Showing that a stopping time is finite for a biased random walk. If we consider $X_i$ iid with $\mathbb{P}(X_i=1) = p$ and $\mathbb{P}(X_i=-1)=1-p$. Where $p \in (1/2,1)$. The random walk is then given by,
$$S_n=\sum_{i=1}^n X_i. $$
We also define the stopping time $\tau = \inf\{k:S_k \in \{-\alpha,\beta\} \}$ with $\alpha,\beta>0$ in the natural numbers.
How to prove that $P(\tau<\infty)=1$? Intuitively it's clear that for sure at some point $S_n$ will hit $\beta$ (because the random walk has a tendincy to move upwards.). I was thinking about showing that $P(S_n=\infty \text{ i.o.})=1$. But I'm not really sure how to make a rigorous argument.
Could someone help me with this?

Consider $\tau_\beta:=\inf\{n>0:S_n = \beta\}$. We construct the martingale $S_n-n(2p-1)$ which is s.t.
$$E[S_n-n(2p-1)|\mathcal{F}_k]=S_k-k(2p-1)$$
Now by optional stopping
$$E[S_{\tau_\beta \wedge n}-(\tau_\beta \wedge n)(2p-1)]=0$$
and by monotone convergence
$$E[\tau_\beta]=\frac{1}{2p-1}\lim_{n \to \infty}E[S_{\tau_\beta \wedge n}]\leq \frac{\beta}{2p-1}$$
So
$$P(\tau_\beta \geq k)\leq \frac{1}{k}E[\tau_\beta]\leq \frac{\beta}{k(2p-1)}\to 0\implies P(\tau_\beta = \infty )=0$$
and so $P(\tau=\infty)=0$.
