Is there a definite value for the sum:

$S=\displaystyle\sum_{r=1}^{\infty} \frac{1}{r^3+1}$

And if so, how would I arrive at finding this sum?

I have tried reducing the above into partial fractions, however I can't seem to arrive at any definitive answer (preferably in terms of elementary function).

  • $\begingroup$ Compare with this question: (not a duplicate, but connected) math.stackexchange.com/questions/331850/… $\endgroup$ – Dennis Gulko Mar 21 '13 at 8:56
  • $\begingroup$ Mathematica gives no closed form result. This doesnt mean there is no closed form, however it means that there probably is none. Numerically, we get 0.6865033423... $\endgroup$ – CBenni Mar 21 '13 at 8:59
  • $\begingroup$ @CBenni: WolframAlpha for Sum[1/(r^3+1),{r,1,Infinity}] extresses it in terms of the digamma function. en.wikipedia.org/wiki/Digamma_function $\endgroup$ – Nikolaj-K Mar 21 '13 at 9:06
  • $\begingroup$ @NickKidman: Yes, but that is not a closed form, as the digamma function is not a elementary function. Also interesting to see that Wolfram|Alpha gives a completely different result than Mathematica :/ $\endgroup$ – CBenni Mar 21 '13 at 9:09
  • $\begingroup$ @Dennis Gulko Thank you for the link, I see how they introduced the Zeta function. However I still would rather have the solution in terms of elementary function. $\endgroup$ – Sy123 Mar 21 '13 at 9:09

Partial fractions gives $$ \frac1{k^3+1}=\frac13\left(\frac1{k+1}-\frac\alpha{k-\alpha}-\frac\beta{k-\beta}\right)\tag{1} $$ where $\alpha+\beta=1$ and $\alpha\beta=1$. Set $\alpha=\dfrac{1+i\sqrt3}{2}$ and $\beta=\dfrac{1-i\sqrt3}{2}$.

The digamma function is $$ \psi(z+1)=-\gamma+\sum_{k=1}^\infty\left(\frac1k-\frac1{k+z}\right)\tag{2} $$ where $\gamma$ is the Euler-Mascheroni constant.

So what Wolfram-Alpha is returning is simply $$ \begin{align} \sum_{k=1}^\infty\frac1{k^3+1} &=\sum_{k=1}^\infty\frac13\left(\frac1{k+1}-\frac\alpha{k-\alpha}-\frac\beta{k-\beta}\right)\\ &=-\frac13\left(-\gamma+\sum_{k=1}^\infty\left(\frac1k-\frac1{k+1}\right)\right)\\ &\hphantom{=}+\frac\alpha3\left(-\gamma+\sum_{k=1}^\infty\left(\frac1k-\frac1{k-\alpha}\right)\right)\\ &\hphantom{=}+\frac\beta3\left(-\gamma+\sum_{k=1}^\infty\left(\frac1k-\frac1{k-\beta}\right)\right)\\ &=\frac\alpha3\psi(1-\alpha)+\frac\beta3\psi(1-\beta)-\frac13(1-\gamma)\tag{3} \end{align} $$ Plugging $(3)$ into Mathematica yields $0.686503342338623885964605212187$.

Note that $(3)$ and this answer sum to $\zeta(3)-\frac12$. Thus, $$ \sum_{k=1}^\infty\frac1{k^3+1}=\frac12+\sum_{k=1}^\infty(-1)^{k-1}(\zeta(3k)-1)\tag{4} $$ which as commented in the other answer, converges over $0.9$ digits per term.

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  • $\begingroup$ Thank you for the comprehensive solution. $\endgroup$ – Sy123 Mar 22 '13 at 13:56

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