Sum $\sum\limits_{n,m=1}^\infty \frac{1}{(n+m)!},$ I am looking at: 
$$\sum_{n,m=1}^\infty \dfrac{1}{(n+m)!},$$ 
my task is to show that it is absolutely convergent and to find its sum. 
I have found the sum doing the following: 
$$\sum_{m,n=1}^\infty \frac{1}{(m+n)!}=\sum_{k=2}^\infty\left(\sum_{m+n=k}\frac{1}{(m+n)!}\right)
= \sum_{k=2}^\infty\left(\sum_{m+n=k}\frac{1}{k!}\right) =\sum_{k=2}^\infty \left(\frac{1}{k!}\left(\sum_{m+n=k}1\right)\right)$$
from which it seems that the sum equals $e-1-\frac{1}{e}$. I believe that this is correct, however when I try the ratio test I keep getting failed convergence. I am not sure how to approach the absolute convergence of this problem. Any hints would be great! 
 A: You had done all the work. We have $\sum_{m+n=k} 1=k-1$.  So our sum is equal to
$$\sum_{k=2}^\infty\frac{k-1}{k!}.\tag{1}$$
This series can be rewritten as 
$$\sum_{k=2}^\infty \left(\frac{1}{(k-1)!}-\frac{1}{k!}\right),$$
which telescopes with sum $1$.
Remark: The convergence of (1) has already been shown implicitly by the telescoping argument. If we wish we could Ratio Test. If $a_n$ is the $n$-th term, we have
$$\frac{a_{n+1}}{a_n}=\frac{n/(n+1)!}{(n-1)/n!}=\frac{n}{n-1}\cdot \frac{n!}{(n+1)!}=\frac{n}{(n-1)(n+1)}.$$
This has limit $0$ as $n\to\infty$.  
A: Provided that both $m$ and $n$ are big enough, $(m+n)!\geq (mn)^2$ holds$^{(*)}$, hence absolute convergence is granted and we may rearrange the double sum as we like. For instance:
$$ \sum_{m,n=1}^{+\infty}\frac{1}{(m+n)!}=\sum_{s=2}^{+\infty}\sum_{n=1}^{s-1}\frac{1}{s!}=\sum_{s\geq 2}\frac{s-1}{s!}=\sum_{s\geq 1}\frac{1}{s!}-\sum_{s\geq 2}\frac{1}{s!}=\color{red}{\huge{1}},$$
sic et simpliciter.
(*) It is enough to combine the inequality $a!\geq a^4$ with the AM-GM inequality $(m+n)\geq 2\sqrt{mn}$.
