How to calculate the sum of the following series? I would like to find the sum of the following series:
$$\sum_{n=0}^{\infty} \binom{2n+1}{n+1} (p(1-p))^n$$
where  $0<p<\frac{1}{2}$. I put it on Wolfram and it said that this series converges to $$\frac{\sqrt{(2p-1)^2}-1}{2p(p-1)\sqrt{(2p-1)^2}}$$
I initially thought it was a Geometric progression, but turns out its not.
I am guessing it could work with binomial series, where we have $\sum_{k=0}^{\infty}\binom{\alpha}{k}x^k=(1+x)^{\alpha}$. But here, my $\alpha$ depends on $k$, so I'm slightly stuck. I am aware that finding the sum of a series is hard and am unsure of how else to go about finding the sum of this series. Any help would be appreciated!
 A: Let your sum be $s(p)$.
Consider a random walk starting at 0 and moving right with probability $p$ and left with probability $1-p$. Then $ps(p)$ is the expected time the walk spends at $1$.
If you never reach $1$ then the time at $1$ is zero. Thus you need to reach $1$, the probability of which is $\frac{p}{(1-p)}$ (see Theorem 1, page 4). Then after that, the walk "restarts" and the expected time spent at $1$ is the same as the expected time spent at $0$ for the original walk. This expected time spent is $\frac{1}{(1-2p)}$ : the probability of ever returning to $0$ is $2p$ (see Theorem 3, page 5), so the expected time spent at $0$ is $1+(2p)+(2p)^2+...=\frac{1}{1-2p}$
(Problem 3, page 6).
In summary, the expected time at $1$ is $\frac{p}{(1-p)(1-2p)}$ and $s(p)=\frac{1}{(1-p)(1-2p)}$
A: Here is an alternate proof of the binomial identity by Markus Scheuer. We
seek to show that
$$\sum_{q=0}^n {q\choose n-q} (-1)^{n-q} {2q+1\choose q+1}
= 2^{n+1}-1.$$
The LHS is
$$\frac{(-1)^n}{2\pi i} \int_{|z|=\varepsilon}
\frac{1}{z^{n+1}}
\frac{1}{2\pi i} \int_{|w|=\gamma}
\sum_{q=0}^n (-1)^q z^q (1+z)^q
\frac{(1+w)^{2q+1}}{w^{q+2}} \; dw \; dz.$$
There is no contribution when $q\gt n$ and we may extend $q$ to
infinity:
$$\frac{(-1)^n}{2\pi i} \int_{|z|=\varepsilon}
\frac{1}{z^{n+1}}
\frac{1}{2\pi i} \int_{|w|=\gamma}
\frac{1+w}{w^2}
\frac{1}{1+z(1+z)(1+w)^2/w} \; dw \; dz
\\ = \frac{(-1)^n}{2\pi i} \int_{|z|=\varepsilon}
\frac{1}{z^{n+1}}
\frac{1}{2\pi i} \int_{|w|=\gamma}
\frac{1+w}{w}
\frac{1}{(1+z+wz)(z+(1+z)w)} \; dw \; dz.$$
Now we determine $\varepsilon$ and $\gamma$ so that the geometric
series converges and the pole at $w=-z/(1+z)$ is inside $|w|=\gamma$
while the pole at $w=-(1+z)/z$ is not. For the series we require
$|z(1+z)(1+w)^2/w| \lt 1.$ With $|z(1+z)| \le \varepsilon
(1+\varepsilon)$ and $|w/(1+w)^2| \ge \gamma/(1+\gamma)^2$ we need
$\varepsilon(1+\varepsilon) \lt \gamma/(1+\gamma)^2.$ Observe that
on $[0,1]$ we have $\gamma/(1+\gamma)^2 \ge \gamma/4$ since $4\ge
(1+\gamma)^2.$ For $\gamma/4 \gt \varepsilon(1+\varepsilon)$ we choose
$\gamma=8\varepsilon$ with $\varepsilon \ll 1$ and we have our pair.
Now for the pole at $-z/(1+z)$ we need for the maximum norm
$\varepsilon/(1-\varepsilon) \lt \gamma = 8\varepsilon$ which holds
with $\varepsilon \lt 7/8$ which we will enforce. The second pole under
consideration is $-(1+z)/z = -1 - 1/z.$ The closest this comes to the
origin is $-1+\varepsilon = -1 + \gamma/8.$ To see that this is outside
$|w|=\gamma$ we need $-1+\gamma/8 \lt -\gamma$ or $\gamma\lt 8/9.$
This means we instantiate $\varepsilon$ to $\varepsilon \lt 1/9$, which
completes the discussion of the contour. 
Now residues sum to zero and the residue at infinity in $w$ is zero by
inspection which means that the inner integral is minus the residue at
$w= -(1+z)/z$, as it is equal to the sum of the residues at zero and at
$w=-z/(1+z)$. We write
$$- \frac{1}{z} \frac{1}{2\pi i} \int_{|w|=\gamma}
\frac{1+w}{w}
\frac{1}{((1+z)/z+w)(z+(1+z)w)} \; dw.$$
We get from this being a simple pole the contribution
(here $(1+w)/w = 1/(1+z) $)
$$- \frac{1}{z} \frac{1}{1+z}
\frac{1}{z-(1+z)^2/z}
= \frac{1}{1+z} \frac{1}{1+2z}$$
which combined with the integral in $z$ gives
$$(-1)^n [z^n] \frac{1}{1+z} \frac{1}{1+2z}
= (-1)^n \sum_{q=0}^n (-1)^q 2^q (-1)^{n-q}
= \sum_{q=0}^n 2^q.$$
This is indeed
$$\bbox[5px,border:2px solid #00A000]{
2^{n+1}-1}$$
as claimed.
