= Solution
First assume $0<q<p<1$. Stop the nearest-neighbour walk on reaching zero or a finite upper level $L>i$. Its probability $h_i^{(L)}$ of reaching zero first solves $h_i^{(L)}=ph_{i+1}^{(L)}+qh_{i-1}^{(L)}$, with boundary values one and zero. Solving this second-order recurrence gives
$$
h_i^{(L)}=\frac{(q/p)^i-(q/p)^L}{1-(q/p)^L}.
$$
Absorption in the finite interval occurs almost surely: from every interior state there is a uniformly positive probability of an absorbing run within $L$ steps. As $L\to\infty$, the events of hitting zero before $L$ increase to the event of ever hitting zero, since a path reaching zero at a finite time has a finite maximum before then. Consequently
$$
\boxed{h_i=(q/p)^i}.
$$
This limiting argument selects the actual <hitting probability>, rather than just an arbitrary bounded solution of the infinite recurrence. If $q=0$, the walk moves deterministically upwards from positive states and the probability is zero there; from state zero it is one. These agree with the formula interpreted separately at $i=0$.
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