4.2 解答

1. 正項級数 $\sum a_{n}$ において， $\lim_{n \to \infty}\frac{a_{n+1}}{a_{n}} = \rho$ が存在するとき， $0 \leq \rho < 1$ ならば， $\sum a_{n}$ は収束．

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

$\displaystyle \lim_{n \to \infty}\frac{a_{n+1}}{a_{n}}$	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^{n}}{n!}$
	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}\left(\frac{n}{n+1}\right)^{n} = \lim_{n \to \infty}\left(\frac{1}{1 + \frac{1}{n}}\right)^{n}$
	$\displaystyle =$	$\displaystyle \frac{1}{e} < 1$

$\displaystyle \lim_{n \to \infty}\frac{a_{n+1}}{a_{n}}$	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}\frac{(n+1)^{n+1}}{(n+1)!}\cdot\frac{n!}{n^{n}}$
	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}\left(\frac{n+1}{n}\right)^{n} = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^{n}$
	$\displaystyle =$	$\displaystyle e > 1$

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

$\displaystyle \lim_{n \to \infty}\frac{a_{n+1}}{a_{n}}$	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}\frac{2^{n+1}}{(n+1)!}\cdot\frac{n!}{2^{n}}$
	$\displaystyle =$	$\displaystyle \lim_{n \to \infty}\frac{2}{n+1} = 0$

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

$\displaystyle \sum (\sqrt[3]{n+1} - \sqrt[3]{n}) = \lim_{n \to \infty}S_{n} = \lim_{n \to \infty}(n+1)^{1/3} = \infty$