Central limit problem and converges

Let $\left(X_{j}\right)_{j \geq 1}$ be i.i.d. with $E\left\{X_{j}\right\}=1$ and $\sigma_{X_{j}}^{2}=\sigma^{2} \in$ $(0, \infty)$. Show that $$|\frac{2}{\sigma}\left(\sqrt{S}_{n}-\sqrt{n}\right)-\frac{S_{n}-n}{\sigma \sqrt{n}}|$$ Converges to $0$ in probability

I know that $\frac{S_{n}-n}{\sigma \sqrt{n}} \rightarrow Z$ and $\frac{2}{\sigma}\left(\sqrt{S}_{n}-\sqrt{n}\right)$ too by the central limit theorem and making: $\begin{aligned}\left|Z_{n}-Y_{n}\right| &=\left|\frac{2}{\sigma}\left(\sqrt{S}_{n}-\sqrt{n}\right)-\frac{S_{n}-n}{\sigma \sqrt{n}}\right| \\ &=\left|\frac{2}{\sqrt{\frac{S_{n}}{n}}+1} \frac{S_{n}-n}{\sigma \sqrt{n}}-\frac{S_{n}-n}{\sigma \sqrt{n}}\right| \\ & \leq\left|\frac{2}{\sqrt{\frac{S_{n}}{n}}+1}-1\right|\left|\frac{S_{n}-n}{\sigma \sqrt{n}}\right| . \end{aligned}$

what can i do next to proof what they ask me?

Solution 1:

A "by hand" way to finish the argument is to use

$$P(\{ |XY|<\varepsilon \}) \geq P(\{ |X|<\varepsilon^2 \} \cap \{ |Y|<1/\varepsilon \}),$$ then treat the first factor as your $X$ (since it is small with high probability) and the second factor as your $Y$ (since it is not generally small, but it is only large with low probability).