[Math] How to find sum of infinite series of non-geometric series

Question

I am familiar with radius of convergence, power series, Taylor series/Maclaurin series, and fundamental infinite series convergence tests (ratio, root, integral, comparison, etc.) introduced in a first semester college Calculus class. I have always had to determine convergence or divergence for most non-geometric or arithmetic series but I do not know a process to find $$\sum\limits_{k=1}^\infty \frac{(-1)^{k+1}k^2}{k^3+1}. $$ $\DeclareMathOperator{\sech}{sech}$
I have put the answer through Wolfram|Alpha and got $$\frac{1}{3}\left ( 1-\ln(2)+\pi \sech\left ( \frac{\sqrt3}{2}\pi \right ) \right ).$$

Answer 1

$\newcommand{\sech}{\operatorname{sech}}$If we set $\alpha=\frac{-1+i\sqrt3}2$, we get $$ \frac{1/3}{k+1}+\frac{1/3}{k+\alpha}+\frac{1/3}{k+\overline\alpha}=\frac{k^2}{k^3+1}\tag{0} $$ Therefore, $$ \begin{align} \sum_{k=1}^\infty(-1)^{k+1}\frac{k^2}{k^3+1} &=\frac13\sum_{k=1}^\infty(-1)^{k+1}\left(\color{#C00000}{\frac1{k+1}}+\color{#00A000}{\frac1{k+\alpha}+\frac1{k+\overline\alpha}}\right)\tag{1}\\ &=\frac13\left(\color{#C00000}{1-\log(2)}+\color{#00A000}{\sum_{k\in\mathbb{Z}}\frac{(-1)^{k+1}}{k-\frac12+i\frac{\sqrt3}2}}\right)\tag{2}\\ &=\frac13\left(1-\log(2)+\pi\csc\!\left(\pi\left(\frac12-i\frac{\sqrt3}2\right)\right)\right)\tag{3}\\[6pt] &=\frac13\left(1-\log(2)+\pi\sech\!\left(\pi\frac{\sqrt3}2\right)\right)\tag{4} \end{align} $$ Explanation:
$(1)$: use the partial fractions from $(0)$
$(2)$: sum of the alternating harmonic series is $\log(2)$
$\hphantom{\text{(2):}}$ and rewrite two unidirectional sums as a bidirectional sum
$(3)$: use $(6)$ from this answer
$(4)$: $\sec\left(\pi i\frac{\sqrt3}2\right)=\sech\left(\pi\frac{\sqrt3}2\right)$