高専の数学解説

第2章 2.4

(1)

∑n=1∞(12)ncos⁡nπ2=−14+116−164+⋯\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n \cos\frac{n\pi}{2} = -\frac{1}{4} + \frac{1}{16} - \frac{1}{64} + \cdots

これは初項 a=−14a = -\dfrac{1}{4}、公比 r=−14r = -\dfrac{1}{4} の無限等比級数なので

∑n=1∞(12)ncos⁡nπ2=a1−r=−141−(−14)=−14−(−1)=−15\sum_{n=1}^{\infty} \left(\frac{1}{2}\right)^n \cos\frac{n\pi}{2} = \frac{a}{1-r} = \frac{-\frac{1}{4}}{1-\left(-\frac{1}{4}\right)} = \frac{-1}{4-(-1)} = -\frac{1}{5}

(2)

∑n=1∞(14n−15n)=∑n=1∞14n−∑n=1∞15n=141−14−151−15=14−1−15−1=13−14=112\sum_{n=1}^{\infty} \left(\frac{1}{4^n} - \frac{1}{5^n}\right) = \sum_{n=1}^{\infty} \frac{1}{4^n} - \sum_{n=1}^{\infty} \frac{1}{5^n} = \frac{\frac{1}{4}}{1-\frac{1}{4}} - \frac{\frac{1}{5}}{1-\frac{1}{5}} = \frac{1}{4-1} - \frac{1}{5-1} = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}

(3)

∑n=1∞1(2n+1)2−1=∑n=1∞14n2+4n+1−1=14∑n=1∞1n(n+1)=14∑n=1∞(1n−1n+1)\sum_{n=1}^{\infty} \frac{1}{(2n+1)^2 - 1} = \sum_{n=1}^{\infty} \frac{1}{4n^2+4n+1-1} = \frac{1}{4}\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \frac{1}{4}\sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1}\right)

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

(4)

∑n=1∞11+2+3+⋯+n=∑n=1∞1n(n+1)2=∑n=1∞2n(n+1)=2∑n=1∞(1n−1n+1)=2\sum_{n=1}^{\infty} \frac{1}{1+2+3+\cdots+n} = \sum_{n=1}^{\infty} \frac{1}{\frac{n(n+1)}{2}} = \sum_{n=1}^{\infty} \frac{2}{n(n+1)} = 2\sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1}\right) = 2

解説: r31bn1z