I'm reading Durrett's book on Probability and in the example of the asymmetric 1D random walk with $P(X_1=1)=p>1/2$, when trying to compute the expectation of the hitting time $T_{b}:=\inf\{n: S_n=b\}$ using optional stopping, in order to use the dominated convergence theorem to prove that $E[S_{T_b\wedge n}]\to E[S_{T_b}]$ he makes the following statement: $$E[\min_{n} S_n]>-\infty.$$
First of all, I think the $\min$ should be replaced by $\inf$ and intuitively I agree with the result, but I can't seem to be able to prove it.
I know that my random walk $S_n$ converges almost surely to $\infty$ by the law of large numbers and the fact that $p>1/2$ and I was thinking that this should imply that there are finitely many negative $S_n$, but this finite number is $\omega$-dependent (i.e. random). My idea was then to say that $$E[\inf_{n}S_n]\geq \sum_{k\in\mathbb{N}} (-k)P(\inf_{n}S_n=-k)$$ and then to use the fact the the above probabilities should be zero for all but finitely many $k$, but I don't know how to get rid of the $\omega$-dependance.
Any hints or (other) ideas?
Let $q$ be the probability that starting from $0$ the walk ever reaches $-1$. Then the probability $a_n$ that starting from $0$ the walk ever reaches $-n$ satisfies
$$ a_{n+1}=qa_n\;, $$
since the walk reaches $-(n+1)$ if and only if it first reaches $-n$ and then with probability $q$ reaches $-(n+1)$. Thus $a_n=q^n$. We have $q\ne1$, so the desired expectation can be computed as a finite expression in terms of $q$.