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 $S(n)$?
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 $S(n)$?
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 $\Gamma(1/2) = \sqrt{\pi}$
According to Maple closed form of $S(n)$ is :
$\frac{2^n}{n}$