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 $\frac{2}{3}e$

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
$$\begin{align}
\sum_{a=1}^\infty\sum_{b=1}^\infty \frac{ab}{(a+b)!}
&= \sum_{k=2}^\infty \frac{1}{k!}\sum_{a+b=k, a, b \ge 1}ab
= \frac16 \sum_{k=2}^\infty \frac{(k-1)k(k+1)}{k!}\\
&= \frac16 \sum_{k=2}^\infty \frac{k+1}{(k-2)!}
= \frac16 \sum_{k=0}^\infty \frac{k+3}{k!}
= \frac16 \left( \sum_{k=1}^\infty \frac{1}{(k-1)!} + \sum_{k=0}^\infty \frac{3}{k!}\right)\\
&= \frac16(e+3e) = \frac23 e
\end{align}
$$
