高専の数学解説

第2章 2.16

(1)

∑n=1∞1n2+2n=∑n=1∞1n(n+2)=12∑n=1∞(1n−1n+2)=12(1−13+12−14+13+15−⋯ )\sum_{n=1}^{\infty} \frac{1}{n^2+2n} = \sum_{n=1}^{\infty} \frac{1}{n(n+2)} = \frac{1}{2}\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+2}\right) = \frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}+\frac{1}{5}-\cdots\right)

=12(1+12)=34= \frac{1}{2}\left(1+\frac{1}{2}\right) = \frac{3}{4}

(2)

∑n=1∞1n(n+1)(n+2)=12∑n=1∞(1n−2n+1+1n+2)=12∑n=1∞(1n−1n+1)+12∑n=1∞(−1n+1+1n+2)\sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)} = \frac{1}{2}\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{2}{n+1}+\frac{1}{n+2}\right) = \frac{1}{2}\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right) + \frac{1}{2}\sum_{n=1}^{\infty}\left(-\frac{1}{n+1}+\frac{1}{n+2}\right)

=12(1−12+12−13+13−14+⋯ )+12(−12+13−13+14−14+15−⋯ )= \frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots\right) + \frac{1}{2}\left(-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+\frac{1}{5}-\cdots\right)

12+12⋅(−12)=14\frac{1}{2}+\frac{1}{2}\cdot\left(-\frac{1}{2}\right) = \frac{1}{4}

解説: r31bn1z