# Sum the infinite series of $\frac{1}{r^3+1}$

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).

• Compare with this question: (not a duplicate, but connected) math.stackexchange.com/questions/331850/… – Dennis Gulko Mar 21 '13 at 8:56
• 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... – CBenni Mar 21 '13 at 8:59
• @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 – Nikolaj-K Mar 21 '13 at 9:06
• @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 :/ – CBenni Mar 21 '13 at 9:09
• @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. – 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.