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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 $S(n)$?

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I think that $n$ has to be greater than $0$ –  pedja Apr 23 '12 at 6:04

2 Answers 2

up vote 4 down vote accepted

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 $\Gamma(1/2) = \sqrt{\pi}$

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According to Maple closed form of $S(n)$ is :

$$\frac{2^n}{n}$$

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What if $n=0$? Can you show that $\sum_{k=0}^{\infty}\frac{\binom{2k}{k}}{2k}\frac{1}{2^{2k}}$ goes to $\infty$? –  draks ... Apr 23 '12 at 8:48

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