Is there a closed-form for $\sum_{n=0}^{\infty}\frac{n}{n^3+1}$? I'm reading a book on complex variables (The Theory of Functions of a Complex Variable, Thorn 1953) and the following is shown:

Let $f(z)$ be holomorphic and single valued in $\mathbb{C}$ except at a finite number of points $a_1,\ldots, a_k$ which are not integers or half-integers, where $f(z)$ has poles or essential singularities. Further, suppose there is some $d$ for which if $|z|>m(d)$,
$
|z^2 f(z)|<d
$. Then
$$
\sum_{n=-\infty}^{\infty} f(n) = -\pi \sum_{n=1}^{k} r_n \cot(\pi a_n),
$$where $r_n$ is the residue of $f(z)$ at $z=a_n$.

Here is a sketch of the proof I supplied: the condition of no $a_n$ integers or half-integers is to prevent 'coupling' between $f$ and cotangent. Let $C_m$ be the counterclockwise rectangular contour with vertices $\pm(m+1/2)\pm m i$; choose $m>1$ big enough so that each $a_n$ is contained in the interior of $C_m$ and the growth condition is satisfied. Then
$$
\frac{1}{2\pi i}\int _{C_m} \pi \cot(\pi z) f(z)\,dz = \sum_{n=-m}^{m}f(n) + \pi\sum_{n=1}^{k} \pi r_n \cot(\pi a_n)
$$A brute force argument shows that along $C_m$, $|\cot(\pi z)|<\coth(\pi m)<2$. Now we use the growth condition:
$$
\left|\frac{1}{2\pi i}\int _{C_m} \pi \cot(\pi z) f(z)\,dz\right|
\leq \frac{1}{2}\int _{C_m} \left| \cot(\pi z) f(z)\right|\,dz
$$
$$
<\int _{C_m} \left|f(z)\right|\,dz \leq (8m+2) \max_{z\in C_m} |f(z)| < \frac{(8m+2)d}{m^2}
$$So as $m\to \infty$, the LHS approaches $0$ and the result follows.
Fine. Now, the series in my title is the second part of an exercise; the first part has you evaluate $\displaystyle{\sum_{n=0}^{\infty} \frac{1}{n^2+1}}$, which is done by writing:
$$
\sum_{n=0}^{\infty} \frac{1}{n^2+1} = \frac{1}{2}\left(\sum_{n=-\infty}^{\infty} \frac{1}{n^2+1}+1\right)
$$
$$
= \frac{1}{2}\left(-\pi \left(\frac{-i}{2}\cot(\pi i)+\frac{i}{2}\cot(-\pi i)\right)+1\right) = \frac{1+\pi\coth(\pi)}{2}
$$But then the book claims  $\displaystyle{\sum_{n=0}^{\infty}\frac{n}{n^3+1}}$ can be evaluated as well. I am skeptical because the singularity occurs at $n=-1$, as opposed to $\pm i$ in the previous problem, and I don't think the same symmetry trick can work; further, this is the only relevant section of the chapter to this exercise. Mathematica gives the value
$$
\frac{-1}{3}\sum_{k=0}^2 \frac{\psi(-\exp(2\pi i (2k+1)/6))}{\exp(2\pi i (2k+1)/6)},
$$where $\psi$ is the digamma function, which is less than transparent, though I could buy how we could get there from the cotangent sum. I'm not sure if the sum has a nicer closed-form or if this is as best as can be done.
 A: Wolfram is more elegant:
$$ \sum_{n=0}^\infty \frac{n}{n^3+1} = -3^{-1}\sum_{\{\omega \in \mathbb{C}| \omega^3 +1 =0\}} \psi (-\omega)\omega^{-1} $$
where
$$\{\omega \in \mathbb{C}| \omega^3 +1 =0\} =\{-1, 1/2+i\sqrt{3}/2, 1/2-i\sqrt{3}/2 \}$$
Also: $$ \psi(z) -\psi(1-z) = - \pi \cot (\pi z) $$
and
$$ \psi(1+z) -\psi(z) = 1/z $$
More digamma function properties on wiki.
Residues (Wolfram again) are: $-1/3$ at $-1$, $(-i+\sqrt{3})/(3(i+\sqrt{3}))$ at $1/2+i\sqrt{3}/2$, and $(i+\sqrt{3})/(3(-i+\sqrt{3}))$ at $1/2-i\sqrt{3}/2$.
It turns out that:
$$ \cot (\pi(1/2+i\sqrt{3}/2)) = -i\tanh(\sqrt{3}\pi/2)$$
and
$$ \cot (\pi(1/2-i\sqrt{3}/2)) = i\tanh(\sqrt{3}\pi/2)$$
So:
$$ (-i+\sqrt{3})/(3(i+\sqrt{3}))\cdot [-i\tanh(\sqrt{3}\pi/2)]
+ (i+\sqrt{3})/(3(-i+\sqrt{3})) \cdot [i\tanh(\sqrt{3}\pi/2)]$$
$$ =-\tanh(\sqrt{3}\pi/2) /\sqrt{3}\ $$
Edit: I'm not sure how we ended up multiplying residues of $f$ at poles and $\cot$ at those poles (times $\pi$). From proofwiki (Marsden and Hoffman book), we need:
$$-\pi \sum_{z_0 {\rm \; pole \; of} \; f(z)} {\rm Res}(\cot (\pi z)f (z), z_0)$$
In any case, this is necessary to deal with pole $-1$ and $f(z)=z/(z^3+1)$:
$${\rm Res}(\cot (\pi z)f (z), -1) = 0 $$
A: Maple says $$ \sum_{n=1}^\infty \frac{n}{n^3+1} = {\frac {\pi\,\tanh \left( {\frac {\pi\,\sqrt {3}}{2}} \right)  \left( 
i+\sqrt {3} \right) }{6}}-{\frac {\gamma}{3}}-{\frac {\Psi \left( {
\frac{1}{2}}+{\frac {i}{2}}\sqrt {3} \right) }{3}}+{\frac{1}{3}}
$$
A: C0mment
A question with a much simpler answer is
$$
\sum \frac{n}{n^3+1}
 = \frac{\pi}{\sqrt 3}\tanh\frac{\pi\sqrt{3}}{2}
$$
where the sum is over all $n \in \mathbb Z$ except $n=-1$.
As an exercise you could try doing this using your contour cotangent method.
