第3章 3.10

(1)

$x + \frac{1}{2}\cdot\frac{x^3}{3} + \frac{1}{2}\cdot\frac{3}{4}\cdot\frac{x^5}{5} + \cdots + \frac{1\cdot3\cdots\cdots(2n-1)}{2\cdot4\cdots\cdots2n}\cdot\frac{x^{2n+1}}{2n+1} + \cdots$

$x$は２乗ずつ増えているので

$r^2 = \lim_{n\to\infty}\left|\frac{a_n}{a_{n+2}}\right| = \lim_{n\to\infty}\left|\frac{\dfrac{1\cdot3\cdots\cdots(2n-1)}{2\cdot4\cdots\cdots2n}\cdot\dfrac{1}{2n+1}}{\dfrac{1\cdot3\cdots\cdots(2n-1)\cdot2n}{2\cdot4\cdots\cdots2n\cdot(2n+1)}\cdot\dfrac{1}{2n+2}}\right| = \lim_{n\to\infty}\left|\frac{(2n+1)\cdot(2n+2)}{2n\cdot(2n+1)}\right|$

$= \lim_{n\to\infty}\left|\frac{2n+2}{2n}\right| = 1$

よって $r=1$

(2)

$\frac{x}{\log 2} - \frac{x^2}{\log 3} + \frac{x^3}{\log 4} - \cdots + (-1)^{n-1}\frac{x^n}{\log(n+1)} + \cdots$

$r = \lim_{n\to\infty}\left|\dfrac{\dfrac{1}{\log(n+1)}}{\dfrac{1}{\log(n+2)}}\right| = \lim_{n\to\infty}\left|\frac{\log(n+2)}{\log(n+1)}\right| = 1$