Proving the Double Sum of Exponentials Equals ae^a-e^a+1

[alyafey22]

Prove the following

\(\displaystyle \sum_{n=1}^\infty \sum_{m=1}^\infty\frac{a^{n+m}}{(n+m)!} = ae^a-e^a+1\)​

 

[polygamma]

Spoiler

$$ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{a^{n+m}}{(n+m)!} = \sum_{n=1}^{\infty} \sum_{m=n}^{\infty} \frac{a^{m+1}}{(m+1)!}$$

$$ = \sum_{m=1}^{\infty} \sum_{n=1}^{m} \frac{a^{m+1}}{(m+1)!} = \sum_{m=1}^{\infty} \frac{m a^{m+1}}{(m+1)!} = \sum_{m=2}^{\infty} \frac{(m-1) a^{m}}{m!}$$

$$ = \sum_{m=2}^{\infty} \frac{m a^{m}}{m!} - \sum_{m=2}^{\infty} \frac{a^{m}}{m!} = (ae^{a} - a) - (e^{a} - 1 - a) = ae^{a}-e^{a}+1$$