数学 - 無限の世界への一歩 - 数列の極限、無限級数 – 極限の計算 - 無限等比数列{r^n}の極限(漸化式)

学習環境

数学読本〈4〉数列の極限，無限級数/順列・組合せ/確率/関数の極限と微分法(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)、14.2(極限の計算)、無限等比数列{r^n}の極限、問18.を取り組んでみる。

問18

$\begin{array}{l} x = \frac{1}{3} x + 1 \\ x = \frac{3}{2} \\ a_{n + 1} - \frac{3}{2} = \frac{1}{3} (a_{n} - \frac{3}{2}) \\ b_{n} = a_{n} - \frac{3}{2} \\ b_{1} = 1 - \frac{3}{2} = - \frac{1}{2} \\ b_{n} = - \frac{1}{2} {(\frac{1}{3})}^{n - 1} \\ a_{n} = b_{n} + \frac{3}{2} \\ \lim_{n \to \infty} a_{n} = \frac{3}{2} \\ 別解 \\ y = \frac{1}{3} x + 1 \\ y = x \\ x = \frac{x}{3} + 1 \\ x = \frac{3}{2} \\ \lim_{n \to \infty} a_{n} = \frac{3}{2} \end{array}$
$\begin{array}{l} x = 9 - \frac{x}{2} \\ x = 6 \\ a_{n + 1} - 6 = - \frac{1}{2} (a_{n} - 6) \\ b_{n} = a_{n} - 6 \\ b_{1} = a_{1} - 6 = 3 - 6 = - 3 \\ b_{n} = - 3 {(- \frac{1}{2})}^{n - 1} \\ a_{n} = - 3 {(- \frac{1}{2})}^{n - 1} + 6 \\ \lim_{n \to \infty} a_{n} = 6 \end{array}$
$\begin{array}{l} x = \frac{3}{5} x - 2 \\ x = - 5 \\ a_{n + 1} + 5 = \frac{3}{5} (a_{n} + 5) \\ b_{n} = a_{n} + 5 \\ b_{1} = 2 + 5 = 7 \\ b_{n} = 7 {(\frac{3}{5})}^{n - 1} \\ a_{n} = 7 {(\frac{3}{5})}^{n - 1} - 5 \\ \lim_{n \to \infty} a_{n} = - 5 \end{array}$
$\begin{array}{l} x = 2 x + 1 \\ x = - 1 \\ a_{n + 1} + 1 = 2 (a_{n} + 1) \\ b_{n} = a_{n} + 1 \\ b_{1} = a_{1} + 1 = 2 \\ b_{n} = 2 \cdot 2^{n - 1} = 2^{n} \\ a_{n} = b_{n} - 1 \\ \lim_{x \to \infty} 2^{n} - 1 = + \infty \end{array}$

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