Let $X_1,X_2,\ldots$ be an iid sequence such that $P\{X_1 = 1\} = p$, $P\{X_1 = -1\} =p$ and $P\{X_1 = 0\} = 1-2p$. We have that $E[X_1] = 0$ and $E[X_1^2] = 2p$.
Define $S_n = \sum_{i=1}^nX_i$ and $S_0 = 0$ and consider the natural filtration. Define the stopping time $\tau = \inf\{n\geq 0: S_n = a \text{ or } S_n = b\}$ where $a < 0 < b$ and $a,b \in \mathbb{Z}$. I need to show that $\tau$ is almost surely finite. For this I am expected to use the central limit theorem.
To make life simpler, I define $\sigma = \inf\{n\geq 0: S_n = -c \text{ or } S_n = c\}$ where $c = -a \vee b$. $\tau \leq \sigma$ clearly. So I will show $\sigma$ is a.s. finite.
From the CLT, I have
$$\lim_{n\to\infty}P\left\{S_n \leq c\sqrt{2pn}\,\right\} = \Phi(c)$$ where $z\mapsto \Phi(z)$ is the distribution function of a standard normal r.v. I also have that
$$\{\sigma < \infty\} = \bigcup_{n=1}^\infty \{S_n = c\text{ or } S_n = -c \}$$
I am guessing I need to use the estimates coming from the CLT but there is a $\sqrt{n}$ in there, which confuses me. Can someone give a hint on how I should continue?
$$ \{\tau<\infty\}=\bigcup_{n}\{\tau\le n\} \text{ and } \{\tau\le n\}\subseteq\{\tau\le n+1\}; $$
$$ \therefore P\{\tau<\infty\}=\lim_{n\to\infty}P\{\tau\le n\}\ge \lim_{n\to\infty}P\{S_n\notin (a,b)\}=1. $$
To see that the last limit is indeed $1$ use the CLT. Let $c_{n,1}=(a+1)/\sqrt{2pn}, c_{n,2}=(b-1)/\sqrt{2pn}$ and note that $$ P\{S_n\notin (a,b)\}=1-P\{c_{n,1}\le S_n/\sqrt{2pn}\le c_{n,2}\}. $$
For any $\epsilon>0$, $\exists n_0$ s.t. $\epsilon\ge(-c_{n,1}\vee c_{n,2})$ for all $n\ge n_0$. Thus, $$ \limsup_{n\to\infty}P\{c_{n,1}\le S_n/\sqrt{2pn}\le c_{n,2}\}\le \Phi(\epsilon)-\Phi(-\epsilon). $$ Now send $\epsilon\downarrow 0$.