The basic idea is that if you win less than half the time your expected value goes to negative infinity linearly and your variance increases sublinearly ($O(\sqrt{n})$), so you almost surely are negative (go broke) eventually.
One approach is to estimate the property of every going one lower than where you started. Let's call this $z$. Note that there's a $q$ probability that it happens right away and a $p$ probability that you're two away. When you're two away, you first need to drop back down to one away (probability $z$ and then zero (another probability $z$), so that happens with probability $z^2$. Thus, we have that the probability of ever dropping down one below your initial position is $z=q+pz^2$. Solving for $z$ gives $0=(pz-q)(z-1)$, so either $z=1$ or $z=q/p$. When $q\geq p$, then you're guaranteed to hit 0, so $z=1$ and when $q<p$, then there's some chance of escaping, so $z=q/p$.
If you start at position $i$, then you need to drop down $i$ times, and each drop occurs with probability $z$ (otherwise you go off to infinity), so the chance you hit zero and go broke is $z^i=(q/p)^i$. Therefore, the chance of escaping is $1-(q/p)^i$.
Another approach is to let $X_n$ be the position after $n$ terms and $Y_n=(q/p)^{X_n}$. Taking the expected value of $Y_{n+1}$ gives that it's equal to $Y_n$, so $Y_n$ is effectively a fair game (the technical term is a martingale) whose future expectancy equals it's current value. When you're broke $Y_n$ is $(q/p)^0=1$ and when you escape to infinity $Y_n$ goes to 0, so your initial value of $(q/p)^i$ equals your expected winnings which evaluates to the probability that you go broke. If you want to make this rigorous, you have to be a little careful to show that you almost surely either go to infinity or go broke.
