4.1 級数の定義

|bf 1.

(a) 等比数列の和

$\displaystyle \sum_{n=0}^{\infty}\frac{3}{10^n}$	$\displaystyle =$	$\displaystyle 3\sum_{n=0}^{\infty}(\frac{1}{10})^{n}$
	$\displaystyle =$	$\displaystyle 3\cdot \frac{1}{1 - \frac{1}{10}} = 3 \cdot \frac{10}{9} = \frac{10}{3}$

(b)

$\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{n}}{5^n}$	$\displaystyle =$	$\displaystyle 1 - \frac{1}{5} + \frac{1}{25} - \frac{1}{36} + \cdots$
$\displaystyle S_{n}$	$\displaystyle =$	$\displaystyle 1 - \frac{1}{5} + \frac{1}{25} - \frac{1}{36} + \cdots + \frac{(-1)^{n}}{5^{n}}$
$\displaystyle +\frac{1}{5}S_{n}$	$\displaystyle =$	$\displaystyle \frac{1}{5} - \frac{1}{25} + \cdots + \frac{(-1)^{n+1}}{5^{n+1}}$
$\displaystyle \hline \frac{6}{5}S_{n}$	$\displaystyle =$	$\displaystyle 1 + \frac{(-1)^{n}}{5^{n+1}}$
$\displaystyle S_{n}$	$\displaystyle =$	$\displaystyle \frac{5}{6} + \frac{(-1)^{n}}{6\cdot 5^{n}}, n \geq 1$
$\displaystyle S$	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}S_{n} = \frac{5}{6}$

(c)

$\displaystyle \sum_{n=0}^{\infty}\frac{1 - 2^{n}}{3^n}$	$\displaystyle =$	$\displaystyle \sum_{n=0}^{\infty}\frac{1}{3^n} - \sum_{n=0}^{\infty}(\frac{2}{3})^{n}$
	$\displaystyle =$	$\displaystyle \frac{1}{1 - \frac{1}{3}} - \frac{1}{1 - \frac{2}{3}}$
	$\displaystyle =$	$\displaystyle \frac{3}{3-1} - \frac{3}{3-2} = \frac{3}{2} - 3 = \frac{-3}{2}$

2.

(a)

$\displaystyle 1.3333\ldots$	$\displaystyle =$	$\displaystyle 1 + \frac{3}{10} + \frac{3}{10^2} + \frac{3}{10^3} + \cdots$
	$\displaystyle =$	$\displaystyle 1 + \sum_{n=1}^{\infty}(\frac{3}{10})^{n} = 1 + \frac{\frac{3}{10}}{1 - \frac{3}{10}}$
	$\displaystyle =$	$\displaystyle 1 + \frac{3}{7} = \frac{10}{7}$

(b)

$\displaystyle 2.4141\ldots$	$\displaystyle =$	$\displaystyle 2 + \frac{41}{100} + \frac{41}{100^2} + \frac{41}{100^3} + \cdots$
	$\displaystyle =$	$\displaystyle 2 + \sum_{n=1}^{\infty}(\frac{41}{100})^{n} = 2 + \frac{\frac{41}{100}}{1 - \frac{41}{100}}$
	$\displaystyle =$	$\displaystyle 2 + \frac{41}{59} = \frac{159}{59}$

(c)

$\displaystyle 0.9999\ldots$	$\displaystyle =$	$\displaystyle \frac{9}{10} + \frac{9}{10^2} + \frac{9}{10^3} + \cdots$
	$\displaystyle =$	$\displaystyle \sum_{n=1}^{\infty}(\frac{9}{10})^{n} = \frac{\frac{9}{10}}{1 - \frac{9}{10}}$
	$\displaystyle =$	$\displaystyle \frac{9}{9} = 1$