Closed form to an interesting series: $\sum_{n=1}^\infty \frac{1}{1+n^3}$

Question

Intutitively, I feel that there is a closed form to $$\sum_{n=1}^\infty \frac{1}{1+n^3}$$

I don't know why but this sum has really proved difficult. Attempted manipulating a Mellin Transform on the integral solution:

$$\int_0^\infty \frac{\text{d}x}{1+x^3}=\frac{\pi}{3}\csc \frac{\pi}{3}$$ But to little avail.

Checking W|A gives the austere solution: $$\frac{1}{3}\sum_{\{x|x^3+1=0\}} x \space\text{digamma}(1-x) $$

Which I completely don't understand. Thank you for any help.

Answer 1

Hint. You may use the following series representation of the digamma function $$ \psi(z+1) + \gamma = \sum_{n=1}^{\infty}\left( \frac{1}n - \frac{1}{n+z}\right).\tag1 $$ Then your goal is to rewrite the general term of your series in a form allowing to use $(1)$. You may start with $$ \begin{align} \frac{1}{1+n^3}=\frac{1}{(n+1)(n-z_0)(n-\bar{z}_0)} \end{align} $$ where $\displaystyle z_0=\frac{1+i\sqrt{3}}2$, then make a partial fraction decomposition giving

$$ \frac{1}{1+n^3}=a_1\left(\frac{1}n - \frac{1}{n+1}\right)+a_2\left(\frac{1}n - \frac{1}{n-z_0}\right)+a_3\left(\frac{1}n - \frac{1}{n-\bar{z}_0}\right). \tag2 $$

By summing $(2)$ you get a closed form of your initial series.