Closed expressions for harmonic-like multiple sums

Question

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.

1. 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

$$w(n,1)=\sum _{i_{1}=1}^{\infty } \sum _{i_{2}=1}^{\infty }... \sum _{i_{n}=1}^{\infty } \frac{1}{i_{1} i_{2}... i_{n} (i_{1}+i_{2}+...+i_{n})}$$

2. Calculate, if possible, closed forms for

$$w(2,2,1) = \sum _{i=1}^{\infty } \sum _{j=1}^{\infty } \frac{1}{i^2 j^2 (i+j)} $$

$$w(2,2,2) = \sum _{i=1}^{\infty } \sum _{j=1}^{\infty } \frac{1}{i^2 j^2 (i+j)^2} $$

Answer 1

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 $$ \sum_{m,n\geq 1}\frac{1}{m^2 n^2(m+n)^2} =\frac{22}{3}\zeta(6)-4\zeta(2)\zeta(4).$$

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} \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}

Answer 3

The question of the OP was completely answered by Jack D'Aurizio. Now, as an extension, I'd like to study more general problems of a similar type.

1. Let us consider first sums of the type

$$w(p,q,r)=\sum _{n=1}^{\infty } \sum _{m=1}^{\infty } \frac{1}{m^p n^q (m+n)^r}\tag{1}$$

and start with the case $p = q$ general $r$ and see how far we can come.

We will consider small values in order to see a pattern.

The solution method will be direct summation, exchange of order of summation and partial fraction decomposition. I discarded use of integrals which did not lead me to results.

2. Important definitions and formulas are

$$H_k=\sum _{i=1}^k \frac{1}{i}$$ $$H_k^{(m)}=\sum _{i=1}^k \frac{1}{i^m}$$ $$\zeta (s)=\sum _{n=1}^{\infty } \frac{1}{n^s}$$

$$S(1,q)/;(q\ge 2) = \sum _{k=1}^{\infty } \frac{H_k}{k^q}= \left(\frac{q}{2}+1\right) \zeta (q+1)-\frac{1}{2} \sum _{k=1}^{q-2} \zeta (k+1) \zeta (q-k)\tag{2}$$

$$S(p,q)/;(p+q\; odd)=\sum _{k=1}^{\infty } \frac{H_k^{(p)}}{k^q}=(-1)^p \sum _{k=1}^{\left\lfloor \frac{q}{2}\right\rfloor } \zeta (2 k) \binom{-2 k+p+q-1}{p-1} \zeta (-2 k+p+q)+(-1)^p \sum _{k=1}^{\left\lfloor \frac{p}{2}\right\rfloor } \zeta (2 k) \binom{-2 k+p+q-1}{q-1} \zeta (-2 k+p+q)+\left(-\frac{1}{2} (-1)^p \binom{p+q-1}{p}-\frac{1}{2} (-1)^p \binom{p+q-1}{q}+\frac{1}{2}\right) \zeta (p+q)+\frac{1}{2} \left(1-(-1)^p\right) \zeta (p) \zeta (q)\tag{3}$$

3. For $p \ge 1, r\ge 1$ we can write

$$ \begin{eqnarray*} w(p,p,r)&=&\sum _{n=1}^{\infty } \sum _{m=1}^{\infty } \frac{1}{m^p n^p (m+n)^r}\\ &=&\sum _{n} \sum _{s}\frac{1}{(s-n)^p n^p s^r}\\ &=&\sum _{s \ge 2} \frac{1}{s^r}{\sum _{n=1}^{r-1} \frac{1}{(n (s-n))^p}}\\ &=&\sum _{n} \sum _{s}\frac{1}{(s-n)^p n^p s^r}\\ &=&\sum _{s \ge 2} \frac{1}{s^{r+p}}{\sum _{n=1}^{r-1} (\frac{1}{n}+\frac{1}{ (s-n)})^p}\tag{4a}\\ &=&\sum _{s \ge 2} \frac{1}{s^{p+r}} \sum _{n=1}^{r-1} \left(\sum _{t=0}^p \binom{p}{t} \frac{1}{n^t (s-n)^{p-t}}\right)\tag{4b}\\ \end{eqnarray*} $$

Now let us calculate the first few p-sums.

$p= 1$

$$ \begin{align} w(1,1,r) &= \sum _{s=2}^{\infty } \frac{1}{s^{r+1}}\sum _{n=1}^{r-1} \left(\frac{1}{s-n}+\frac{1}{n}\right)=2\sum _{s=2}^{\infty } \frac{ H_{s-1}}{s^{r+1}}\\ &= 2 \sum _{s=2}^{\infty } \frac{H_s-\frac{1}{s}}{s^{r+1}}=2 \sum _{s=2}^{\infty } \frac{H_s}{s^{r+1}}-2 \;\zeta(r+2)\\ &= \left(r+1\right) \zeta (r+2)- \sum _{k=1}^{r-1} \zeta (k+1) \zeta (-k+r+1)\tag{5} \end{align} $$

Hence for $p=1$ the sum reduces to zeta values for any $r\ge 2$.

For the first 5 values of r we have

$$ \begin{array}{l} w(1,1,1)=2 \zeta (3) \\ w(1,1,2)= 3 \zeta(4)-\zeta(2)^2=\frac{\pi ^4}{180} \\ w(1,1,3) = 4 \zeta(5)-2 \zeta(2) \zeta(3)=4 \zeta (5)-\frac{\pi ^2 \zeta (3)}{3} \\ w(1,1,4) = 5 \zeta(6)-\zeta(3)^2-2 \zeta(2) \zeta(4)=\frac{\pi ^6}{630}-\zeta (3)^2 \\ w(1,1,5) = 6 \zeta(7) -2 \zeta(3) \zeta(4)-2 \zeta(2) \zeta(5)=-\frac{\pi ^4 \zeta (3)}{45}-\frac{\pi ^2 \zeta (5)}{3}+6 \zeta (7) \\ \end{array} $$

$p= 2$

Since

$$\left(\frac{1}{s-n}+\frac{1}{n}\right)^2=\frac{1}{n^2}+\frac{2}{n (s-n)}+\frac{1}{(s-n)^2}=\frac{1}{n^2}+\frac{2}{s} \left(\frac{1}{s-n}+\frac{1}{n}\right)+\frac{1}{(s-n)^2}$$

the n-sum becomes

$$\sum _{n=1}^{s-1} \left(\frac{1}{n^2}+\frac{2}{s} \left(\frac{1}{s-n}+\frac{1}{n}\right)+\frac{1}{(s-n)^2}\right)=\frac{4 H_{s-1}}{s}+2 H_{s-1}^{(2)}$$

Hence

$$ \begin{align} w(2,2,r)&=4 \sum _{s=1}^{\infty } \frac{H_{s-1}}{s^{3+r}}+2 \sum _{s=1}^{\infty } \frac{H_{s-1}^{(2)}}{s^{2+r}}\\ &=-6 \zeta(4+r)+4 \sum _{s=1}^{\infty } \frac{H_{s}}{s^{3+r}}+2 \sum _{s=1}^{\infty } \frac{H_{s}^{(2)}}{s^{2+r}}\tag{6a}\\ &= -6 \zeta(4+r)+ 4 S(1,3+r)+ 2 S(2,2+r)\tag{6b} \end{align} $$

Notice that in $(6b)$ the explicit formula $(3)$ for the sums $S(p,q)$ are only valid for odd $r$.

Hence for $p=2$ and odd $r$ we have a solution in terms of zeta values.

For even $r$ there seems to be no general formula for S(2,2m) (more generally for even p+q) and we have to take $(6a)$

The "problem child" is the sum

$$\sum _{s=1}^{\infty } \frac{H_{s}^{(2)}}{s^{2+r}}/; r \; even$$

I have only seen these two cases which the even sums reduce to zeta values (formulas (40) and (42) resp. in [1])

$$S(2,4)=\zeta (3)^2-\frac{\pi ^6}{2835}$$ $$S(4,2)=\frac{37 \pi ^6}{11340}-\zeta (3)^2$$

$p=3$

to be done.

References

[1] http://mathworld.wolfram.com/HarmonicNumber.html
[2] http://algo.inria.fr/flajolet/Publications/FlSa98.pdf
[3] http://mathworld.wolfram.com/EulerSum.html