# Change double sum limits

I am trying to change a double sum limits but I am not sure if I am doing correctly. The sum is and my solution is

$\sum_{r=0}^n \sum_{m=0}^\infty \frac{a^m\sqrt{(m+r)!}}{m!}=\sum_{r=0}^{m'}\sum_{m'=0}^\infty \frac{a^{m'-r}\sqrt{m'!}}{(m'-r)!};\quad with\quad m'=m+r$

Currently, I am struggling to find the solution to an equation of which this double sum is part, and I think this double sum could be the problem. So I would be really grateful if some could help me to understand if my change of limits is correct. Thanks in advance.

• the sum is not defined for m=0 – gimusi May 22 '18 at 15:57
• Sorry, I changed that. A factorial was missing. – mors May 22 '18 at 16:00
• in the second sum $m'$ should start from $m'=r$ – gimusi May 22 '18 at 16:02
• why did you change $\sum_{r=0}^n$ with $\sum_{r=0}^{m'}$? – gimusi May 22 '18 at 16:03
• I did $\sum_{r=0}^n \sum_{m=0}^\infty \rightarrow \sum_{r=0}^n \sum_{m'=r}^\infty \rightarrow \sum_{r=0}^{m'} \sum_{m'=0}^\infty$. This because I want to get rid of the second sum througth a kronecker delta – mors May 22 '18 at 16:07

## 1 Answer

For $m'=m+r$ we have that

$$\sum_{r=0}^n \sum_{m=0}^\infty \frac{a^m\sqrt{(m+r)!}}{m!}=\sum_{r=0}^{n}\sum_{m'=r}^\infty \frac{a^{m'-r}\sqrt{m'!}}{(m'-r)!}$$