# Question about calculating a series involving zeta functions

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 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 $$a$$?

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

• Nice question and problem ! $\to +1$ Commented Sep 19, 2021 at 6:04
• I would want to find this as this is part of the expansion of $(\zeta(3))^2$. Commented Sep 19, 2021 at 7:07
• Why $(\zeta(3))^2$ ? Commented Sep 19, 2021 at 7:52
• @ClaudeLeibovici I've included this above. Commented Sep 19, 2021 at 12:05
• "We can easily know that the sum is in the form of $a+b\pi^2$ and $b=\frac{1}{n^4}$" First, I think you mean $k$ instead of $n$, and, more importantly, there is good reason to doubt this behaviour because asymptotically we have $k^3 s(k) \simeq \zeta(3) -\frac{\pi^2}{2k}+O(\frac{\log(k)}{k^2})$ which is not compatible with your assumed behaviour. Commented Sep 19, 2021 at 12:29

§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

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

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

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 $$q=p+2$$ $$\begin{array}{c} \left\{\{0,2\},\frac{\zeta(3)}{d}-\frac{\zeta(2)}{d^2}\right\} \\ \left\{\{1,3\},\frac{4 \zeta(2)}{d^4}-\frac{2 \zeta(3)}{d^3}+\frac{\zeta(4)}{d^2}\right\} \\ \left\{\{2,4\},-\frac{15 \zeta(2)}{d^6}+\frac{5 \zeta(3)}{d^5}-\frac{3 \zeta(4)}{d^4}+\frac{\zeta(5)}{d^3}\right\} \\ \left\{\{3,5\},\frac{56 \zeta(2)}{d^8}-\frac{14 \zeta(3)}{d^7}+\frac{11 \zeta(4)}{d^6}-\frac{4 \zeta(5)}{d^5}+\frac{\zeta(6)}{d^4}\right\} \\ \end{array}$$ • Very interesting generalization. Commented Sep 19, 2021 at 13:11 • Mathematica gives the fully simplified result\sum\limits_{n=1}^\infty\frac{1}{n^3(n+k)^3}=\frac{1}{2 k^5}\left(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\right)$which seems inconsistent with formula (1) above but consistent with the answer posted by @ClaudeLeibovici. Commented Sep 19, 2021 at 16:03 • @Steven Clark Just for clarity, pease explain what you mean by "(your expression) seems inconstent with formula (1)" (now (1.1)) Commented Sep 20, 2021 at 11:49 • It seems the$-\frac{\pi^4}{k^4}$term in formula (1.1) should really be$-\frac{\pi ^2}{k^4}$. Commented Sep 20, 2021 at 14:02 • Oh yes, you are right. Thank you for pointing this typo out. I have corrected it. Commented Sep 20, 2021 at 15:03 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 $$a_1=\zeta (3)+\frac{1}{4}+6 \gamma +\frac{\pi ^2}{2}=9.85015\cdots$$ $$a_2=\frac{64 \zeta (3)-21+96 \gamma +16 \pi ^2+96 \log (2)}{512} =0.65586\cdots$$ • Good job, and numerically valid, but notice that there is no$\zeta(3)$in the exact result (cf. my solution). All terms except$\frac{\pi^2}{k^4}\$ are rational quantities. Commented Sep 22, 2021 at 21:38
• Notice that the non trivial coefficients {6,3,1} of the harmonic numbers in my formula (1.1) can be found in OEIS A092392, in fact they are just C(2n-k,n) with n=2, k=0..2 Commented Sep 22, 2021 at 21:44

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 $$\frac{1}{k^3}\left(H_k^{(3)}-\frac{3}{k}\left(2\zeta(2)-H_k^{(2)}-\frac{2H_k^{(1)}}{k}\right)\right)$$ $$\Rightarrow\frac{1}{k^3}\left(H_k^{(3)}-\frac{6\zeta(2)}{k}+\frac{3H_k^{(2)}}{k}+\frac{6H_k^{(1)}}{k^2}\right)$$ $$\Rightarrow\color{red}{\underbrace{\frac{H_k^{(3)}}{k^3}+\frac{3H_k^{(2)}}{k^4}+\frac{6H_k^{(1)}}{k^5}}_{a}}-\frac{\pi^2}{k^4}$$