<s>[INST] Evaluate $\sum_{k=2}^\infty\frac{1}{k^3-1}$ I was considering a specialization of the Cauchy product 
$$ \left(\sum_{n=1} ^\infty x^n \right) \left(\sum_{n=1}^\infty (-1)^n  x^n \right)=\frac{-x^2}{1-x^2},$$
that converges for $0<x<1$.
The cited specialization is $x=\frac{1}{j^{3/2}}$ for integers $j\geq 2$. Then since  $ \sum_{k=1}^{n} (-1)^k=\frac{(-1)^n-1}{2}$ the Cauchy product is computed (I don't know if these kind of calculations were in the literature, my intention was find an identity for $\zeta(3)$, where $\zeta(s)$ is the Riemann Zeta function, which was failed; on the other hand I believe that it is possible get some generalization of the following by the nature of the Cauchy's products, then if is such the case feel free to study this, if it is interesting) as 
$$\frac{1}{1-j^3}=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{j^{\frac{3}{2}(n+1)}} \left( \frac{(-1)^n-1}{2} \right) .$$
Now I compute $$\frac{1}{j^3}=\frac{1}{1-j^3}+\text{something}=\text{RHS}+\text{something},$$
where $\text{something}=\frac{1-2j^3}{j^3(1-j^3)}$, and thus taking the sum from $j=2$ to infinite, one has 
$$\zeta(3)=1+\sum_{j=2}^\infty\frac{1-2j^3}{j^3(1-j^3)}+\sum_{n=1}^\infty(-1)^{n+1} \frac{(-1)^n-1}{2}\sum_{j=2}^\infty\frac{1}{j^{\frac{3}{2}(n+1)}}. $$
Thus if there are no mistakes I can write first this simplification $$\sum_{j=2}^\infty\frac{1}{1-j^3}=\sum_{n=1}^\infty\frac{-1-(-1)^{n+1}}{2} \left( \zeta(\frac{3}{2}(n+1))-1 \right) ,$$ and after that $$\sum_{j=2}^\infty\frac{1}{j^3-1}=\sum_{m=1}^\infty(\zeta(3m+1)-1).$$
The online calculator of Wolfram Alpha knons how compute 

sum 1/(k^3-1) from k=2 to infinite

and 

Question. Can you explain how obtain $$\sum_{k=2}^\infty\frac{1}{k^3-1}?$$
  What is the meaning of such calculations  from the online calculator? Thanks in advance.

 [/INST] The key is to use  the identity:
$$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}.\tag{1}$$
that leads to:
$$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)(n+c)} = \frac{1}{b-c}\left(\frac{\psi(a)-\psi(c)}{a-c}-\frac{\psi(a)-\psi(b)}{a-b}\right).\tag{2} $$
We have $k^3-1 = (k-1)(k-\omega)(k-\omega^2)$, hence by taking $a=1$, $b=2-\omega$, $c=2-\omega^2$:
$$ \sum_{k\geq 2}\frac{1}{k^3-1} = \sum_{n\geq 0}\frac{1}{(n+a)(n+b)(n+c)} = \frac{i}{\sqrt{3}}\left(\frac{\psi(1)-\psi(2-\omega^2)}{\omega^2-1}-\frac{\psi(1)-\psi(2-\omega)}{\omega-1}\right)\tag{3} $$
or:

$$ \sum_{k\geq 2}\frac{1}{k^3-1} = \color{red}{\frac{2}{\sqrt{3}}\,\text{Im}\left(\frac{H_{1-\omega}}{1-\omega}\right)}=0.2216893951\ldots\tag{4} $$

The same can be achieved by noticing that the LHS of $(4)$ equals
$$ \sum_{k\geq 2}\left(\frac{1}{k^3}+\frac{1}{k^6}+\frac{1}{k^9}+\ldots\right) = \sum_{n\geq 1}\left(\zeta(3n)-1\right) \tag{5} $$
and by applying a discrete Fourier transform to the generating function for $\zeta(n)-1$.
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