How to find the sum of this series: $\sum_{n=1}^{\infty}(-1)^{n+1} \frac{n^2}{n^3+1}$?

Question

$$\frac{1^2}{1^3+1}-\frac{2^2}{2^3+1}+\frac{3^2}{3^3+1}-\frac{4^2}{4^3+1}+\cdots$$ 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 $1-\log(2)$

Answer 1

$$ \begin{align} &\sum_{n=1}^\infty(-1)^{n-1}\frac{n^2}{n^3+1}\\ &=\frac13\sum_{n=1}^\infty(-1)^{n-1}\left(\frac1{n+1}+\frac1{n+e^{2\pi i/3}}+\frac1{n+e^{-2\pi i/3}}\right)\tag{1}\\ &=\frac13\sum_{n=1}^\infty(-1)^{n-1}\left(\frac1{n+1}+\frac1{n+e^{2\pi i/3}}+\frac1{n-1-e^{2\pi i/3}}\right)\tag{2}\\ &=\frac13\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n+1} -\frac13\sum_{n=-\infty}^\infty\frac{(-1)^n}{n+e^{2\pi i/3}}\tag{3}\\ &=\frac{1-\log(2)}3-\frac\pi3\csc\left(\pi e^{2\pi i/3}\right)\tag{4}\\ &=\frac{1-\log(2)}3-\frac\pi3\csc\left(-\frac\pi2+i\frac{\pi\sqrt3}2\right)\tag{5}\\ &=\bbox[5px,border:2px solid #C0A000]{\frac{1-\log(2)}3+\frac\pi3\operatorname{sech}\left(\frac{\pi\sqrt3}2\right)}\tag{6} \end{align} $$ 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

Answer 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.

Answer 3

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.