Let $X(i)$ be a random walk where starting at $i > 0$, at each step it increases by a positive integer $c > 1$ or goes down by $1$, with equal probability. Let $p_i$ be the probability that it goes to $0$. What I would like to know is how can I show that $\lim_{i \to \infty} p_i = 0$?
I have been explained this holds because it is a random walk with mean $ > 0$, and I can understand it intuitively, but I am looking for a more rigorous explanation. Thank you
For some $i$,
\begin{align*} p_i &= \mathbb{P}[\text{we arrive at $0$}] \\ &= \sum_{j=0}^\infty \mathbb{P}[\text{we arrive at $0$} \, | \, \text{we go forward a total of $j$ times}] \\ &\leq \sum_{j=0}^\infty \binom{j(c+1)+i}{j}\bigg(\frac{1}{2}\bigg)^{j(c+1)+i} \\ &= \bigg(\frac{1}{2}\bigg)^{i} \sum_{j=0}^\infty \binom{j(c+1)+i}{j}\bigg(\frac{1}{2}\bigg)^{j(c+1)} \\ & \leq \bigg(\frac{1}{2}\bigg)^{i} \sum_{j=0}^\infty \bigg(\frac{c+1}{c}\bigg)^{i} \binom{j(c+1)}{j}\bigg(\frac{1}{2}\bigg)^{j(c+1)} \\ & = \bigg(\frac{c+1}{2c}\bigg)^{i} \sum_{j=0}^\infty \binom{j(c+1)}{j}\bigg(\frac{1}{2}\bigg)^{j(c+1)}, \end{align*} which tends to $0$, as $i \to \infty$.