What does $\sum_{k=0}^{\infty}\binom{2k+1}{k+1}p^{k+1}q^{k}$ converge to when $p+q=1$ and $0 \leq p,q \leq 1$? What does $\sum_{k=0}^{\infty}\binom{2k+1}{k+1}p^{k+1}q^{k}$ converge to when $p+q=1$ and $0 \leq p,q \leq 1$?
I'm trying to solve a problem involving the expected number of steps to reach an integer on an asymmetric random walk on the integers, but I can't figure out what the above converges to. Putting it into wolfram alpha and trying out different values for p and q not only shows that indeed converges so long as p isn't 0.5, but that there's some strong pattern in how it converges. Can't quite figure it out though.
 A: For $0 < p < 1$ and $p \ne 1/2$, let $$S(p) = \sum_{k=0}^\infty \binom{2k+1}{k+1} p^{k+1} (1-p)^k.$$  Then $$S(p) = \frac{1}{2(1-p)} \left( \frac{1}{|2p-1|} - 1 \right).$$
Evaluation of this sum is a bit tricky, but not too bad.  First, write
$$\begin{align*}
S(p) &= p \sum_{k=0}^\infty \binom{2k+1}{k+1} (p(1-p))^k \\
&= p \sum_{k=0}^\infty \left( 2 - \frac{1}{k+1} \right) \binom{2k}{k} (p(1-p))^k.
\end{align*}$$
Now define the function $$f(z) = \sum_{k=0}^\infty \binom{2k}{k} z^k = \frac{1}{\sqrt{1-4z}},$$ which we can evaluate using a number of methods, such as the binomial series, or by generating functions, or even by combinatorial argument.  Then $$F(z) = \int_{t=0}^z f(t) \, dt = \sum_{k=0}^\infty \frac{1}{k+1} \binom{2k}{k} z^{k+1} = \frac{1 - \sqrt{1-4z}}{2};$$ consequently, $$\frac{F(z)}{z} = \sum_{k=0}^\infty \frac{1}{k+1} \binom{2k}{k} z^k = \frac{1 - \sqrt{1 - 4z}}{2z}.$$  Putting this together with $f(z)$ and setting $z = p(1-p)$ yields $$S(p) = p \left( 2f(p(1-p)) - \frac{F(p(1-p))}{p(1-p)} \right),$$ and the rest is simplification.  I leave the computational details to you as an exercise.
