Limit of the probability of an event (sum of i.i.d. r.v.'s)

Let $X_1,X_2,\dots$ be iid random variables with mean zero and finite variance, and let $S_n = \sum_{k=1}^n X_k$. For $b>0$, find the limit $$\lim_{n\to\infty}\mathbb{P}\!\left(\left\lvert\frac{1}{\sqrt{n}}S_n\right\rvert<b\right)$$

I know to use central limit theorem to solve this, but I just don't know how to approach it.

• You mean $P(S_n/\sqrt{n}<b)$ or $P(|S_n/\sqrt{n}|<b)$? – Jimmy R. Feb 16 '16 at 15:36
• @JimmyR. second one sorry for my poor coding – Molizard Feb 16 '16 at 15:37
• No problem. I edited it, so can you please tell me if it is correct know? Thanks. – Jimmy R. Feb 16 '16 at 15:39
• Also to be clear: $S_n$ stands for $S_n=X_1+X_2+\dots+X_n$? – Jimmy R. Feb 16 '16 at 15:40
• Yeah, that's fine. It does. – Molizard Feb 16 '16 at 15:42

The CLT indeed implies that $$\frac{S_n}{\sqrt{n}} \overset{d}\to N(0,σ^2)$$ where $σ^2$ denotes the finite variance of the $X_i$'s, i.e. $σ^2=Var(X_i)$, for $i\in \mathbb N$. So, \begin{align}P\left(\left|\frac{1}{\sqrt{n}}S_n\right|<b\right)&=P\left(-b< \frac{1}{\sqrt{n}}S_n<b\right)\\[0.3cm]&=P\left(\frac{1}{\sqrt{n}}S_n<b\right)-\left(\frac{1}{\sqrt{n}}S_n<-b\right)\\[0.3cm]&\to Φ(b/σ)-Φ(-b /σ)=2Φ(b/σ)-1\end{align} as $n\to +\infty$.