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

I'm trying to compute:

$$ \sum_{m=0}^{\infty} \sum_{n=0}^{\infty}\frac{1}{\Gamma \left(\frac{m+n}{2}\right)\Gamma \left(\frac{1+m+n}{2} \right)}$$

(From CMJ)

Using the duplication formula:

$$ \Gamma(x)\Gamma \left(x+\frac{1}{2} \right)=\frac{\sqrt{\pi}}{2^{2x-1}}\Gamma(2x)$$

$$ \frac{1}{\Gamma \left(\frac{m+n}{2}\right)\Gamma \left(\frac{1+m+n}{2} \right)}=\frac{1}{\sqrt{\pi}}\frac{2^{m+n-1}}{\Gamma(m+n)}=\frac{1}{\sqrt{\pi}}\frac{2^{m+n-1}}{(m+n-1)!}$$


$$ \sum_{m=0}^{\infty}\frac{1}{\Gamma \left(\frac{m+n}{2}\right)\Gamma \left(\frac{1+m+n}{2} \right)}=\frac{1}{\sqrt{\pi}}\sum_{m=0}^{\infty}\frac{2^{m+n-1}}{(m+n-1)!}=\frac{1}{\sqrt{\pi}}\sum_{m=n-1}^{\infty} \frac{2^m}{m!}$$

$$ \frac{1}{\sqrt{\pi}}\sum_{m=n-1}^{\infty} \frac{2^m}{m!} \sim_{n\rightarrow\infty} \frac{1}{\sqrt{\pi}} \frac{2^{n-1}}{(n-1)!}$$

The series $$ \sum_{n\geq1} \frac{1}{\sqrt{\pi}} \frac{2^{n-1}}{(n-1)!}$$ is convergent so:

$$ \sum_{m=0}^{\infty} \sum_{n=0}^{\infty}\frac{1}{\Gamma \left(\frac{m+n}{2}\right)\Gamma \left(\frac{1+m+n}{2} \right)}=\frac{1}{\sqrt{\pi}} \sum_{n=1}^{\infty} \sum_{m=n-1}^{\infty} \frac{2^m}{m!}$$

Is there a simple way to compute this quantity?

share|cite|improve this question
up vote 10 down vote accepted

Rearranging the terms and putting those with $k = m+n$ together (and leaving out the one where $k=0$), we get

$$\sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{1}{\Gamma(\frac{m+n}{2})\Gamma(\frac{1+m+n}{2})} = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{2^{m+n-1}}{\Gamma(m+n)\sqrt{\pi}}$$ $$= \sum_{k=1}^{\infty} \frac{(k+1)2^{k-1}}{(k-1)!\sqrt{\pi}} = \sum_{k=1}^{\infty} \frac{(k-1)2^{k-1}}{(k-1)!\sqrt{\pi}} + 2\sum_{k=1}^{\infty} \frac{2^{k-1}}{(k-1)!\sqrt{\pi}}$$$$= \frac{2}{\sqrt{\pi}}\sum_{k=2}^{\infty} \frac{2^{k-2}}{(k-2)!} + \frac{2}{\sqrt{\pi}}e^2=\frac{4e^2}{\sqrt{\pi}}.$$

share|cite|improve this answer
So the factor $(k+1)$ in $\sum_{k=1}^{\infty} \frac{(k+1)2^{k-1}}{(k-1)!\sqrt{\pi}}$ comes from the fact that there are $k+1$ $(m,n)$ such that $m+n=k$? – Chon Nov 21 '12 at 18:23
@Chon Yes. ${}$ – Cocopuffs Nov 21 '12 at 18:29
OK, thank you very much for your answer Cocopuffs! – Chon Nov 21 '12 at 19:14

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.