第2章 2.3

(1)

初項 $a = 1$ 公比 $r = \dfrac{2}{3}$ なので $a_n = \left(\dfrac{2}{3}\right)^{n-1}$

$|r| < 1$ なので収束

$\sum_{n=1}^{\infty} a_n = \frac{a}{1-r} = \frac{1}{1-\dfrac{2}{3}} = 3$

(2)

$a = 1$ $r = -\dfrac{1}{2}$ なので $a_n = \left(-\dfrac{1}{2}\right)^{n-1}$

$|r| < 1$ なので収束

$\sum_{n=1}^{\infty} a_n = \frac{1}{1-\left(-\dfrac{1}{2}\right)} = \frac{2}{3}$

(3)

$a = \dfrac{1}{4}$ $r = \dfrac{4}{3}$ なので $a_n = \dfrac{1}{4}\left(\dfrac{4}{3}\right)^{n-1}$

$|r| > 1$ なので発散

(4)

$a = 1$ $r = \dfrac{1}{\sqrt{2}}$ なので $a_n = \dfrac{1}{\sqrt{2}^{n-1}}$

$|r| < 1$ なので収束

$\sum_{n=1}^{\infty} a_n = \frac{1}{1-\dfrac{1}{\sqrt{2}}} = \frac{\sqrt{2}}{\sqrt{2}-1} = \sqrt{2}(\sqrt{2}+1) = 2+\sqrt{2}$