Consider the following stochastic process, $$b(i+1) = b(i) + \xi_i (b_i),$$ where $\xi_i(b_i) \in \{-1, k \}$ are the independent increments having the following distribution: $$\begin{align} P (\xi = k) &= c / b(i),\\ P (\xi = -1) &= 1 - P (\xi = k), \end{align}$$ where $k$ is some positive integer and $c$ a parameter $0<c<1$. If $b(i) = 0$ for some $i$, then the process stops. Take a positive initial position $b(0)$. Is the probability that the process will never stop equal to zero? How to prove it?
Observe that there exists a position $b^*=b^*(c,k)$ such that if $b>b^*$ (resp. $b<b^*$) it will be more likely for the increment to be negative (resp. positive). This means that the process tends to be close to this value. This position is simply the solution of $k/b^{*} = 1 - 1/{b^{*}}$.
As you noted, away from zero the bias is uniformly negative. This suffices to guarantee recurrence, that is, that the process hits zero with full probability.
In a nutshell, choose $N\gt2(k+1)c$, and replace the dynamics on $b\geqslant N$ by $b\to b+k$ with probability $c/N$ and $b\to b-1$ with probability $1-c/N$. Starting from any $b(0)\geqslant N$, this modified process performs a homogenous random walk with steps of expectation $k\cdot c/N-1\cdot(1-c/N)\leqslant-1/2$ hence it hits $N$ almost surely. The modified transition probabilities put more weight on the positive steps and less on the negative steps hence, by coupling, the modified paths stay above those of the original process, as long as one stays above $N$. In particular, starting from every $b(0)\geqslant N$, the original process hits $N$ almost surely, that is, $N$ is recurrent.
Recurrence being a class property and the process being irreducible, this shows that the original process hits the origin with full probability.