Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $S(n)$?

share|cite|improve this question
I think that $n$ has to be greater than $0$ – pedja Apr 23 '12 at 6:04
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 $\Gamma(1/2) = \sqrt{\pi}$

share|cite|improve this answer

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


share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.