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).
 A: 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.
