Let $(X_i)$ be iid. random variables with:
$$\mathbb{P}(X_i = 1) = p > (1-p) = \mathbb{P}(X_i = -1)$$
and set $S_n = X_1 + \cdots X_n$. We will write $T_k$ for $\inf \{n : S_n = k\}$ i.e. the expected hitting time of $k$.
My aim is to prove that $\mathbb{E}[T_1]$ is finite, and I would like to know if my reasoning is accurate. I've only written in detail the parts I'm unsure about, and I state the queries I have at the end of the post.
Proof
Let $U_k$ be the hitting time given by $T_{-k} \wedge T_1$. Then, it is fairly simple to show that $\mathbb{E}[U_k] \le (p-q)^{-1}$.
It is then tempting to apply monotone convergence, by saying that $U_k \uparrow T_1 \implies \mathbb{E}[U_k] \to \mathbb{E}[T_1]$. However, I do not think that it is immediate that $U_k \uparrow T_1$.
Therefore, instead, let $U$ be the limit of the $U_k$ for now. We wish to prove that $U = T$ a.s..
It is clear that $U \le T$. Further note that:
$$\{ \omega : U(\omega) < T(\omega)\} = \{ \omega : (S_n(\omega)) \text{ hits arbitrarily large negative values before hitting } 1\}$$
But we clearly have:
$$\{ \omega : (S_n(\omega)) \text{ hits ... before hitting } 1\} \subseteq \{ \omega : (S_n(\omega)) \text{ hits arbitrarily large negative values}\}$$
so, we may relax the equality to obtain:
$$\mathbb{P}(U < T) \le \mathbb{P}((S_n) \text{ hits arbitrarily large negative values})$$
Now let $m < 0$, and note that:
$$\mathbb{P}(\exists n: S_n = m) = \mathbb{P}(\exists n: S_{n} = m \;| \; \exists n': S_{n'} = m+1)\cdot \mathbb{P}(\exists n': S_{n'} = m+1) $$
But then, by translation invariance and the Markov property:
$$\mathbb{P}(\sim\exists n: S_n = m \;| \exists n': S_{n'} = m+1) = \mathbb{P}(\sim\exists n: S_n = -1 \;| \exists n': S_{n'} = 0) $$
$$ = \mathbb{P}(\sim\exists n: S_n = -1)$$
$$ = \mathbb{P}(S_n \ge 0 \; \forall n)$$ $$>0$$
by the answer given here.
$$\implies \mathbb{P}(\exists n: S_{n} = m \;| \; \exists n': S_{n'} = m+1) < 1$$
In particular, we deduce that $\mathbb{P}(\exists n: S_n = m) \to 0$ as $m \to \infty$, so by the downwards continuity of the probability measure:
$$\mathbb{P}((S_n) \text{ hits arbitrarily large negative values}) = 0$$
$$\implies \mathbb{P}(U < T) =0$$
Thus, $U = T$ a.s., and the result follows by the monotone convergence theorem.
Question: Am I right to believe that it is not trivial that $U_k \uparrow T$?