Let $p\in (0,1)$ be fixed, and let $q=1-p$. A frog performs a (discrete time) random walk on the 1-dimensional lattice $\mathbb{Z}$ the following way:
The initial position is $X_0=0$. The frog jumps $1$ step right with probability $p$ and jumps 1 step left with probaibility $q$ at each time step, independently of what happened before, until it reaches either the point $a=-10$ or the point $b=+30$, which are sticky: if the frog reaches one of them, it stays there forever.
Let $X_n$ denote the position of the frog after n steps (for $n=0,1,2,...$). $Y_n:=(\frac{q}{p})^{X_n}$.
Show that $Y_n$ converges almost surely to some limiting random variable $Y_{\infty}$. What are the possible values of $Y_{\infty}$?
My solution: We can first show that $X_n$ converges almost surely. It is obvious that $X_n$ will converge to $$X_{\infty}=\begin{cases} 30\\ -10 \end{cases}$$
I need to show $P(X_n=30)\rightarrow c_1$($c_1$ is some constant), and $P(X_n=-10)\rightarrow c_2$($c_2$ is some constant).
So let $a,b$ be the times of steps the frog jumps to right and left in the first n-th steps.
Suppose $n$ is large enough, to make $X_n=30$ we need $a-b\geqslant 30$, since $a+b=n$ so we get the inequality that $n-2b\geqslant 30$, so $b\leqslant \frac{n}{2}-15$. So $$P(X_n=30)=\sum_{i=0}^{[\frac{n}{2}]-15}\binom{n}i p^{n-i}q^i,$$ and
$$P(X_n=-10)=\sum_{i=[\frac{n}{2}]+5}^n \binom{n}i p^{n-i}q^i.$$
My question is: Is this method right and how to show that these two summations converge?