 Solution

Let $r=q/p>1$ and $H_k$ be the first hitting time of level $k$. Nearest-neighbour steps imply $S_{H_k}=k$ whenever $H_k<\infty$: the walk cannot overshoot its first upward crossing. On this event the stopped martingale eventually has the constant value $r^k$. On $\{H_k=\infty \}$, it is $Z_n$ and tends to zero by part (i). ThereforeFor every $n$, $0\le Z_{n\wedge H_k}\le r^k$. Before the hit the walk is below $k$, and at the hit it equals $k$. The optional stopping theorem for the bounded stopping time $n\wedge H_k$ gives $\mathbb{E}{Z}_{n\wedge H_k}=\mathbb{E}{Z}_0=1$. The dominated convergence theorem now yields $1=r^k\mathbb{P}(H_k<\infty )$. ConsequentlyThe uniform bound on the stopped variables justifies the limit; applying optional stopping directly at the possibly infinite $H_k$ without this argument would not be valid.