How to show that these two numbers are equal? i.e.
$$\sum_{n=0}^\infty\binom{2n}{n}p^n q^n = \frac{1}{1-2q}$$
This is how this equation is derived:
Suppose we have a simple random walk $\{R_n, n\geq 0\}$ with positive drift, that is $p+q=1$ and $p>0.5$. Here we aim to calculate $f_0$, the probability of ever returning to state 0 given that the walk started at state 0.
Which is $f_0 = P[T_{0,0} < \infty]$, where $T_{0,0}$ is the first time returning to 0 when starting at 0.
First we can show that $\frac{1}{1-f_0} = \sum_{n=0}^\infty\binom{2n}{n}p^n q^n$ by using the geometric random variable:
Let $N_0 = \sum_{i=0}^\infty \mathbb{1}(x_i = 0) $ be the total number of visiting state 0. Then $N_0$ follows a geometric distribution with parameter $1-f_0$.
Take the expectation of geom r.v. we have $E[N_0] = \frac{1}{1-f_0}$.
On the other hand, the walk can only return to zero point in $i=2k$ steps, hence $E[N_0] = \sum_{i=0}^\infty E[ \mathbb{1}(x_i = 0)] = \sum_{n=0}^\infty\binom{2n}{n}p^n q^n = \frac{1}{1-f_0}$
Secondly, we can show $f_0 = 2q$ by using the one-step ahead method:
Suppose $f_{i,j}$ is the probability of returning ever returning to $j$ given that the walk starts at $i$. By conditioning on the first step, we have $f_0 = q f_{-1,0} + p f_{1,0}$.
By the positive drift, we have $f_{-1,0} = 1$ and $f_{1,0} = q/p$, this yields $f_0 = 2q$
