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.


One can recognize an instance of the binomial series: $$\frac{1}{n}\binom{2n}{n+1}2^{-2n}=2(-1)^n\binom{1/2}{n+1},$$ hence $S=\displaystyle\lim_{x\to1^-}S(x)$ where $S(x)=\color{red}{-}2\displaystyle\sum_{n\geq 1}\binom{1/2}{n+1}(-x)^{n\color{red}{+1}}=2(1-\sqrt{1-x})-x$.
