Sum of factorial fractions Find the sum $$\sum\limits_{a=0}^{\infty}\sum\limits_{b=0}^{\infty}\sum\limits_{c=0}^{\infty}\frac{1}{(a+b+c)!}$$ I tried making something like a geometric series but couldn't. Then I couldn't think of anything.
 A: We have:
$$S=\sum_{a,b,c\geq 0}\frac{1}{(a+b+c)!}=\sum_{n\geq 0}\frac{r(n)}{n!}$$
where:
$$ r(n)=\#\{(a,b,c)\in\mathbb{N}^3:a+b+c=n\}=[x^n]\frac{1}{(1-x)^3}=\frac{(n+2)(n+1)}{2} $$
leads to:
$$ S=\frac{1}{2}\left.\frac{d^2}{dz^2}\left(z^2 e^z\right)\right|_{z=1}=\left.\frac{z^2+4z+2}{2}e^z\right|_{z=1}=\color{red}{\frac{7e}{2}}. $$
A: Consider an integer $m$ where $a+b+c=m$ $\lbrace m,a,b,c\in \Bbb N_0\rbrace$.
The number of possible combinations of $a, b, c$ is given by
$$\begin{align}
N(m)&=\sum_{a=0}^{m}\sum_{b=0}^{m-a}1\\
&=\sum_{a=0}^{m}(m-a+1)\\
&=\sum_{a=1}^{m+1}a\quad \text{ or } {a\choose 1}\\
&={m+2\choose 2}
\end{align}$$
The above can also be derived using the "stars and bars" method to count the number of ways to partition a set with $m$ elements into $n$ groups (here $n=3$), which is given as 
$$N(m)=\binom {m+n-1}{n-1}\Biggr|_{n=3}=\binom {m+2}2$$
.
Hence
$$\begin{align}
\sum_{a=0}^{\infty}\sum_{b=0}^{\infty}\sum_{c=0}^{\infty}\frac 1{(a+b+c)!}
&=\sum_{m=0}^\infty \frac {N(m)}{m!}\\
&=\sum_{m=0}^\infty \frac {1}{m!}{m+2\choose 2}\\
&=\sum_{m=0}^\infty \frac {1}{m!}\frac{(m+2)(m+1)}2\\
&=\frac 12 \sum_{m=0}^\infty \frac {m^2+3m+2}{m!}\\
&=\frac 12 \sum_{m=0}^\infty \frac {m(m-1)+4m+2}{m!}\\
&=\frac 12 \sum_{m=0}^\infty \frac 1{(m-2)!}+\frac 4{(m-1)!}+\frac 2{m!}\\
&=\frac 72 \sum_{m=0}^\infty \frac 1{m!}\\
&=\frac 72e\qquad \blacksquare
\end{align}$$
