<s>[INST] How to find the sum of this series: $\sum_{n=1}^{\infty}(-1)^{n+1} \frac{n^2}{n^3+1}$? $$\frac{1^2}{1^3+1}-\frac{2^2}{2^3+1}+\frac{3^2}{3^3+1}-\frac{4^2}{4^3+1}+\cdots$$
in terms of summation i can write it as 
$$S_{n}=\sum_{n=1}^{\infty}(-1)^{n+1} \frac{n^2}{n^3+1}$$
How to continue from this point? 
used partial fraction:
$$\frac{n^2}{n^3+1}=\frac{1}{3}\cdot\frac{2n-1}{n^2-n+1}+\frac{1}{3}\cdot \frac{1}{n+1}$$
I'm stuck with the first term in the partial fraction , the second term simply yields $1-\log(2)$
 [/INST] From partial fractions, 
$$ \frac{n^2}{n^3 + 1} = \frac{1}{3(n+1)} + \frac{1}{3(n-r)} + \frac{1}{3(n-\overline{r})} $$
where $r$ is a root of $z^2 - z + 1$.
Now $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n +1} = 1 - \log(2) $$
while
$$ \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n - r} = \frac{1}{2} \Psi\left(1 - \frac{r}{2}\right) - \frac{1}{2} \Psi\left(\frac{1}{2} -\frac{r}{2}\right) $$ 
Thus your sum becomes
$$\frac{1-\log(2)}{3} + \frac{1}{6}\left(\Psi \left(\frac{3+i \sqrt{3}}{4}\right)+\Psi \left(\frac{3-i \sqrt{3}}{4}\right) - \Psi \left(\frac{1+i \sqrt{3}}{4}\right)  - \Psi \left(\frac{1-i \sqrt{3}}{4}\right)\right) 
$$
Now using the identity $\Psi(1-x) = \Psi(x)+\pi \cot(\pi x)$, the sum becomes
$$ \frac{1-\log(2)}{3} - \frac{\pi}{6} \cot\left(\frac{\pi}{4} (3 + i \sqrt{3})\right) - \frac{\pi}{6} \cot\left(\frac{\pi}{4} (3 - i \sqrt{3})\right) $$
and this simplifies to Dr. Graubner's answer.
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