 Solution

For a finite upper barrier $L>i$, let $h_i^{(L)}$ be the probability of hitting zero before $L$. It satisfies $h_0^{(L)}=1$, $h_L^{(L)}=0$, and $h_i^{(L)}=p{h}_{i+1}^{(L)}+q{h}_{i-1}^{(L)}$. For $0<q<p$, solving the second-order recurrence with roots $1$ and $r=q/p$ givesAs $L\to \infty$, the events increase to eventual hitting of zero: any finite path reaching zero has a finite maximal height. Since $r<1$, this yields $\boxed{h_i=(q/p{)}^i}$, with $h_0=1$. In the endpoint case $q=0$, a positive initial state only moves upward and the hitting probability is zero, agreeing with the positive-state formula. The finite-barrier argument selects the correct solution rather than the also-harmonic constant one.