Evalute triple sum $ \sum_{m\geq 0}\sum_{n \geq 0}\sum_{p\geq 0} \frac{m!n!p!}{(m+n+p+2)!}$

Question

I had a triple $T$ sum to evaluate

$$\sum_{m\geq 0}\sum_{n \geq 0}\sum_{p\geq 0} \frac{m!n!p!}{(m+n+p+2)!}$$ where $!$ denotes factorials.

The managed to find the closed form of it, $\displaystyle \frac{\pi^2}{4}$ however, my work is quite tedious. My work follows as

The infinite triple sum $T$ can be reduced to $$T=\sum_{k\geq 1}\left(\frac{1}{k^2} +\frac{1}{k(k+1)^2}+\cdots\right)\\=\sum_{k\geq1}\sum_{l\geq 0}\left(\prod_{j\geq 0}(k+j)\right)^{-1}\frac{l!}{k+l}$$ where $l,j,k$ are some dummy variable. Further by partial fraction decomposition, the latter expression can be reduced to the following expression $$\sum_{k\geq 1}\sum_{l\geq 0} \sum_{q=0}^l{l\choose q}\frac{(-1)^q}{(k+q)(k+l)}=\\ \sum_{k\geq 1}\left(\sum_{q=0 ,q\neq l}{l\choose q}\frac{(-1)^q}{(k+q)(k+l)}+\sum_{l\geq 0}\frac{(-1)^l}{(k+l)^2}\right)$$ Using the linearity and we see that the last sum we have $$\sum_{k\geq 1}\sum_{l\geq 0}\frac{(-1)^l}{(k+l)^2}=\sum_{r\geq 0}\frac{1}{(2r+1)^2}=\frac{3}{4}\zeta(2)=\frac{\pi^2}{8}\cdots(1)$$ and the former sum $$\sum_{k\geq 1}\sum_{l\geq 1}\sum_{q=0, q\neq l}\frac{(-1)^q}{l-q}\left(\frac{1}{k+q}-\frac{1}{k+l}\right){l\choose q}=\sum_{l\geq 1}\sum_{q=0, q\neq l}\frac{(-1)^q(H_l-H_q)}{l-q}{l\choose q}$$ expanding the sum we obtained the telescoping series giving us $$1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots =\frac{\pi^2}{8}\cdots(2)$$ Adding $(1)$ and $(2)$ we obtained $T=\frac{\pi^2}{4}$

My first query, How can we resolve this sum in others way?

Moreover,with this motivation I wish to generalize the triple sum $T(k)$ for one parameter $k\geq 0$ as

$$\sum_{m\geq 0}\sum_{n \geq 0}\sum_{p\geq 0} \frac{m!n!p!}{(m+n+p+k)!}=?$$

Following work as above I find no fine clues to handle it.

My second query, How to handle the generalized sum?

Thank you !!

Answer 1

Given the poles and residues of the $\Gamma$ function, or just by creative telescoping, we have $$\sum_{p\geq 0}\frac{p!}{(p+K)!}=\frac{1}{(K-1)\Gamma(K)}\tag{1} $$ hence $$\sum_{m,n,p\geq 0}\frac{m!n!p!}{(p+m+n+2)!}=\sum_{m,n\geq 0}\frac{\Gamma(m+1)\Gamma(n+1)}{(m+n+1)\Gamma(m+n+2)}=\sum_{m,n\geq 0}\frac{1}{(m+n+1)}\int_{0}^{1}x^n(1-x)^m\,dx\tag{2}$$ and by rearranging $$\sum_{m,n,p\geq 0}\frac{m!n!p!}{(p+m+n+2)!}=\int_{0}^{1}\sum_{m,n\geq 0}\frac{x^n(1-x)^m}{(m+n+1)}\,dx=\int_{0}^{1}\frac{\log(1-x)-\log(x)}{1-2x}\,dx\tag{3}$$ equals $$ 2\int_{0}^{1/2}-\log\left(\frac{x}{1-x}\right)\frac{dx}{1-2x} \stackrel{x\mapsto\frac{z}{1+z}}{=}2\int_{0}^{1}\frac{-\log(z)}{1-z^2}\,dz=2\sum_{n\geq 0}\frac{1}{(2n+1)^2}=\frac{\pi^2}{4}.\tag{4} $$

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This approach can be generalized. For instance $$\sum_{m,n,p\geq 0}\frac{m!n!p!}{(p+m+n+3)!}=\sum_{m,n\geq 0}\frac{\Gamma(m+1)\Gamma(n+1)}{(m+n+2)^2\Gamma(m+n+2)}=\sum_{m,n\geq 0}\frac{1}{(m+n+2)^2}\int_{0}^{1}x^n(1-x)^m\,dx\tag{2'}$$ and by rearranging $$\sum_{m,n,p\geq 0}\frac{m!n!p!}{(p+m+n+3)!}=\int_{0}^{1}\sum_{m,n\geq 0}\frac{x^n(1-x)^m}{(m+n+2)^2}\,dx=\int_{0}^{1}\frac{x\text{Li}_2(1-x)-(1-x)\text{Li}_2(x)}{x(1-3x+2x^2)}\,dx\tag{3'}$$ equals $$ 2\int_{0}^{1/2}\frac{x\text{Li}_2(1-x)-(1-x)\text{Li}_2(x)}{x(1-3x+2x^2)} \stackrel{x\mapsto\frac{z}{1+z}}{=}2\int_{0}^{1}\frac{z\text{Li}_2\left(\frac{1}{1+z}\right)-\text{Li}_2\left(\frac{z}{1+z}\right)}{z(1-z)}\,dz\tag{4'} $$ or $$ 2\int_{0}^{1}\left[z\,\text{Li}_2\left(\frac{1}{1+z}\right)-\text{Li}_2\left(\frac{z}{1+z}\right)+(1-z)\,\text{Li}_2\left(\frac{1}{2-z}\right)-\text{Li}_2\left(\frac{1-z}{2-z}\right)\right]\frac{dz}{z}.$$ By integration by parts this expression boils down to Euler sums with weight $\leq 3$.
After a massive amount of computations we have $$\boxed{\sum_{a,b,c\geq 0}\frac{a!b!c!}{(a+b+c+3)!}=\color{red}{\frac{13}{4}\zeta(3)-\frac{\pi^2}{2}\log(2)}.}$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[15px,#ffd]{\sum_{m\ \geq\ 0}\sum_{n\ \geq\ 0}\sum_{p\ \geq\ 0} {m!\, n!\, p! \over \pars{m + n + p + 2}!}} \\[5mm] = &\ \sum_{m = 0}^{\infty}\sum_{n = 0}^{\infty} {m!\, n! \over \pars{m + n + 1}!} \sum_{p = 0}^{\infty} {\Gamma\pars{m + n + 2}\Gamma\pars{p + 1} \over \Gamma\pars{m + n + p + 3}} \\[5mm] = &\ \sum_{m = 0}^{\infty}\sum_{n = 0}^{\infty} {m!\, n! \over \pars{m + n + 1}!} \sum_{p = 0}^{\infty}\int_{0}^{1}x^{m + n + 1}\pars{1 - x}^{\, p}\,\dd x \\ = &\ \int_{0}^{1}\sum_{m = 0}^{\infty}\sum_{n = 0}^{\infty} {\Gamma\pars{m + 1}\Gamma\pars{n + 1} \over \Gamma\pars{m + n + 2}}\, x^{m + n + 1}\ \overbrace{\sum_{p = 0}^{\infty}\pars{1 - x}^{\, p}}^{\ds{1 \over x}}\ \dd x \\[5mm] = &\ \int_{0}^{1}\sum_{m = 0}^{\infty}\sum_{n = 0}^{\infty} \bracks{\int_{0}^{1}y^{m}\pars{1 - y}^{n}\,\dd y}x^{m + n}\,\dd x \\[5mm] = &\ \int_{0}^{1}\int_{0}^{1}\bracks{\sum_{m = 0}^{\infty}\pars{xy}^{m}} \braces{\sum_{n = 0}^{\infty} \bracks{x\pars{1 - y}}^{\, n}}\,\dd x\,\dd y \\[5mm] = &\ \int_{0}^{1}\int_{0}^{1}{\dd x\,\dd y \over \pars{1 - xy}\pars{1 - x + xy}} \\[5mm] = &\ \int_{0}^{1}\bracks{\pars{y - 1} \int_{0}^{1}{\dd x \over \pars{y - 1}x + 1} + y\int_{0}^{1}{\dd x \over 1 - xy}}\,{\dd y \over 2y - 1} \\[5mm] = &\ \int_{0}^{1}{\ln\pars{y} - \ln\pars{1 - y} \over 2y - 1}\,\dd y = \int_{0}^{1}\ln\pars{y \over 1 - y} \,{\dd y \over 2y - 1} \\[5mm] \stackrel{y/\pars{1 - y}\ =\ t}{=}\,\,\,& -\int_{0}^{\infty}{\ln\pars{t} \over 1 - t^{2}}\,\dd t = -\int_{0}^{1}{\ln\pars{t} \over 1 - t^{2}}\,\dd t - \int_{1}^{0}{\ln\pars{1/t} \over 1 - \pars{1/t}^{2}} \,\pars{-\,{\dd t \over t^{2}}} \\[5mm] = &\ -2\int_{0}^{1}{\ln\pars{t} \over 1 - t^{2}}\,\dd t = \bbox[15px,#ffd,border:1px solid navy]{\pi^{2} \over 4}\ \approx\ 2.4674 \end{align}

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Note that \begin{align} -2\int_{0}^{1}{\ln\pars{t} \over 1 - t^{2}}\,\dd t & = -\int_{0}^{1}{\ln\pars{t} \over 1 - t}\,\dd t - \int_{0}^{1}{\ln\pars{t} \over 1 + t}\,\dd t \\[5mm] & = -\int_{0}^{1}{\ln\pars{1 - t} \over t}\,\dd t + \int_{0}^{-1}{\ln\pars{-t} \over 1 - t}\,\dd t \\[5mm] & \stackrel{\mrm{IBP}}{=}\,\,\, -\int_{0}^{1}{\ln\pars{1 - t} \over t}\,\dd t + \int_{0}^{-1}{\ln\pars{1 - t} \over t}\,\dd t \\[5mm] & = \mrm{Li}_{2}\pars{1} - \mrm{Li}_{2}\pars{-1} \quad\pars{\ DiLogarithm\ } \\[5mm] & = {\pi^{2} \over 6} - \pars{-\,{\pi^{2} \over 12}} = {\pi^{2} \over 4} \end{align}

Answer 3

I am treating the case $k=4$ in a separate answer since the rendering times are a bit long. We have $$ \sum_{n\geq 0}\frac{a!b!c!}{(a+b+c+4)!}=\sum_{b,c\geq 0}\frac{B(b+1,c+1)}{(3+b+c)^2(2+b+c)}=\int_{0}^{1}\sum_{b,c\geq 0}\frac{x^b(1-x)^c}{(3+b+c)^2(2+b+c)}\,dx $$ and the RHS equals $$\small \int_{0}^{1}\sum_{s\geq 0}\frac{1}{(s+3)^2(s+2)}\sum_{b=0}^{s}x^b(1-x)^{s-b}\,dx =\int_{0}^{1}\sum_{s\geq 0}\left[\frac{1}{s+2}-\frac{1}{s+3}-\frac{1}{(s+3)^2}\right]\frac{x^{s+1}-(1-x)^{s+1}}{2x-1}\,dx$$ which Mathematica is able to evaluate as $$ \sum_{n\geq 0}\frac{a!b!c!}{(a+b+c+4)!}=\color{red}{6-\frac{\pi^2}{2}+\pi^2\log(2)-\frac{13}{2}\zeta(3)}\approx 0.0929163927751. $$