4.1 解答

級数 $\sum a_{n}$ は $\lim_{n \to \infty}{a_{n}} \neq 0$ のとき発散する．

$\displaystyle \lim_{n \to \infty}a_{n} = \lim_{n \to \infty}\frac{2n + 1}{3n + 1} = \frac{2}{3}$

$\displaystyle \lim_{n \to \infty}a_{n}$	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}\frac{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}}{n}$
	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}\frac{n\sqrt{1 + \frac{1}{n^2}} + n\sqrt{1- \frac{1}{n^2}}}{n}$
	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}\left(\sqrt{1 + \frac{1}{n^2}} + \sqrt{1 - \frac{1}{n^2}}\right) = \sqrt{2}$

$\displaystyle \lim_{n \to \infty}a_{n} = \lim_{n \to \infty}\cos{\frac{\pi}{n}} = 1$

$\displaystyle \frac{1}{4n^2 - 1} = \frac{1}{2}\left(\frac{1}{2n -1} - \frac{1}{2n + 1}\right)$

$\displaystyle S_{n} = \frac{1}{2}\left(1 - \frac{1}{3} + \frac{1}{3} - \frac{1}... ...ts + \frac{1}{2n - 1} - \frac{1}{2n+1}\right) = \frac{1}{2}(1 - \frac{1}{2n+1})$

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{4n^2 - 1} = \lim_{n \to \infty}S_{n} = \frac{1}{2}$

$\displaystyle \frac{n}{(n+1)!} = \frac{1}{n!} - \frac{1}{(n+1)!}$

$\displaystyle S_{n} = 1 - \frac{1}{2!} + \frac{1}{2!} - \frac{1}{3!} + \cdots + \frac{1}{n!} - \frac{1}{(n+1)!} = 1 - \frac{1}{(n+1)!}$

$\displaystyle \sum_{n=1}^{\infty}\frac{n}{(n+1)!} = \lim_{n \to \infty}S_{n} = 1$

$\displaystyle \frac{1}{n(n+1)(n+2)} = \frac{1}{2}\left(\frac{1}{n} - \frac{2}{n... ...1}{2}\left(\frac{1}{n} - \frac{1}{n+1} - (\frac{1}{n+1} - \frac{1}{n+2})\right)$

$\displaystyle 2S_{n}$	$\displaystyle =$	$\displaystyle 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \cdots + \frac{1}{n} - \frac{1}{n+1}$
	$\displaystyle -$	$\displaystyle \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{n+1} - \frac{1}{n+2}$
	$\displaystyle =$	$\displaystyle 1 - \frac{1}{n+1} - (\frac{1}{2} - \frac{1}{n+2})$

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)} = \lim_{n \to \infty}S_{n} = \frac{1}{4}$