概念問題 解答

微積分学　極限

問題4の解答

${\displaystyle 0.999\cdots }$

${\displaystyle =0.9+0.09+0.009+\cdots }$

${\displaystyle =\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\cdots }$

${\displaystyle =\frac{9}{10}+\frac{9}{10}\cdot \frac{1}{10}+\frac{9}{10}{\left(\frac{1}{10}\right)}^2+\cdots }$

${\displaystyle =\underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\frac{9}{10}}{\left(\frac{1}{10}\right)}^{k-1}}$

よって，${\displaystyle 0.999\cdots }$は 初項が${\displaystyle \frac{9}{10}}$，公比が${\displaystyle \frac{1}{10}}$ の等比数列の和（級数）になる．また,

${\displaystyle \underset{n\to \infty }{\lim }{\displaystyle \sum _{k=1}^n\frac{9}{10}}{\left(\frac{1}{10}\right)}^{k-1}}$${\displaystyle =\underset{n\to \infty }{\lim }\frac{\frac{9}{10}\left\{1-{\left(\frac{1}{10}\right)}^n\right\}}{1-\frac{1}{10}}}$${\displaystyle =\frac{\frac{9}{10}}{1-\frac{1}{10}}=1}$

である．

以上より，答えは3となる．