Application of The Classical Central Limit Theorem (2)

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Let $\{X_j,\,j\geq1\}$ be independent, identically distributed r.v.'s with mean $0$ and variance $1$. Prove that both $$\frac{\displaystyle\sum_{j=1}^nX_j}{\sqrt{\displaystyle\sum_{j=1}^nX^2_j}}\quad
\mbox{ and }\quad\frac{\displaystyle{\sqrt{n}\sum_{j=1}^nX_j}}{\displaystyle\sum_{j=1}^nX^2_j}$$converge in distribution to $\Phi$.

$\bullet$ Proof.
Since $\{X_j,\,j\geq1\}$ are \textit{i.i.d.}, we have by the Classical Central Limit Theorem, $$
\frac{\sum_{j=1}^nX_j-n\mathscr{E}(X_1)}{\sigma(X_1)\sqrt{n}}
    = \frac{1}{\sqrt{n}}\sum_{j=1}^nX_j
    \overset{d}{\longrightarrow}\boldsymbol{\Phi},
$$where $\boldsymbol{\Phi}$ is normal distribution with mean 0 and variance 1.  By Weak Law of Large Numbers, the sample variance converge to 1 in pr., i.e. $$s_n^2=\frac{1}{n}\sum_{j=1}^n\left(X_j-\mathscr{E}(X_j)\right)^2=\frac{1}{n}\sum_{j=1}^nX_j^2\overset{p}{\rightarrow}\sigma^2=1\,\Rightarrow\,
\frac{1}{\sqrt{n}}\sqrt{\sum_{j=1}^nX^2_j}\overset{p}{\longrightarrow}1.$$Thus, we have both results as follows. $$\frac{\sum_{j=1}^nX_j}{\sqrt{\sum_{j=1}^nX^2_j}}
    = \frac{\frac{1}{\sqrt{n}}\sum_{j=1}^nX_j}{\frac{1}{\sqrt{n}}\sqrt{\sum_{j=1}^nX^2_j}}
    \overset{d}{\longrightarrow}\boldsymbol{\Phi},$$and, $$\frac{{\sqrt{n}\sum_{j=1}^nX_j}}{\sum_{j=1}^nX^2_j}
    = \frac{\frac{1}{\sqrt{n}}\sum_{j=1}^nX_j}{\frac{1}{n}\sum_{j=1}^nX^2_j}
    \overset{d}{\longrightarrow}\boldsymbol{\Phi}.$$

$\Box$