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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)!}$

So:

$ \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?

1 Answers 1

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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}}.$

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    OK, thank you very much for your answer Cocopuffs!2012-11-21