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!

• In the inner sum, there are $k-1$ copies of $\frac{1}{k!}$. So we want $\sum_{k=2}^\infty \frac{k-1}{k!}$. Nice cancellations. – André Nicolas Sep 9 '15 at 14:19
• So, youre saying that this should just equal 1 right – ToddNas Sep 9 '15 at 14:20
• Yes, it telescopes. – André Nicolas Sep 9 '15 at 14:24
• ahh that makes perfect sense, thank you. Any idea on the absolute convergence? – ToddNas Sep 9 '15 at 14:30
• It converges, of course absolutely, all the terms are positive. If you apply Ratio Test to $\sum \frac{k-1}{k!}$ it will tell you convergence, the ratio $a_{n+1}/a_n$ has limit $0$. But we don't need Ratio Test, telescoping argument shows convergence, it even gives an estimate of the "truncation error". – André Nicolas Sep 9 '15 at 14:47

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 $n\to\infty$.
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 $(m+n)\geq 2\sqrt{mn}$.