# Show $\lim_{n\to\infty}\sqrt{n}\bigg(\frac{\sum_{j=1}^{n}X_j}{\sum_{j=1}^{n}X_j^2}\bigg)=Z$

Let $(X_j)_{j\ge 1}$ be independent, double exponential with parameter $1$. Show that; $$\displaystyle\lim_{n\to\infty}\sqrt{n}\bigg(\frac{\sum_{j=1}^{n}X_j}{\sum_{j=1}^{n}X_j^2}\bigg)=Z$$ where $\mathcal L(Z)=\mathcal N(0,\frac12),$ and the convergence is in distribution.

So common density is $\displaystyle\frac12e^{-|x|}$ for $-\infty<x<\infty$ and

$\mu=0,\ \sigma^2=2$ for all $X_j$

If I write $S_n=\sum_{j=1}^{n}X_j$

$\displaystyle\sqrt{n}\bigg(\frac{\sum_{j=1}^{n}X_j}{\sum_{j=1}^{n}X_j^2}\bigg)=\sqrt{n}\Bigg(\frac{\frac{S_n}{\sqrt{2n}}}{\frac{\sum_{j=1}^{n}X_j^2}{\sqrt{2n}}}\Bigg)$

The numerator should converge to the standard normal random variable, but how can I manipulate the denominator to get the correct result ?

By the strong law of large numbers we have $$\frac{\sum_{j=1}^n X_j^2}{\sqrt{n}\sqrt{2n}}=\frac{1}{\sqrt{2}}\frac1n\sum_{j=1}^n X_j^2\to \frac{1}{\sqrt{2}}{\rm E}[X_1^2]=\sqrt{2}$$ almost surely, and hence the result follows from Slutsky's theorem.