Summing a binomial series Consider the following  sum: 
$$S(n)=\sum_{k=0}^{\infty}\frac{\binom{2k+n}{k}}{2k+n}\frac{1}{2^{2k}};n=0,1,2,3,...$$ Is there a closed form for $S(n)$? 
 A: The sum at hand is a hypergeometric series. Let
$$
   c_k  = \frac{1}{n+2k} \binom{n+2k}{k} \frac{1}{2^{2k}} = \frac{(n-1+2k)!}{k! (n+k)!} \frac{1}{4^k}
$$
Indeed, the hypergeometric certificate is:
$$
   \frac{c_{k+1}}{c_k} = \frac{1}{4} \frac{(n+2k)(n+2k+1)}{(n+1+k) (k+1)}
$$ 
Meaning that
$$
   \sum_{k=0}^\infty c_k = c_0 \sum_{k=0}^\infty \frac{\left(n/2\right)_k \left(n/2+1/2\right)_k}{(n+1)_k} \frac{1}{k!} = \frac{1}{n} \cdot {}_2 F_1 \left(\frac{n}{2}, \frac{n+1}{2} ; n+1 ; 1 \right)
$$
where $(a)_k$ denotes Pochhammer symbol.
Using Gauss's theorem, applicable for $\Re(c-a-b)>0$
$$
   {}_2 F_{1} \left(a,b; c; 1\right) = \frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}  
$$
we have
$$
  \sum_{k=0}^\infty c_k = \frac{1}{n} \frac{ \Gamma(n+1) \Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{n}{2} + 1 \right) \Gamma\left(\frac{n+1}{2} \right)  } \stackrel{\text{duplication}}{=} \frac{1}{n} \frac{ 2^{n} \pi^{-1/2} \Gamma\left(\frac{n+1}{2}\right) \Gamma\left(\frac{n}{2}\right) \Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{n}{2} \right) \Gamma\left(\frac{n+1}{2} \right)  } = \frac{2^n}{n}
$$
Since $\Gamma(1/2) = \sqrt{\pi}$
A: According to Maple closed form of $S(n)$ is :
$$\frac{2^n}{n}$$
