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 $$\frac1{k^3n^3}-\frac1{k^3(n+k)^3}=\frac1{k^3}\bigg(\frac1{n^3}-\frac1{(n+k)^3}\bigg)=\frac1{k^3}\sum_{i=1}^{k}\frac1{i^3}=\zeta(6)+\sum_{i<j}\frac1{i^3j^3}=\sum_{n=1}^\infty\frac1{n^3(n+k)^3}$$
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 $a$?

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 $(\zeta(3))^2\approx1.44494079841...$
 A: 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
$$a_1=\zeta (3)+\frac{1}{4}+6 \gamma +\frac{\pi ^2}{2}=9.85015\cdots$$
$$a_2=\frac{64 \zeta (3)-21+96 \gamma +16 \pi ^2+96 \log (2)}{512} =0.65586\cdots$$
A: 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 $$\frac{1}{k^3}\left(H_k^{(3)}-\frac{3}{k}\left(2\zeta(2)-H_k^{(2)}-\frac{2H_k^{(1)}}{k}\right)\right)$$ $$\Rightarrow\frac{1}{k^3}\left(H_k^{(3)}-\frac{6\zeta(2)}{k}+\frac{3H_k^{(2)}}{k}+\frac{6H_k^{(1)}}{k^2}\right)$$ $$\Rightarrow\color{red}{\underbrace{\frac{H_k^{(3)}}{k^3}+\frac{3H_k^{(2)}}{k^4}+\frac{6H_k^{(1)}}{k^5}}_{a}}-\frac{\pi^2}{k^4}$$
