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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.

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  • $\begingroup$ If some user is interesting in this, or other of my questions in previous post s he/she is welcome to ask in this site new questions, study it or share it with his/her colleagues. Many thanks. $\endgroup$ – user243301 Jul 8 '16 at 22:05
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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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    $\begingroup$ He used notation $\omega$ for the cube root of $1$ in the second quadrant, $\omega = (1+i\sqrt{3})/2$. $\endgroup$ – GEdgar Jul 9 '16 at 19:41
  • $\begingroup$ Tomorrow I am going to study in detail your answer, very thanks much for your identities what now is a reference for us in this site. $\endgroup$ – user243301 Jul 9 '16 at 20:57

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