分母に3次式の総和

\begin{align*} \sum_{k=1}^{\infty}\frac{1}{\left(4k\right)^{3}-4k} & =\sum_{k=1}^{\infty}\frac{1}{4k\left(\left(4k\right)^{2}-1\right)}\\ & =\sum_{k=1}^{\infty}\frac{1}{4k\left(4k-1\right)\left(4k+1\right)}\\ & =\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{4k}\left(\frac{1}{4k-1}-\frac{1}{4k+1}\right)\\ & =\frac{1}{2}\sum_{k=1}^{\infty}\left(\frac{1}{4k-1}-\frac{1}{4k}-\frac{1}{4k}+\frac{1}{4k+1}\right)\\ & =\frac{1}{2}\sum_{k=1}^{\infty}\left(\frac{1}{4k-1}-\frac{1}{4k}+\frac{1}{4k+1}-\frac{1}{4k+2}-\frac{1}{4k}+\frac{1}{4k+2}\right)\\ & =\frac{1}{2}\sum_{k=1}^{\infty}\left(\frac{1}{4k-1}-\frac{1}{4k}+\frac{1}{4k+1}-\frac{1}{4k+2}\right)-\frac{1}{2}\sum_{k=1}^{\infty}\left(\frac{1}{4k}-\frac{1}{4k+2}\right)\\ & =\frac{1}{2}\sum_{k=3}^{\infty}\frac{\left(-1\right)^{k+1}}{k}-\frac{1}{4}\sum_{k=1}^{\infty}\left(\frac{1}{2k}-\frac{1}{2k+1}\right)\\ & =\frac{1}{2}\left(\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}-\left(1-\frac{1}{2}\right)\right)-\frac{1}{4}\sum_{k=2}^{\infty}\left(\frac{\left(-1\right)^{k}}{k}\right)\\ & =\frac{1}{2}\left(\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}-\left(1-\frac{1}{2}\right)\right)-\frac{1}{4}\sum_{k=1}^{\infty}\left(\frac{\left(-1\right)^{k}}{k}+1\right)\\ & =\left(\frac{1}{2}+\frac{1}{4}\right)\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k}-\frac{1}{2}\left(1-\frac{1}{2}\right)-\frac{1}{4}\\ & =\frac{3}{4}\log2-\frac{1}{2} \end{align*}