Infinite sum with binomial coeffcient - Hypergeometric function ${}_2F_1$ and gauss theorem From an urn and balls problem, I end up with the need to compute the following sum
$$S = \sum_{n\geq 1} \frac{1}{n}\binom{2n}{n+1}2^{-2n}$$
Using Maple I discovered $S=1$. Starting with some basics transformations, I get
$$S= \sum_{n=1}^{\infty} \frac{2^{-n}(2n-1)(2n-3)\ldots 3}{(n+1)!}$$
Therefore I can write $S'$ using hypergeometric function taken at point $z=1$,
$$S+1 = {}_2F_1(\frac{1}{2},1,2,1)$$
Then using Gauss Hypergeometric theorem 
$$S+1 = {}_2F_1(\frac{1}{2},1,2,1) = \frac{\Gamma(2)\Gamma(1/2)}{\Gamma(1)\Gamma(3/2)}=\frac{\Gamma(1/2)}{\Gamma(3/2)}$$
And using $\Gamma(z+1)=z\Gamma(z)$ I get $S=1$.

My question relates to Gauss Hypergeometric theorem. I couldn't find an online proof, or an explanation. And I was wandering if I could find a more direct approach for ${}_2F_1(\frac{1}{2},1,2,1) = 2$ and directly for my sum $S$.
  Gauss Theorem is quite generic, and it feels kinda using a bazooka to kill a bird. With the specific value I have, there might be an easier approach.


The given series is telescopic. If we set $a_n=\frac{1}{4^n}\binom{2n}{n}$ we have
$$ \frac{a_{n+1}}{a_n} = \frac{(2n+2)(2n+1)}{4(n+1)^2} = \frac{2n+1}{2n+2}=1-\frac{1}{2n+2}$$
hence
$$ a_n-a_{n+1} = \frac{1}{2n+2}\binom{2n}{n}2^{-2n}= \frac{1}{2}\cdot\frac{1}{n}\binom{2n}{n+1}2^{-2n}$$ 
and since $a_n\to 0$ we have
$$ S=\sum_{n\geq 1}\frac{1}{n}\binom{2n}{n+1}2^{-2n}=2a_1 = \color{red}{1}.$$
As an alternative approach, by exploiting $a_n=\frac{2}{\pi}\int_{0}^{\pi/2}(\cos\theta)^{2n}\,d\theta $ we have
$$S=\frac{2}{\pi}\sum_{n\geq 1}\frac{1}{n+1}\int_{0}^{\pi/2}(\cos\theta)^{2n}\,d\theta=-\frac{2}{\pi}\int_{0}^{\pi/2}\frac{2\log\sin\theta+\cos^2\theta}{\cos^2\theta}\,d\theta$$
and it is not difficult to compute the last integral. 
