Show that the series $\sum_{n,m=1}^\infty 1/(n+m)!$ is absolutely convergent and find its sum

Question

Show that the series $$\sum_{n,m=1}^\infty \dfrac{1}{(n+m)!}$$ is absolutely convergent and find its sum.

This comes from a chapter called interchange of limit operations. I tried using the ratio test but wasn't sure if this was the correct route.

Answer 1

Notice that: $$\sum_{m,n=1}^\infty\frac{1}{(m+n)!}=\sum_{k=2}^\infty\sum_{\array{m+n=k\\m,n\geq 1}}\frac{1}{k!} = \sum_{k=2}^\infty\frac{k-1}{k!}\leq\sum_{k=2}^\infty \frac{k(k-1)}{k!}=\sum_{k=2}^\infty \frac{1}{(k-2)!}<\infty$$

Answer 2

While this doesn't address the issue of absolute convergence, it's easy to find the sum's value once you've justified the operations. Consider introducing an auxiliary variable $k$ for $m+n$, and rewrite the double-infinite sum as a sum of 'diagonal sums' : $$\sum_{m,n=1}^\infty \frac{1}{(m+n)!}=\sum_{k=2}^\infty\left(\sum_{m+n=k}\frac{1}{(m+n)!}\right) = \sum_{k=2}^\infty\left(\sum_{m+n=k}\frac{1}{k!}\right) =\sum_{k=2}^\infty \left(\frac{1}{k!}\left(\sum_{m+n=k}1\right)\right)$$

Now, the value of the inner sum should be easy to find; then use that to simplify the expression. You should find a familiar value for the outer sum...

Answer 3

Consider the top right quadrant of a graph with axes of $m$ and $n$. With each integer point not on the axes, give it a value of $\frac{1}{(m+n)!}$. Now draw rings around points with the same value, and we realise the original sum is equal to $\sum_{r=1}^\infty \frac{r}{(r+1)!}$. Define $S_n=\sum_{r=1}^n \frac{r}{(r+1)!}$ and calculating the first few $S_n$, we can conjecture that: $S_n=1-\frac{1}{(n+1)!}$, which is very easy to prove by induction. Letting $n$ tend to infinity we obtain the original sum, which is equal to 1.

Answer 4

$\displaystyle{\large% {\cal Q}\left(z\right) \equiv \sum_{n, m = 1}^{\infty}{z^{m + n} \over \left(m + n\right)!}\,, \qquad {\cal Q}\left(1\right)\ =\ ?}$

\begin{align} {\cal Q}'\left(z\right) &= \sum_{n, m = 1}^{\infty}{z^{m + n - 1} \over \left(m + n - 1\right)!} = \sum_{n = 1}^{\infty}\sum_{m = 0}^{\infty}{z^{m + n} \over \left(m + n\right)!} = \sum_{n = 1}^{\infty}\left[% {z^{n} \over n!} + \sum_{m = 1}^{\infty}{z^{m + n} \over \left(m + n\right)!} \right] \\[3mm]&= {\rm e}^{z} - 1 + {\cal Q}\left(z\right) \quad\Longrightarrow\quad {{\rm d}\left[{\rm e}^{-z}\,{\cal Q}\left(z\right)\right] \over {\rm d}z} = 1 - {\rm e}^{-z}\,,\quad {\cal Q}\left(0\right) = 0 \\[1mm]-&----------------------------- \end{align}

$$ {\rm e}^{-z}\,{\cal Q}\left(z\right) - {\rm e}^{-0}\,{\cal Q}\left(0\right) = \left(z + {\rm e}^{-z}\right) - \left(0 + {\rm e}^{-0}\right) = z + {\rm e}^{-z} - 1 $$

$$ {\cal Q}\left(z\right) = z\,{\rm e}^{z} + 1 - {\rm e}^{-z} \quad\Longrightarrow\quad {\cal Q}\left(1\right) = {\rm e} + 1 - {\rm e}^{-1} $$

$$ \begin{array}{|c|}\hline\\ \color{#ff0000}{\large\quad% \sum_{n, m = 1}^{\infty}{1 \over \left(m + n\right)!} \color{#000000}{\ =\ } {\rm e} + 1 - {\rm e}^{-1} \quad} \\ \\ \hline \end{array} $$