Question about calculating a series involving zeta functions

Question

On this page it had shown that the sum of $\frac{1}{n^3(n+1)^3}=10-\pi^2$. I'm curious about, what is the value of $$\sum_{n=1}^\infty\frac1{n^3(n+k)^3}$$ For some positive integer $k$.
According to partial fraction expansion, we can show that $$\frac1{n^3(n+k)^3}= 6\bigg(\frac1{nk^5}-\frac1{(n+k)k^5}\bigg)-3\bigg(\frac1{k^4n^2}+\frac1{k^4(n+k)^2}\bigg)+\frac1{k^3n^3}-\frac1{k^3(n+k)^3}$$

It is obvious to show that the first part and the last part are telescoping series, and for the last part, we can see that $$\frac1{k^3n^3}-\frac1{k^3(n+k)^3}=\frac1{k^3}\bigg(\frac1{n^3}-\frac1{(n+k)^3}\bigg)=\frac1{k^3}\sum_{i=1}^{k}\frac1{i^3}=\zeta(6)+\sum_{i<j}\frac1{i^3j^3}=\sum_{n=1}^\infty\frac1{n^3(n+k)^3}$$ Which leads to the original question.
The particular values of the sum are

$k$	$$\sum_{n=1}^\infty\frac1{n^3(n+k)^3}$$
$1$	$10-\pi^2$
$2$	$\frac {21}{32}-\frac1{16}\pi^2$
$3$	$\frac {809}{5832}-\frac1{81}\pi^2$

We can easily know that the sum is in the form of $a+b\pi^2$ and $b=\frac1{k^4}$. So what about the value of $a$?

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Edit: Some notes on $\zeta(3)$:

By squaring $\zeta(3)$, $$(\zeta(3))^2=\zeta(6)+\sum_{i\ne j}\frac1{i^3j^3}$$.

Note that $i$ and $j$ are both integers and we can assume that $i$ is strictly larger than $j$, or we could say that $i=n$, $j=n+k$ for some positive integer $k$. Hence
$$(\zeta(3))^2=\zeta(6)+2\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac1{n^3(n+k)^3}$$

Assume $\sum_{n=1}^\infty\frac1{n^3(n+k)^3} = a_k-\frac{\pi^2}{k^4}$.
Thus we can know $$\begin{align}(\zeta(3))^2&=\zeta(6)+2\sum_{k=1}^{\infty}\bigg(a_k-\frac{\pi^2}{k^4}\bigg)\\&=\frac{\pi^6}{945}+2\sum_{k=1}^{\infty}a_k-2\pi^2\zeta(4)\\&=2\sum_{k=1}^{\infty}a_k-\frac{4\pi^6}{189}\end{align}$$

For $\sum_{k=1}^{10}a_k$, we can calculate that $$\begin{align}(\zeta(3))^2&\approx 2\sum_{k=1}^{8}a_k-\frac{4\pi^6}{189}\\&\approx 1.42163941214...\end{align}$$

And $(\zeta(3))^2\approx1.44494079841...$

Answer 1

§1. Applying the general theory developed in §2 we find the following expression for the sum (see §3):

$$s(k) :=\sum_{n\ge1} \frac{1}{n^3(n+k)^3} = -\frac{\pi^2}{k^4}+\frac{6}{k^5}H_{k}+\frac{3}{k^4}H_{k,2}+\frac{1}{k^3}H_{k,3}\tag{1.1} $$

where $H_{k,p} = \sum_{i=1}^{k} \frac{1}{i^p}$ is the generalized harmonic number.

$(1)$ shows that your numbers $a_k$ defined more correctly with an index $k$ by the relation

$$s(k) = a_k-\frac{\pi^2}{k^4}\tag{1.2}$$

are given explicitly by

$$a_k = \frac{6}{k^5}H_{k}+\frac{3}{k^4}H_{k,2}+\frac{1}{k^3}H_{k,3}\tag{1.3}$$

The first few numbers are

$$a_k|_{k=01}^{k=10}=\{10, \frac{21}{32}, \frac{809}{5832}, \frac{2615}{55296},\frac{ 112831}{5400000}, \frac{168791}{15552000}, \frac{ 17769701}{2823576000},\\ \frac{ 22201623}{5619712000},\frac{ 30715230979}{11666192832000}, \frac{ 29416735711}{16003008000000}\}\tag{1.4}$$

Notice that these numbers have been given by Claude Leibovici earlier.

§2. For the derivation of even more general formulae for sums of the type

$$s(j,p,k,q) :=\sum_{n\ge 1} \frac{1}{(n+j)^p (n+k)^q}\tag{2.1}$$

we start with the expression

$$\frac{1}{(n+j)(n+k)} = \frac{1}{k-j}(\frac{1}{n+j} -\frac{1}{n+k} )=\frac{1}{k-j}\left(\left(\frac{1}{n+j}-\frac{1}{n}\right) -\left(\frac{1}{n+k}+\frac{1}{n}\right) \right)$$

which after summing over $n$ gives

$$s(j,1,k,1) = \sum_{n\ge 1} \frac{1}{(n+j)(n+k)} \\ =\frac{1}{k-j}\sum_{n\ge 1} \left(\left(\frac{1}{n+j}-\frac{1}{n}\right) -\left(\frac{1}{n+k}+\frac{1}{n}\right) \right)=\frac{1}{k-j}(H_k - H_j)\tag{2.2}$$

Here we have used the representation of the harmonic number

$$H_k = \sum_{n\ge1}(\frac{1}{n}-\frac{1}{n+k})\tag{2.3}$$

Raising the (negative) powers $p$ and $q$ can be easily done by differentiating, viz.

$$\frac{\partial}{\partial j}s(j,p,k,q) = - p\; s(j,p+1,k,q)\tag{2.4a}$$

$$\frac{\partial}{\partial k}s(j,p,k,q) = - q\; s(j,p,k,q+1)\tag{2.4b}$$

and (from $(2.3))

$$\frac{\partial}{\partial k}H_k = \sum_{n\ge1} (\frac{1}{n+k})^2=\sum_{n\ge1} \left((\frac{1}{n+k})^2-\frac{1}{n^2}\right)+ \sum_{n\ge1}\frac{1}{n^2} = - H_{k,2}+\zeta(2)\tag{2.5}$$

The higher derivatives are given recursively by

$$\frac{\partial}{\partial k}H_{k,r} = r\left(\zeta(r+1) -H_{k,r+1} \right)\tag{2.6}$$

and explicitly by

$$\frac{\partial^r}{\partial k^r}H_k =(-1)^{r+1} r!(\zeta(r+1)-H_{k,r+1}), r\ge 1\tag{2.7}$$

which follow from the definition

$$\sum_{n\ge 1}(\frac{1}{n^r} - \frac{1}{(n+k)^r} ) = H_{k,r}\tag{2.8}$$

Hence we find that the sums $s(i,p,k,q)$ can be expressed by $\zeta$-functions and (generalized) harmonic functions.

Wait! What happens when $j=k$ (a case we have tacidly excluded)? This case is just the simpler sum $s(j,p,k\to j,q) = s(j,p+q,j,0)$.

§3. Application to the specific case of the OP.

We can write

$$s(k) = \frac{1}{4} \frac{\partial^2}{\partial j^2}\frac{\partial^2}{\partial k^2}\sum_{n\ge 1} \frac{1}{(n+j)(n+k)}|_{j \to 0}\\ =\frac{1}{4} \frac{\partial^2}{\partial j^2}\frac{\partial^2}{\partial k^2}\frac{1}{k-j}(H_k - H_j)|_{j \to 0}\tag{3.1}$$

where in the last equality we have used $(2.2)$.

Now, using $(2.7)$, this can be easily converted into our main result $(1.1)$.

§4. Explicit expression for the general sum (2.1)

The shifted sum defined by $s_x(j,p,k,q) = s(j,p+1,k,q+1)$ is given by

$$\begin{align}s_x(j,p,k,q) := \sum_{n=1}^{\infty}\frac{1}{(n+j)^{p+1} (n+k)^{q+1}}\\= +(-1)^q \binom{p+q}{q} \frac{H_j}{(j-k)^{p+q+1}}\\ +(-1)^p \binom{p+q}{q} \frac{H_k}{(k-j)^{p+q+1}}\\ +(-1)^{q} \sum _{m=1}^p \binom{p+q-m}{q} \frac{H_{j,m+1}-\zeta (m+1)}{(j-k)^{-m+p+q+1}}\\ +(-1)^{p} \sum _{n=1}^q \binom{p+q-n}{p} \frac{H_{k,n+1}-\zeta (n+1)}{(k-j)^{-n+p+q+1}}\\ \end{align}\tag{4.1}$$

Derivation

Using

$$\begin{align}& \frac{1}{(j+n)^{p+1} (k+n)^{q+1}}\\ =&\frac{(-1)^{p+q}}{p! p!} \frac{\partial ^{(p+q)}}{\partial j^{p}\, \partial k^{q}}\left(\frac{1}{(j+n) (k+n)}\right)\end{align}\tag{4.2}$$

we have

$$\begin{align}sx(j,p,k,q) =&\frac{(-1)^{p+q}}{p! p!} \frac{\partial ^{(p+q)}}{\partial j^{p}\, \partial k^{q}}\left(\frac{H_{j}-H_{k}}{j-k}\right)\\ = &\frac{(-1)^{p+q}}{p! p!} D_{j}^{p} D_{k}^{q}\left(\frac{H_{j}-H_{k}}{j-k}\right)\\ \end{align}\tag{4.3}$$

We can write

$$\begin{align} &D_{j}^{p} D_{k}^{q}\left(\frac{H_{j}-H_{k}}{j-k}\right) =D_{j}^{p} D_{k}^{q}\left(\frac{H_{j}}{j-k}\right)-D_{j}^{p} D_{k}^{q}\left(\frac{H_{k}}{j-k}\right) \end{align}\tag{4.4} $$

Carefully carrying out the derivatives with the first term

$$\begin{align} &D_{j}^{p} D_{k}^{q}\left(\frac{H_{j}}{j-k}\right)=D_{j}^{p}\left( H_{j} D_{k}^{q}\left(\frac{1}{j-k}\right)\right)\\ =&q! D_{j}^{p}\left( H_{j} \frac{1}{(j-k)^{q+1}}\right)\\ =&q!\sum _{m=0}^p \binom{p}{m} \left( D_{j}^{m} H_{j}\right) \left( D_{j}^{p-m}\frac{1}{(j-k)^{q+1}}\right)\\ =&q!\sum _{m=0}^p \binom{p}{m} \left( D_{j}^{m} H_{j}\right) \left(\frac{(-1)^{p-m} (-m+p+q)!}{q!}\frac{1}{ (j-k)^{p+q+1-m}} \right)\\ =&q!\left( H_{j}\right) \left(\frac{(-1)^{p} (p+q)!}{q!}\frac{1}{ (j-k)^{p+q+1}} \right)\\ +&q!\sum _{m=1}^p \binom{p}{m} \left( D_{j}^{m} H_{j}\right) \left(\frac{(-1)^{p-m} (-m+p+q)!}{q!}\frac{1}{ (j-k)^{p+q+1-m}} \right)\\ =& (-1)^{p} (p+q)!\frac{H_{j}}{ (j-k)^{p+q+1}}\\ +&\sum _{m=1}^p \binom{p}{m} \left( D_{j}^{m} H_{j}\right) \left(\frac{(-1)^{p-m} (-m+p+q)!}{ (j-k)^{p+q+1-m}} \right)\\ =& (-1)^{p} (p+q)!\frac{H_{j}}{ (j-k)^{p+q+1}}\\ +&\sum _{m=1}^p \binom{p}{m}(-1)^{m+1} m!\left(\zeta(m+1)-H_{j,m+1}\right) \left(\frac{(-1)^{p-m} (p+q-m)!}{ (j-k)^{p+q+1-m}} \right) \end{align} $$

The second term is transformed in a similar manner. Putting things togehther and simplifying gives $(4.1)$ as requested.

§5. Discussion

The general sum is composed of a transcendental part and a rational part.

Notice that this structure might be conceiled if polygamma functions are used.

The transcendental part is a linear combination of $\zeta$-functions with rational coefficients, the rational part is a similar linear combination of (generalized) harmonic numbers.

This structure is exhibited already in $(1.1)$.

The transcendental part TP of the sum

$$s_x(j,p,k,q)=\sum_{n=1}^{\infty}\frac{1}{(n+j)^{p+1} (n+k)^{q+1}}$$

for some values of $p$ and $q$ in the format $\{\{p,q\},TP(s_x)\}$ are (here $d=j-k$)

For $q=p$

$$ \begin{array}{c} \{\{0,0\},0\} \\ \left\{\{1,1\},\frac{2 \zeta(2)}{d^2}\right\} \\ \left\{\{2,2\},-\frac{6 \zeta(2)}{d^4}\right\} \\ \left\{\{3,3\},\frac{20 \zeta(2)}{d^6}+\frac{2 \zeta(4)}{d^4}\right\} \\ \end{array} $$

As mentioned before, for $p=q$ only even $\zeta$-functions appear.

For $q=p+1$

$$ \begin{array}{c} \left\{\{0,1\},\frac{\zeta(2)}{d}\right\} \\ \left\{\{1,2\},\frac{\zeta(3)}{d^2}-\frac{3 \zeta(2)}{d^3}\right\} \\ \left\{\{2,3\},\frac{10 \zeta(2)}{d^5}+\frac{\zeta(4)}{d^3}-\frac{2 \zeta(3)}{d^4}\right\} \\ \left\{\{3,4\},-\frac{35 \zeta(2)}{d^7}+\frac{5 \zeta(3)}{d^6}+\frac{\zeta(5)}{d^4}-\frac{5 \zeta(4)}{d^5}\right\} \\ \end{array} $$

For $q=p+2$

$$\begin{array}{c} \left\{\{0,2\},\frac{\zeta(3)}{d}-\frac{\zeta(2)}{d^2}\right\} \\ \left\{\{1,3\},\frac{4 \zeta(2)}{d^4}-\frac{2 \zeta(3)}{d^3}+\frac{\zeta(4)}{d^2}\right\} \\ \left\{\{2,4\},-\frac{15 \zeta(2)}{d^6}+\frac{5 \zeta(3)}{d^5}-\frac{3 \zeta(4)}{d^4}+\frac{\zeta(5)}{d^3}\right\} \\ \left\{\{3,5\},\frac{56 \zeta(2)}{d^8}-\frac{14 \zeta(3)}{d^7}+\frac{11 \zeta(4)}{d^6}-\frac{4 \zeta(5)}{d^5}+\frac{\zeta(6)}{d^4}\right\} \\ \end{array}$$

Answer 2

Using partial summation and asymptotics, we have $$a_k=\sum_{n=1}^\infty\frac1{n^3(n+k)^3}+\frac {\pi^2}{k^4}$$ $$a_k=\frac{1}{2k ^5}\Big[k^2 (\psi ^{(2)}(k+1)+2 \zeta (3))+\pi ^2 k-6 k \psi ^{(1)}(k+1)+12 \psi^{(0)}(k+1)+12 \gamma \Big]$$ which generate the sequence $$\left\{10,\frac{21}{32},\frac{809}{5832},\frac{2615}{55296},\frac{112831}{54000 00},\frac{168791}{15552000},\frac{17769701}{2823576000},\frac{22201623}{5619 712000},\dots\right\}$$

I did not find anything in $OEIS$ but the asymptotics is $$a_k=\frac{\zeta (3)}{k^3}+\frac{\pi ^2}{2 k^4}+\frac{12 \log (k)+12 \gamma -7}{2k^5}+\frac 5 {k^6}-\frac 5 {4k^7}+O\left(\frac{1}{k^9}\right)$$ which is not bad even for small values of $k$. Using this truncated series, we have $$a_1=\zeta (3)+\frac{1}{4}+6 \gamma +\frac{\pi ^2}{2}=9.85015\cdots$$ $$a_2=\frac{64 \zeta (3)-21+96 \gamma +16 \pi ^2+96 \log (2)}{512} =0.65586\cdots$$

Answer 3

Here's a smooth and Elementary way of solving the problem .

I'll use the fact that $$\frac{1}{a.b}=\frac{1}{(b-a)}{\left(\frac{1}{a}-\frac{1}{b}\right)}$$

So , $$\sum_{n=1}^{\infty}\left(\frac{1}{n(n+k)}\right)^3=\sum_{n=1}^{\infty}\frac{1}{k^3}\left(\frac{1}{n}-\frac{1}{(n+k)}\right)^3$$ $$\Rightarrow \frac{1}{k^3}\sum_{n=1}^{\infty}\underbrace{\frac{1}{n^3}-\frac{1}{(n+k)^3}}_{H^{(3)}_{k}}-\frac{3}{n(n+k)}\left(\frac{1}{n}-\frac{1}{(n+k)}\right)\tag{*}$$

Now we need to compute this sum .$$\sum_{n=1}^{\infty}\frac{3}{n(n+k)}\left(\frac{1}{n}-\frac{1}{(n+k)}\right)\Rightarrow3\color{red}{\sum_{n=1}^{\infty}\frac{k}{n^2(n+k)^2}}\tag{1}$$ Second sum which we have to compute is $$\sum_{n=1}^{\infty}\left(\frac{1}{n(n+k)}\right)^2=\sum_{n=1}^{\infty}\frac{1}{k^2}\left(\frac{1}{n}-\frac{1}{(n+k)}\right)^2$$ $$\Rightarrow\frac{1}{k^2}\sum_{n=1}^{\infty}\frac{1}{n^2}+\frac{1}{(n+k)^2}-\frac{2}{n(n+k)}$$ $$=\frac{1}{k^2}\left(\sum_{n=1}^{\infty}\frac{1}{n^2}+\color{red}{\underbrace{\sum_{n=1}^{\infty}\frac{1}{(n+k)^2}+\sum_{n=1}^{k}\frac{1}{n^2}}_{\zeta(2)}}-H_{k}^{(2)}-\color{green}{\sum_{n=1}^{\infty}\frac{2}{n(n+k)}}\right)\tag{2}$$ The third sum which we need to compute is $$\Rightarrow\sum_{n=1}^{\infty}\frac{1}{n(n+k)}=\sum_{n=1}^{\infty}\frac{1}{k}\left(\frac{1}{n}-\frac{1}{(n+k)}\right)=\frac{1}{k}\left(\sum_{n=1}^{\infty}\frac{1}{n}-\sum_{n=1}^{\infty}\frac{1}{(n+k)}\right)$$ $$\Rightarrow \frac{1}{k}\sum_{n=1}^{k}\frac{1}{n}=\color{green}{\frac{H_k^{(1)}}{k}}\tag{3}$$

Now finally assembling all the sums we get , $$\sum_{n=1}^{\infty}\left(\frac{1}{n(n+k)}\right)^2=\frac{1}{k^2}\left(2\zeta(2)-H_k^{(2)}-\frac{2H_k^{(1)}}{k}\right)\tag{4}$$ Using the value of sum from equation $(4)$ to equation $(*)$ , we'll get $$\frac{1}{k^3}\left(H_k^{(3)}-\frac{3}{k}\left(2\zeta(2)-H_k^{(2)}-\frac{2H_k^{(1)}}{k}\right)\right)$$ $$\Rightarrow\frac{1}{k^3}\left(H_k^{(3)}-\frac{6\zeta(2)}{k}+\frac{3H_k^{(2)}}{k}+\frac{6H_k^{(1)}}{k^2}\right)$$ $$\Rightarrow\color{red}{\underbrace{\frac{H_k^{(3)}}{k^3}+\frac{3H_k^{(2)}}{k^4}+\frac{6H_k^{(1)}}{k^5}}_{a}}-\frac{\pi^2}{k^4}$$