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?