does anyone know a nice form of the infinite sum $\sum_{n=0, m=0}^{\infty} \frac{a^n b^m}{(n+m)!}$? I was wondering if anyone on here knows of a closed form or special function for this infinite sum:
$$\sum_{n=0, m=0}^{\infty} \frac{a^n b^m}{(n+m)!}$$
Or the sum of any non-trivial subset.
 A: Letting $p=n+m$ this is:
$$\sum_{p=0}^{\infty} \frac{1}{p!}\sum_{n=0}^{p} a^nb^{p-n}$$
But $$\sum_{n=0}^{p} a^nb^{p-n}=\frac{a^{p+1}-b^{p+1}}{a-b}$$
So your sum is:
$$\frac{1}{a-b}\sum_{p=0}^\infty \frac{a^{p+1}-b^{p+1}}{p!} = \frac{ae^a-be^b}{a-b}$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
\color{#f00}{\sum_{n = 0,m = 0}^{\infty}{a^{n}b^{m}\over \pars{n + m}!}} & =
\sum_{n = 0}^{\infty}a^{n}\sum_{m = 0}^{\infty}{b^{m}\over \pars{n + m}!} =
\sum_{n = 0}^{\infty}a^{n}b^{-n}\sum_{m = n}^{\infty}{b^{m}\over m!} =
\sum_{m = 0}^{\infty}{b^{m}\over m!}\sum_{n = 0}^{m}\pars{{a \over b}}^{n}
\\[3mm] & = 
\sum_{m = 0}^{\infty}{b^{m}\over m!}{\pars{a/b}^{m + 1} - 1 \over a/b - 1} =
{b \over a - b}\pars{{a \over b}\sum_{m = 0}^{\infty}{a^{m} \over m!} -
\sum_{m = 0}^{\infty}{b^{m} \over m!}}
\\[3mm] & =
{b \over a - b}\pars{{a \over b}\,\expo{a} - \expo{b}} =
\color{#f00}{{a\expo{a} - b\expo{b} \over a - b}}
\end{align} 
