[2025年東京理科大学理学部数学科]出題ミス

\begin{align*} \sum_{k=2}^{\infty}\frac{2}{k^{2}+3k-2} & =\sum_{k=2}^{\infty}\frac{2}{\left(k+\frac{3}{2}\right)^{2}-\frac{9}{4}-2}\\ & =\sum_{k=2}^{\infty}\frac{2}{\left(k+\frac{3}{2}\right)^{2}-\frac{17}{4}}\\ & =\sum_{k=-1}^{\infty}\frac{2}{\left(k+\frac{3}{2}\right)^{2}-\frac{17}{4}}-\sum_{k=-1}^{1}\frac{2}{\left(k+\frac{3}{2}\right)^{2}-\frac{17}{4}}\\ & =\sum_{k=1}^{\infty}\frac{2}{\left(k-2+\frac{3}{2}\right)^{2}-\frac{17}{4}}-\left(\frac{2}{\left(-1+\frac{3}{2}\right)^{2}-\frac{17}{4}}+\frac{2}{\left(0+\frac{3}{2}\right)^{2}-\frac{17}{4}}+\frac{2}{\left(1+\frac{3}{2}\right)^{2}-\frac{17}{4}}\right)\\ & =\sum_{k=1}^{\infty}\frac{2}{\left(k-\frac{1}{2}\right)^{2}-\frac{17}{4}}-\left(\frac{2}{\frac{1}{4}-\frac{17}{4}}+\frac{2}{\frac{9}{4}-\frac{17}{4}}+\frac{2}{\frac{25}{4}-\frac{17}{4}}\right)\\ & =2\sum_{k=1}^{\infty}\frac{1}{\left(k-\frac{1}{2}\right)^{2}-\left(\frac{\sqrt{17}}{2}\right)^{2}}-\left(-\frac{2\cdot4}{16}-\frac{2\cdot4}{8}+\frac{2\cdot8}{8}\right)\\ & =\frac{2\pi}{\sqrt{17}}\cdot\frac{2\frac{\sqrt{17}}{2}}{\pi}\sum_{k=1}^{\infty}\frac{1}{\left(k-\frac{1}{2}\right)^{2}-\left(\frac{\sqrt{17}}{2}\right)^{2}}-\left(-\frac{1}{2}-1+1\right)\\ & =\frac{2\pi}{\sqrt{17}}\tan\frac{\sqrt{17}\pi}{2}+\frac{1}{2}\cmt{\because\tan\pi z=\frac{2z}{\pi}\sum_{k=1}^{\infty}\frac{1}{\left(k-\frac{1}{2}\right)^{2}-z^{2}}}\\ & =\frac{2\pi}{\sqrt{17}}\tan\frac{\sqrt{17}\pi}{2}+\frac{1}{2}\\ & =\frac{2\pi}{\sqrt{17}}\tan\frac{\sqrt{17}\pi}{2}+\frac{1}{2}\\ & =\frac{1}{2}+\frac{2\pi}{\sqrt{17}}\tan\frac{\sqrt{17}\pi}{2} \end{align*}