How to find sum of infinite series of non-geometric series? 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 $$\frac{1}{3}\left ( 1-\ln(2)+\pi \sech\left ( \frac{\sqrt3}{2}\pi \right ) \right ).$$
 A: $\newcommand{\sech}{\operatorname{sech}}$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
$(4)$: $\sec\left(\pi i\frac{\sqrt3}2\right)=\sech\left(\pi\frac{\sqrt3}2\right)$
A: The first step is to  decompose $ \frac{k^2}{k^3+1}$ as $\frac13\frac{1}{k+1}+\frac13\frac{2k-1}{k^2-k+1}$. 
The first sum is easily evaluated as $$\frac13\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k+1}=\frac13(1-\ln2).$$
The second one,  $$\frac13\sum_{k=1}^{\infty} \frac{(-1)^{k+1}(2k-1)}{k^2-k+1}=\frac13\pi\operatorname{sech}\frac{\sqrt{3}\pi}{2}$$
was already done on this site.
It follows from taking the logarithmic derivative of the Weierstrass product form of cosine : $$\cos(\pi x)=\prod_{k=0}^{\infty}\left( 1-\frac{4x^2}{(2k+1)^2}\right)$$
