How to show that $\sum_{n,m = 1}^{\infty} \frac{1}{\left(n+m\right)!}$ converges absolutely I am working with the series

$$\sum_{n,m = 1}^{\infty} \frac{1}{\left(n+m\right)!}$$

and I would like to show that it converges absolutely. It is easy to see that it actually converges to $1$ as
$$\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{\left(n+m\right)!} = \sum_{k=2}^{\infty}\sum_{n=1}^{k-1}\frac{1}{k!} = \sum_{k=2}^{\infty}\left(\frac{1}{(k-1)!} - \frac{1}{k!} \right) = 1 $$
but does that consitute a formal proof? I thought I could instead show that the partial sums are bounded as in that case the nonnegativity would imply convergence but I have not been able to do that. Could you please give me some advice? Thank you.
 A: Sum along the values of $r=m+n$ As there are $r-1$ solutions to the equation $m+n=r$ with $m,n\ge 1$, we have:
$$\sum_{n,m = 1}^{\infty} \frac{1}{(n+m)!}=\sum_{r= 2}^{\infty} \frac{r-1}{r!}=\sum_{r= 2}^{\infty}\biggl( \frac{1}{(r-1)!}-\frac{1}{r!}\biggr)$$
which is a telescoping series which simplifies to its initial term $1$.
A: I used the fact that the number of ways to sum with $k$ natural addends a number $n\in\Bbb N$ without restriction is $\binom{n-1}{k-1}$[1].
Then we have that $$\sum_{m,n\ge 1} \frac1{(m+n)!}=\sum_{k\ge 2}\frac{\binom{k-1}{2-1}}{k!}=\sum_{k\ge 2}\frac{k-1}{k!}$$
From here it follow the same solution.
A: The given answers assert that this double series converges to the value obtained by summing diagonally:
$$\sum_{n,m \in \mathbb{N}} \frac{1}{(n+m)!} = \sum_{n =2}^\infty \sum_{j+k = n} \frac{1}{(j+k)!} = \sum_{n=2}^\infty\frac{n-1}{n!}= 1.$$
This is indeed true since the terms are non-negative.  In this case, the double series converges if and only if sums by rows or columns (iterated series), diagonal sums, or any other arrangement converge and all such sums will be equal. This fact is so well known that it is often taken for granted. However, this result is not generally true for conditionally convergent series where terms change sign.
For the formal proof you asked about, consider the general treatment of double series $\sum_{m,n} a_{mn}$ with $a_{mn} > 0$.
We define the partial sum
$$S_{mn}  = \sum_{j=1}^m \sum_{k=1}^n a_{jk},$$
and a diagonal sum
$$\sigma_n = \sum_{p=2}^n \sum_{j+k= p}a_{jk}.$$
We say that the double series converges to $S$ if for every $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that $m,n \geqslant N$ implies $|S_{mn} -S| < \epsilon.$
First, the double series converges to $S$ if and only if we can sum by squares:
$$S = \lim_{n \to \infty}S_{nn}.$$
The forward implication is easily proved by observing that with $m=n \geqslant N$ we have $|S_{nn} -S| < \epsilon.$
For the reverse implication, we see that if $S_n \to S$, then given $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that $n \geqslant N$ implies $S- \epsilon < S_{nn} \leqslant S.$ Thus, for any $m,n \geqslant N$ we have $S - \epsilon < S_{NN} \leqslant S_{mn} \leqslant S_{m+n,m+n} \leqslant S.$ 
Finally, using a squeezing argument, we can show that the diagonal sum converges, $\sigma_n \to S,$ if and only if $S_{nn} \to S$. This follows because for $m > 2n$ we have $S_{nn} \leqslant \sigma_m \leqslant S_{mm}$ and $\sigma_n \leqslant S_{mm} \leqslant  \sigma_{2m}.$
