TITLE: How to find the sum of this series: $\sum_{n=1}^{\infty}(-1)^{n+1} \frac{n^2}{n^3+1}$?
QUESTION [11 upvotes]: $$\frac{1^2}{1^3+1}-\frac{2^2}{2^3+1}+\frac{3^2}{3^3+1}-\frac{4^2}{4^3+1}+\cdots$$
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 $1-\log(2)$

REPLY [13 votes]: $$
\begin{align}
&\sum_{n=1}^\infty(-1)^{n-1}\frac{n^2}{n^3+1}\\
&=\frac13\sum_{n=1}^\infty(-1)^{n-1}\left(\frac1{n+1}+\frac1{n+e^{2\pi i/3}}+\frac1{n+e^{-2\pi i/3}}\right)\tag{1}\\
&=\frac13\sum_{n=1}^\infty(-1)^{n-1}\left(\frac1{n+1}+\frac1{n+e^{2\pi i/3}}+\frac1{n-1-e^{2\pi i/3}}\right)\tag{2}\\
&=\frac13\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n+1}
-\frac13\sum_{n=-\infty}^\infty\frac{(-1)^n}{n+e^{2\pi i/3}}\tag{3}\\
&=\frac{1-\log(2)}3-\frac\pi3\csc\left(\pi e^{2\pi i/3}\right)\tag{4}\\
&=\frac{1-\log(2)}3-\frac\pi3\csc\left(-\frac\pi2+i\frac{\pi\sqrt3}2\right)\tag{5}\\
&=\bbox[5px,border:2px solid #C0A000]{\frac{1-\log(2)}3+\frac\pi3\operatorname{sech}\left(\frac{\pi\sqrt3}2\right)}\tag{6}
\end{align}
$$
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