sequences and series - Computing $sum_{m=0}^{infty} sum_{n=0}^{infty}frac{1}{Gamma left(frac{m+n}{2}right)Gamma left(frac{1+m+n}{2} right)}$

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?

Answer

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

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