40. Compute
$$
\sum_{k=1}^{\infty} \frac{3 k+1}{2 k^{3}+k^{2}} \cdot(-1)^{k+1}
$$
Answer: $\frac{\pi^{2}}{12}+\frac{\pi}{2}-2+\ln 2$
Solution: Via partial fraction decomposition we write the sum as
$$
\sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{2}{1+2 k}+\frac{1}{k^{2}}\right)(-1)^{k+1}
$$

Now recall that
$$
\begin{aligned}
\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6} & =S_{1} \\
\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}=\ln (2) & =S_{2} \\
\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{2 k-1}=\frac{\pi}{4} & =S_{3}
\end{aligned}
$$

Manipulating (1), we deduce
$$
\begin{aligned}
\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{2}} & =\left(\sum_{k=1}^{\infty} \frac{1}{k^{2}}\right)-2 \cdot \sum_{k=1}^{\infty} \frac{1}{(2 k)^{2}} \\
=\frac{\pi^{2}}{6}-2 / 4 \cdot \frac{\pi^{2}}{6}=\frac{\pi^{2}}{12} & =S_{4}
\end{aligned}
$$

It is then easily seen that the answer is equal to $S_{2}+2 \cdot S_{3}-2+S_{4}$.