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 $1-\log(2)$
 A: A route. One may recall the standard series representation of the digamma function,
$$
\psi(z+1)+\gamma=\sum_{n=1}^\infty\left(\frac1n-\frac1{n+z}\right),\quad \text{Re}\: z>-1,
$$
giving
$$
2\sum _{n=1}^{\infty } \frac{(-1)^n}{n+z}=\psi\left(\frac{z}{2}+\frac{1}{2}\right)-\psi\left(\frac{z}{2}+1\right),\quad \text{Re}\: z>-1,
$$ then by writting
$$
(-1)^{n+1}\frac{n^2}{n^3+1}=a\cdot\frac{(-1)^{n+1}}{n+1}+b\cdot\frac{(-1)^{n+1}}{n+e^{2i\pi/3}}+\bar{b}\cdot\frac{(-1)^{n+1}}{n+e^{-2i\pi/3}}
$$ one may conclude with special values of the digamma function.
A: 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.
A: $$
\begin{align}
&\sum_{n=1}^\infty(-1)^{n-1}\frac{n^2}{n^3+1}\\
&=\frac13\sum_{n=1}^\infty(-1)^{n-1}\left(\frac1{n+1}+\frac1{n+e^{2\pi i/3}}+\frac1{n+e^{-2\pi i/3}}\right)\tag{1}\\
&=\frac13\sum_{n=1}^\infty(-1)^{n-1}\left(\frac1{n+1}+\frac1{n+e^{2\pi i/3}}+\frac1{n-1-e^{2\pi i/3}}\right)\tag{2}\\
&=\frac13\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n+1}
-\frac13\sum_{n=-\infty}^\infty\frac{(-1)^n}{n+e^{2\pi i/3}}\tag{3}\\
&=\frac{1-\log(2)}3-\frac\pi3\csc\left(\pi e^{2\pi i/3}\right)\tag{4}\\
&=\frac{1-\log(2)}3-\frac\pi3\csc\left(-\frac\pi2+i\frac{\pi\sqrt3}2\right)\tag{5}\\
&=\bbox[5px,border:2px solid #C0A000]{\frac{1-\log(2)}3+\frac\pi3\operatorname{sech}\left(\frac{\pi\sqrt3}2\right)}\tag{6}
\end{align}
$$
Explanation:
$(1)$: partial fractions
$(2)$: $1+e^{2\pi i/3}+e^{-2\pi i/3}=0$
$(3)$: $n\mapsto1-n$ sends $\frac{(-1)^n}{n-1-e^{2\pi i/3}}\mapsto\frac{(-1)^n}{n+e^{2\pi i/3}}$ and $\{1,2,3,\dots\}\mapsto\{0,-1,-2,\dots\}$
$(4)$: Equation $(3)$ from this answer
$(5)$: expand
$(6)$: evaluate
