This is an exercise of Central Limit Theorem (CLT):
Let $(X_j)_{j\geq 1}$ be i.i.d. with $E[X_1]=1$ and $\sigma_{X_1}^2=\sigma^2\in(0,\infty)$($\sigma>0$). Show that $ \frac{2}{\sigma}(\sqrt{S_n}-\sqrt{n})\to Z $ in distribution with $Z\sim N(0,1)$.
What I think is that $ \frac{S_n-n}{\sigma\sqrt{n}}\to Z $ in distribution, which is the CLT. We also have $ \frac{S_n-n}{\sigma\sqrt{n}}=\frac{2(\sqrt{S_n}-\sqrt{n})}{\sigma}\frac{(\sqrt{S_n}+\sqrt{n})}{2\sqrt{n}}. $ Then it suffices to show that $ \frac{\sqrt{S_n}+\sqrt{n}}{2\sqrt{n}}\to 1 $ in probability. How can I go on?