Closed expressions for harmonic-like multiple sums Inspired by Find the sum of the double series $\sum_{k=1}^\infty \sum_{j=1}^\infty \frac{1}{(k+1)(j+1)(k+1+j)} $ here are some related problems.


*

*Prove that


$$w(2,1)  = \sum _{i=1}^{\infty } \sum _{j=1}^{\infty } \frac{1}{i j (i+j)} = 2 \zeta (3)$$
$$w(3,1)=\sum _{i=1}^{\infty } \sum _{j=1}^{\infty } \sum _{k=1}^{\infty } \frac{1}{i j k (i+j+k)}=6 \zeta (4)$$
And can you guess (and prove) the closed form result for
$$w(n,1)=\sum _{i_{1}=1}^{\infty } \sum _{i_{2}=1}^{\infty }... \sum _{i_{n}=1}^{\infty } \frac{1}{i_{1} i_{2}...  i_{n} (i_{1}+i_{2}+...+i_{n})}$$


*Calculate, if possible, closed forms for


$$w(2,2,1)  = \sum _{i=1}^{\infty } \sum _{j=1}^{\infty } \frac{1}{i^2 j^2 (i+j)} $$
$$w(2,2,2)  = \sum _{i=1}^{\infty } \sum _{j=1}^{\infty } \frac{1}{i^2 j^2 (i+j)^2} $$
 A: 1.
$$\begin{eqnarray*}w(n,1)&=&\sum_{a_1,\ldots,a_n\geq 1}\int_{0}^{+\infty}\prod_{k=1}^{n}\frac{e^{-a_kx}}{a_k}\,dx\\&=&\int_{0}^{+\infty}\left[-\log(1-e^{-x})\right]^n\,dx\\&=&\int_{0}^{1}\frac{\left[-\log(1-u)\right]^n}{u}\,du\\&=&\int_{0}^{1}\frac{[-\log v]^n}{1-v}\,dv\\&=&\sum_{m\geq 0}\int_{0}^{1}\left[-\log v\right]^n v^m\,dv\\&=&\sum_{m\geq 0}\frac{n!}{(m+1)^{n+1}}=\color{red}{n!\cdot\zeta(n+1).}\end{eqnarray*}$$
2.1. By classical Euler sums,
$$\begin{eqnarray*} w(2,2,1)&=& \int_{0}^{1}\sum_{i,j\geq 1}\frac{z^{i-1}}{i^2}\cdot\frac{z^{j-1}}{j^2}\cdot z\,dz\\&=&\int_{0}^{1}\frac{\text{Li}_2(z)^2}{z}\,dz\\&=&2\int_{0}^{1}\log(z)\log(1-z)\,\text{Li}_2(z)\frac{dz}{z}\\&=&2\sum_{n\geq 1}\frac{H_n-n\zeta(2)+n H_n^{(2)}}{n^4}\\&=&\color{red}{2\,\zeta(2)\,\zeta(3)-3\,\zeta(5)}.\end{eqnarray*}$$
2.2. In a similar fashion,
$$\begin{eqnarray*} w(2,2,2)&=& \int_{0}^{1}\sum_{i,j\geq 1}\frac{z^{i-1}}{i^2}\cdot\frac{z^{j-1}}{j^2}\cdot (-z\log z)\,dz\\&=&\int_{0}^{1}\frac{-\log(z)\,\text{Li}_2(z)^2}{z}\,dz\\&=&-\int_{0}^{1}\log(z)^2\log(1-z)\,\text{Li}_2(z)\frac{dz}{z}\\&=&\sum_{n\geq 1}\frac{2nH_n-2n^2(\zeta(2)-H_n^{(2)})-2n^3(\zeta(3)-H_n^{(3)})}{n^6}\\&=&\color{red}{\frac{\pi^6}{2835}}.\end{eqnarray*}$$
An alternative way:
$$\begin{eqnarray*} \sum_{m,n\geq 1}\frac{1}{m^2 n^2(m+n)^2}&=&\sum_{s\geq 2}\frac{1}{s^2}\sum_{k=1}^{s-1}\frac{1}{k^2(s-k)^2}\\&=&\sum_{s\geq 2}\frac{1}{s^4}\sum_{k=1}^{s-1}\left(\frac{1}{k}+\frac{1}{s-k}\right)^2\\&=&2\sum_{s\geq 1}\frac{H_{s-1}^{(2)}}{s^4}+4\sum_{s\geq 1}\frac{H_{s-1}}{s^5}\\&=&-6\,\zeta(6)+2\sum_{s\geq 1}\frac{H_s^{(2)}}{s^4}+4\sum_{s\geq 1}\frac{H_s}{s^5}\end{eqnarray*}$$
where $\sum_{s\geq 1}\frac{H_s^{(2)}}{s^4}=\zeta(3)^2-\frac{1}{3}\zeta(6)$ by Flajolet and Salvy's $(b)$ and
$$ 2\sum_{s\geq 1}\frac{H_s}{s^5} = 7\zeta(6)-2\zeta(2)\zeta(4)-\zeta(3)^2$$
follows from Euler's theorem, leading to
$$ \sum_{m,n\geq 1}\frac{1}{m^2 n^2(m+n)^2} =\frac{22}{3}\zeta(6)-4\zeta(2)\zeta(4).$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
\mrm{w}\pars{n,1} & \equiv
\sum _{i_{1} = 1}^{\infty}\ldots\sum _{i_{n} = 1}^{\infty}
{1 \over i_{1}\ldots i_{n} \pars{i_{1} + \cdots +i_{n}}}
\\[5mm] & =
\sum _{i_{1} = 1}^{\infty}\ldots\sum _{i_{n} = 1}^{\infty}
\pars{\int_{0}^{1}x_{1}^{i_{1} - 1}\,\dd x_{1}}\cdots
\pars{\int_{0}^{1}x_{n}^{i_{n} - 1}\,\dd x_{n}}
\pars{\int_{0}^{1}x^{i_{1} + \cdots + i_{n} - 1}\,\dd x}
\\[5mm] & =
\int_{0}^{1}\cdots\int_{0}^{1}\int_{0}^{1}
\sum_{i_{1} = 1}^{\infty}\pars{x_{1}x}^{i_{1}}\ldots
\sum_{i_{n} = 1}^{\infty}\pars{x_{n}x}^{i_{n}}\,
{\dd x_{1}\ldots\dd x_{n}\,\dd x \over x_{1}\ldots x_{n}x}
\\[5mm] & =
\int_{0}^{1}\cdots\int_{0}^{1}\int_{0}^{1}
{x_{1}x \over 1 - x_{1}x}\ldots{x_{n}x \over 1 - x_{n}x}\,
{\dd x_{1}\ldots\dd x_{n}\,\dd x \over x_{1}\ldots x_{n}x}
\\[5mm] & =
\int_{0}^{1}x^{n - 1}\pars{\int_{0}^{1}{\dd\xi \over 1 - x\xi}}^{n}\dd x =
\int_{0}^{1}x^{n - 1}\bracks{\pars{-1}^{n}\ln^{n}\pars{1 - x} \over x^{n}}
\,\dd x
\\[5mm] & =
\pars{-1}^{n}\int_{0}^{1}{\ln^{n}\pars{1 - x} \over x}\,\dd x
\,\,\,\stackrel{x\ \mapsto\ 1 - x}{=}\,\,\,
\pars{-1}^{n}\int_{0}^{1}{\ln^{n}\pars{x} \over 1 - x}\,\dd x
\\[5mm] & \stackrel{\mrm{IBP}}{=}\,\,\,
\pars{-1}^{n}\int_{0}^{1}\ln\pars{1 - x}
\bracks{n\ln^{n - 1}\pars{x}\,{1 \over x}}\dd x =
\pars{-1}^{n + 1}\,n\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln^{n - 1}\pars{x}\,\dd x
\\[5mm] & \stackrel{\mrm{IBP}}{=}\,\,\,
\pars{-1}^{n}\,n\pars{n - 1}\int_{0}^{1}\mrm{Li}_{2}\pars{x}
\ln^{n - 2}\pars{x}\,{1 \over x}\,\dd x
\\[5mm] & =
\pars{-1}^{n}\,n\pars{n - 1}\int_{0}^{1}\mrm{Li}_{3}'\pars{x}
\ln^{n - 2}\pars{x}\,\dd x
\\[5mm] & \stackrel{\mrm{IBP}}{=}\,\,\,
\pars{-1}^{n + 1}\,n\pars{n - 1}\pars{n - 2}\int_{0}^{1}\mrm{Li}_{3}\pars{x}
\ln^{n - 3}\pars{x}\,{1 \over x}\,\dd x
\\[5mm] & =
\pars{-1}^{n + 1}\,n\pars{n - 1}\pars{n - 2}\int_{0}^{1}\mrm{Li}_{4}'\pars{x}
\ln^{n - 3}\pars{x}\,\dd x
\\[5mm] & = \cdots =
\pars{-1}^{2n + 2}\,n\pars{n - 1}\pars{n - 2}\ldots 1\
\underbrace{\int_{0}^{1}\mrm{Li}_{n + 1}'\pars{x}\,\dd x}
_{\ds{\mrm{Li}_{n + 1}\pars{1} = \zeta\pars{n + 1}}}
\\[5mm] & \implies
\bbx{\mrm{w}\pars{n,1} \equiv
\sum _{i_{1} = 1}^{\infty}\ldots\sum _{i_{n} = 1}^{\infty}
{1 \over i_{1}\ldots i_{n} \pars{i_{1} + \cdots +i_{n}}} =
n!\zeta\pars{n + 1}}
\end{align}
