This is the text of the problem: Let $(X_{j})_{j\geq 1}$ be i.i.d. nonnegative with $E\{X_{1}\}=1$ and $\sigma_{X_{1}}^{2}=\sigma^{2} \in (0, \infty)$, and let $S_{n}=\sum_{j=1}^{n}X_{j}$. Show that $\frac{2}{\sigma}(\sqrt{S_{n}}-\sqrt{n})\to^{\mathcal{D}}Z$, with $\mathcal{L}(Z)=N(0,1)$.
I was given the following hint: Notice that $(2\sqrt{x})^{\prime} = x^{-1/2}$. So, $2(\sqrt{S_{n}}-\sqrt{n})=\int_{n}^{S_{n}}\frac{dx}{\sqrt{x}}$. Now, $\frac{S_{n}-n}{\sqrt{n}} \to^{\mathcal{D}} N(0,1)$ by the Central Limit Theorem. So, show that $\int_{n}^{S_{n}}\frac{dx}{\sqrt{x}} \approx \frac{S_{n}-n}{\sqrt{n}}$.
So, this is what I tried to do: $|\int_{n}^{S_{n}}\frac{dx}{\sqrt{x}}-\frac{S_{n}-n}{\sqrt{n}}|=|\int_{n}^{S_{n}}\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{n}}\right)dx|=2 \sqrt{S_{n}}-\sqrt{n}-\frac{S_{n}}{\sqrt{n}}$, and eventually what I would like to do is show that this goes to $0$ as $n$ gets large. However, I am having trouble doing that. Please help!
I made a comment, but now I think I got a answer. After some manipulations you can get this
$$\frac{2}{\sigma}(\sqrt{S_{n}}-\sqrt{n}) = \frac{2\sqrt{n}}{\sqrt{S_n}+\sqrt{n}}\cdot \frac{S_n -n}{\sigma \sqrt{n}}$$
But, as I said intuitively $\frac{2\sqrt{n}}{\sqrt{S_n}+\sqrt{n}}$ goes to 1 because $X_i$ is one on average and then $S_n$ is $n$ on average.
Now, the Law of Large Numbers tells us that $$ \frac{S_n}{n} \rightarrow 1 \text{ a.s}$$. This implies that $$ \frac{2\sqrt{n}}{\sqrt{S_n}+\sqrt{n}} \rightarrow 1 \text{ a.s}$$ and by Slutsky's Theorem you have your result once $ \frac{S_n -n}{\sigma \sqrt{n}} \rightarrow \mathcal{N}(0,1)$ by the Central Limit Theorem.