4.4 解答

4.4

1. ベキ級数 $\sum a_{n}x^{n}$ の収束半径は次の式で与えられる．

$\displaystyle \rho = \lim_{n \to \infty}\vert\frac{a_{n}}{a_{n+1}}\vert$

(a)

$\displaystyle \rho = \lim_{n \to \infty}\vert\frac{a_{n}}{a_{n+1}}\vert = \lim_{n \to \infty}\vert\frac{2n+2}{2n+1}\vert = 1$

(b) $\sum \frac{n}{3^{n}}$ について考える．

$\displaystyle \rho$	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}\vert\frac{a_{n}}{a_{n+1}}\vert = \lim_{n \to \infty}\vert\frac{2^{2n}}{2n!}\frac{2(n+1)!}{2^{2(n+1)}}\vert$
	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}\frac{2^{2n}(n+1)}{2^{2n+2}} = \infty$

(c)

$\displaystyle \rho$	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}\vert\frac{a_{n}}{a_{n+1}}\vert = \lim_{n \to \infty}\vert\frac{n!}{(n+1)!}\vert$
	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}\frac{1}{n+1} = 0$

2. (a)

$\displaystyle \log(\frac{1+x}{1-x})$	$\displaystyle =$	$\displaystyle \log(1+x) - \log(1-x)$
	$\displaystyle =$	$\displaystyle x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots + (x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots$
	$\displaystyle =$	$\displaystyle 2[x + \frac{x^3}{3} + \frac{x^5}{5} + \cdots + ]$

(b) 部分分数分解すると

$\displaystyle \frac{1}{3x^2 - 4x + 1} = \frac{1}{(3x-1)(x-1)} = \frac{A}{3x-1} + \frac{B}{x-1}$

より $A = -\frac{3}{2}, B = \frac{1}{2}$ ．ここで，

$\displaystyle \frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots = \sum_{n=0}^{\infty}x^{n} (\vert x\vert < 1)$

であることに注意すると

$\displaystyle \frac{1}{3x^2 - 4x + 1}$	$\displaystyle =$	$\displaystyle \frac{1}{2}(\frac{1}{x-1} - \frac{3}{3x - 1})$
	$\displaystyle =$	$\displaystyle \frac{1}{2}(-\frac{1}{1-x} + 3\frac{1}{1 - 3x})$
	$\displaystyle =$	$\displaystyle \frac{1}{2}(-\sum_{n=0}^{\infty}x^{n} + 3\sum_{n=0}^{\infty}(3x)^{n})$