数列と関数数列の極限小数展開循環小数、有理数で表現

学習環境

現代数学への入門微分と積分1 初等関数を中心に (青本和彦(著)、岩波書店)の第1章(数列と関数)、1.2(数列の極限)、a(小数展開)の問14の解答を求めてみる。

\begin{array}{l} 0.597373 \dots \\ = \frac{5}{10} + \frac{9}{1 0^{2}} + \frac{7}{1 0^{3}} + \frac{3}{1 0^{4}} + \dots \\ = \frac{5}{10} + \frac{9}{1 0^{2}} + \lim_{n \to \infty} \sum_{k = 0}^{n} \frac{7}{1 0^{3 + 2 k}} + \lim_{n \to \infty} \sum_{k = 0}^{n} \frac{3}{1 0^{4 + 2 k}} \\ = \frac{5}{10} + \frac{9}{100} + \frac{7}{1 0^{3}} \cdot \frac{1}{1 - \frac{1}{1 0^{2}}} + \frac{3}{1 0^{4}} \cdot \frac{1}{1 - \frac{1}{1 0^{2}}} \\ = \frac{5}{10} + \frac{9}{1 0^{2}} + \frac{7}{1 0^{3}} \cdot \frac{1 0^{2}}{99} + \frac{3}{1 0^{4}} \cdot \frac{1 0^{2}}{99} \\ = \frac{5}{10} + \frac{9}{100} + \frac{7}{990} + \frac{1}{3300} \\ = \frac{4950 + 891 + 70 + 3}{9900} \\ = \frac{5914}{9900} \\ = \frac{2957}{4950} \end{array}

\begin{array}{l} 0.4169169 \dots \\ = \frac{4}{10} + \frac{1}{1 0^{2}} + \frac{6}{1 0^{3}} + \frac{9}{1 0^{4}} + \dots \\ = \frac{4}{10} + \lim_{n \to \infty} \sum_{k = 0}^{n} (\frac{1}{1 0^{3 k + 2}} + \frac{6}{1 0^{3 k + 3}} + \frac{9}{1 0^{3 k + 4}}) \\ = \frac{4}{10} + \frac{1}{1 0^{2}} \cdot \frac{1}{1 - \frac{1}{1 0^{3}}} + \frac{6}{1 0^{3}} \cdot \frac{1}{1 - \frac{1}{1 0^{3}}} + \frac{9}{1 0^{4}} \cdot \frac{1}{1 - \frac{1}{1 0^{3}}} \\ = \frac{4}{10} + \frac{1}{1 0^{2}} \cdot \frac{1 0^{3}}{999} + \frac{6}{1 0^{3}} \cdot \frac{1 0^{3}}{999} + \frac{9}{1 0^{4}} \cdot \frac{1 0^{3}}{999} \\ = \frac{4}{10} + \frac{10}{999} + \frac{6}{999} + \frac{9}{9990} \\ = \frac{3996 + 100 + 60 + 9}{9990} \\ = \frac{4165}{9990} \\ = \frac{833}{1998} \end{array}

\begin{array}{l} 0.2199 \dots \\ = \frac{2}{10} + \frac{1}{1 0^{2}} + \sum_{k = 0}^{\infty} \frac{9}{1 0^{k + 3}} \\ = \frac{2}{10} + \frac{1}{1 0^{2}} + \frac{9}{1 0^{3}} \cdot \frac{1}{1 - \frac{1}{10}} \\ = \frac{2}{10} + \frac{1}{1 0^{2}} + \frac{9}{1 0^{3}} \cdot \frac{10}{9} \\ = \frac{2}{10} + \frac{1}{1 0^{2}} + \frac{1}{1 0^{2}} \\ = \frac{22}{100} \\ = \frac{11}{50} \end{array}