How do I evaluate $\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\frac{ab}{(a+b)!}$ 
$$\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\frac{ab}{(a+b)!}$$

I'm not really comfortable with more than 1 sigma's and that's why this question is confusing me. I don't think it's possible to reduce the number of variables to 1 here.
The answer is $\frac{2}{3}e$
 A: Since all terms in the sum are non-negative, one can arbitrary change the order of summation without changing the value of the sum. In particular, we can group terms with same value of $k = a + b$ and sum over them first. Notice
$$\sum_{a+b=k, a, b \ge 1}ab
= \sum_{a=1}^{k-1} a(k-a)
= k \frac{(k-1)k}{2} - \frac{(k-1)k(2k-1)}{6}
= \frac{(k-1)k(k+1)}{6}
$$
We have
$$\begin{align}
\sum_{a=1}^\infty\sum_{b=1}^\infty \frac{ab}{(a+b)!}
&= \sum_{k=2}^\infty \frac{1}{k!}\sum_{a+b=k, a, b \ge 1}ab
= \frac16 \sum_{k=2}^\infty \frac{(k-1)k(k+1)}{k!}\\
&= \frac16 \sum_{k=2}^\infty \frac{k+1}{(k-2)!}
= \frac16 \sum_{k=0}^\infty \frac{k+3}{k!}
= \frac16 \left( \sum_{k=1}^\infty \frac{1}{(k-1)!} + \sum_{k=0}^\infty \frac{3}{k!}\right)\\
&= \frac16(e+3e) = \frac23 e
\end{align}
$$
A: Let $a+b=n$. 
$$\begin{align}
\sum_{a=1}^\infty\sum_{b=1}^\infty\frac {ab}{(a+b)!}
&=\sum_{n=2}^\infty\sum_{a=1}^{n-1}\frac {a(n-a)}{n!}\\
&=\sum_{n=2}^\infty\frac n{n!}\sum_{a=1}^{n-1}a-\sum_{n=2}^\infty\frac 1{n!}\sum_{a=1}^{n-1}a^2\\
&=\sum_{n=2}^\infty \frac 1{(n-1)!}\frac {n(n-1)}2-\sum_{n=2}^\infty \frac 1{n!}\frac 16 (n-1)n(2n-1)\\
&=\frac 16 \sum_{n=2}^\infty \frac 1{(n-2)!}\left[3n-(2n-1)\right]\\
&=\frac 16\sum_{n=2}^\infty \frac {n+1}{(n-2)!}\\
&=\frac 16\sum_{n=2}^\infty \frac {n-2+3}{(n-2)!}\\
&=\frac 16\sum_{n=2}^\infty\frac 1{(n-3)!}+\frac 3{(n-2)!}\\
&=\frac 16(e+3e)\\
&=\color{red}{\frac 23e}\end{align}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{a = 1}^{\infty}\sum_{b = 1}^{\infty}{ab \over \pars{a + b}!} & =
\sum_{a = 1}^{\infty}a\sum_{b = 1}^{\infty}b
\sum_{n = 2}^{\infty}{\bracks{z^{n}}z^{a + b} \over n!} =
\sum_{n = 2}^{\infty}{1 \over n!}\bracks{z^{n}}
\pars{\sum_{a = 1}^{\infty}a\, z^{a}}^{2}
\\[5mm] & =
\sum_{n = 2}^{\infty}{1 \over n!}\bracks{z^{n}}
\bracks{z \over \pars{1 - z}^{2}}^{2} =
\sum_{n = 2}^{\infty}{1 \over n!}\bracks{z^{n - 2}}\pars{1 - z}^{-4}
\\[5mm] & =
\sum_{n = 2}^{\infty}{1 \over n!}{-4 \choose n - 2}\pars{-1}^{n - 2} =
\sum_{n = 2}^{\infty}{1 \over n!}{n + 1 \choose n - 2} =
{1 \over 6}\
\overbrace{\sum_{n = 2}^{\infty}{\pars{n + 1}n\pars{n - 1} \over n!}}
^{\ds{4\expo{}}}
\\[5mm] & = \bbx{{2 \over 3}\expo{}}
\end{align}
