2}\sum_{k=1}^{n}X_{k}}{\sqrt{n^{-1}\sum_{m=1}^{n}X_{m}^{2}}}\Rightarrow \dfrac{\mathcal{N}(0,Var(X_{1})}{\sqrt{Var(X_{1})}}=\mathcal{N}(0,1),\ \text{as desired.}$$
