# How do I evaluate $\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\frac{ab}{(a+b)!}$

$$\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}\frac{ab}{(a+b)!}$$

I'm not really comfortable with more than 1 sigma's and that's why this question is confusing me. I don't think it's possible to reduce the number of variables to 1 here.

The answer is $\frac{2}{3}e$

• Do the summations one at a time. Can you do $$\sum_{b=1}^{\infty}{b\over(a+b)!}$$ – Gerry Myerson Dec 11 '17 at 5:55
• If you know what the sum should be, please mention that in the question. – coffeemath Dec 11 '17 at 5:56
• Looks like $\frac23 e$ to me. – achille hui Dec 11 '17 at 6:05
• @coffeemath, yes the answer by achille hui is correct $\frac{2e}{3}$ – Rick Dec 11 '17 at 6:07
• Looks like $\frac23e$ to wolfram too! – Nosrati Dec 11 '17 at 6:08

Since all terms in the sum are non-negative, one can arbitrary change the order of summation without changing the value of the sum. In particular, we can group terms with same value of $k = a + b$ and sum over them first. Notice $$\sum_{a+b=k, a, b \ge 1}ab = \sum_{a=1}^{k-1} a(k-a) = k \frac{(k-1)k}{2} - \frac{(k-1)k(2k-1)}{6} = \frac{(k-1)k(k+1)}{6}$$ We have \begin{align} \sum_{a=1}^\infty\sum_{b=1}^\infty \frac{ab}{(a+b)!} &= \sum_{k=2}^\infty \frac{1}{k!}\sum_{a+b=k, a, b \ge 1}ab = \frac16 \sum_{k=2}^\infty \frac{(k-1)k(k+1)}{k!}\\ &= \frac16 \sum_{k=2}^\infty \frac{k+1}{(k-2)!} = \frac16 \sum_{k=0}^\infty \frac{k+3}{k!} = \frac16 \left( \sum_{k=1}^\infty \frac{1}{(k-1)!} + \sum_{k=0}^\infty \frac{3}{k!}\right)\\ &= \frac16(e+3e) = \frac23 e \end{align}
• Very nice (+1). Interesting to note that $\displaystyle\sum_{a+b=k; a,b\ge 1}ab=\binom {k+1}3$ – hypergeometric Dec 11 '17 at 15:04
• @hypergeometric: that can be seen as a consequence of stars and bars, since $\sum_{\substack{a+b=k\\a,b\geq 1}}ab$ is the coefficient of $x^k$ in the square of $\sum_{m\geq 1}m x^m=\frac{x}{(1-x)^2}$, and $$\frac{x^2}{(1-x)^4}=\sum_{k\geq 2}\binom{k+1}{3}x^k.$$ – Jack D'Aurizio Dec 11 '17 at 17:05
Let $a+b=n$.
\begin{align} \sum_{a=1}^\infty\sum_{b=1}^\infty\frac {ab}{(a+b)!} &=\sum_{n=2}^\infty\sum_{a=1}^{n-1}\frac {a(n-a)}{n!}\\ &=\sum_{n=2}^\infty\frac n{n!}\sum_{a=1}^{n-1}a-\sum_{n=2}^\infty\frac 1{n!}\sum_{a=1}^{n-1}a^2\\ &=\sum_{n=2}^\infty \frac 1{(n-1)!}\frac {n(n-1)}2-\sum_{n=2}^\infty \frac 1{n!}\frac 16 (n-1)n(2n-1)\\ &=\frac 16 \sum_{n=2}^\infty \frac 1{(n-2)!}\left[3n-(2n-1)\right]\\ &=\frac 16\sum_{n=2}^\infty \frac {n+1}{(n-2)!}\\ &=\frac 16\sum_{n=2}^\infty \frac {n-2+3}{(n-2)!}\\ &=\frac 16\sum_{n=2}^\infty\frac 1{(n-3)!}+\frac 3{(n-2)!}\\ &=\frac 16(e+3e)\\ &=\color{red}{\frac 23e}\end{align}
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{a = 1}^{\infty}\sum_{b = 1}^{\infty}{ab \over \pars{a + b}!} & = \sum_{a = 1}^{\infty}a\sum_{b = 1}^{\infty}b \sum_{n = 2}^{\infty}{\bracks{z^{n}}z^{a + b} \over n!} = \sum_{n = 2}^{\infty}{1 \over n!}\bracks{z^{n}} \pars{\sum_{a = 1}^{\infty}a\, z^{a}}^{2} \\[5mm] & = \sum_{n = 2}^{\infty}{1 \over n!}\bracks{z^{n}} \bracks{z \over \pars{1 - z}^{2}}^{2} = \sum_{n = 2}^{\infty}{1 \over n!}\bracks{z^{n - 2}}\pars{1 - z}^{-4} \\[5mm] & = \sum_{n = 2}^{\infty}{1 \over n!}{-4 \choose n - 2}\pars{-1}^{n - 2} = \sum_{n = 2}^{\infty}{1 \over n!}{n + 1 \choose n - 2} = {1 \over 6}\ \overbrace{\sum_{n = 2}^{\infty}{\pars{n + 1}n\pars{n - 1} \over n!}} ^{\ds{4\expo{}}} \\[5mm] & = \bbx{{2 \over 3}\expo{}} \end{align}