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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?

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Presumably one needs to add the hypothesis that $X_1\geqslant0$ with full probability. –  Did Nov 19 '12 at 21:47
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So we need to show that $\frac{S_n}n$ converges in probability to $1$. As $t\mapsto \sqrt t$ is continuous, it follows from the law of large numbers.

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You have $\frac{\sqrt{S_n} + \sqrt{n}}{2 \sqrt{n}} = \frac{1}{2} \left( 1 + \sqrt{\frac{S_n}{n}}\right)$. What can you say about the limit of $\frac{S_n}{n}$ ?

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