無限の世界への一歩 - 数列の極限、無限級数無限級数の和の計算部分和の極限の和、差

学習環境

新装版数学読本4 (松坂和夫(著)、岩波書店)の第14章(無限の世界への一歩 - 数列の極限、無限級数)、14.3(無限級数)、無限級数の和の計算の問33の解答を求めてみる。

\begin{array}{l} \sum_{n = 1}^{\infty} \frac{2^{n} + 1}{3^{n}} = \sum_{n = 1}^{\infty} ({(\frac{2}{3})}^{n} + {(\frac{1}{3})}^{n}) \\ \sum_{n = 1}^{\infty} {(\frac{2}{3})}^{n} = \frac{2}{3} \cdot \frac{1}{1 - \frac{2}{3}} = 2 \\ \sum_{n = 1}^{\infty} {(\frac{1}{3})}^{n} = \frac{1}{3} \cdot \frac{1}{1 - \frac{1}{3}} = \frac{1}{2} \\ \sum_{n = 1}^{\infty} \frac{2^{n} + 1}{3^{n}} = 2 + \frac{1}{2} = \frac{5}{2} \end{array}

\begin{array}{l} \sum_{n = 1}^{\infty} \frac{5^{n} - 2^{n}}{1 0^{n}} = \sum_{n = 1}^{\infty} ({(\frac{1}{2})}^{n} - {(\frac{1}{5})}^{n}) \\ \sum_{n = 1}^{\infty} {(\frac{1}{2})}^{n} = \frac{1}{2} \cdot \frac{1}{1 - \frac{1}{2}} = 1 \\ \sum_{n = 1}^{\infty} {(\frac{1}{5})}^{n} = \frac{1}{5} \cdot \frac{1}{1 - \frac{1}{5}} = \frac{1}{4} \\ \sum_{n = 1}^{\infty} \frac{5^{n} - 2^{n}}{1 0^{n}} = 1 - \frac{1}{4} = \frac{3}{4} \end{array}

\begin{array}{l} \sum_{n = 1}^{\infty} (3^{n} - 2^{n}) x^{n - 1} = \sum_{n = 1}^{\infty} (3 {(3 x)}^{n - 1} - 2 {(2 x)}^{n - 1}) \\ \sum_{n = 1}^{\infty} 3 {(3 x)}^{n - 1} = \frac{3}{1 - 3 x} \\ \sum_{n = 1}^{\infty} 2 {(2 x)}^{n - 1} = \frac{2}{1 - 2 x} \\ \sum_{n = 1}^{\infty} (3^{n} - 2^{n}) x^{n - 1} = \frac{3}{1 - 3 x} - \frac{2}{1 - 2 x} \\ = \frac{1}{(3 x - 1) (2 x - 1)} \end{array}