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 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.
