4.1 解答

4.1

1.

(a)


$\displaystyle \rho$ $\displaystyle =$ $\displaystyle \lim_{n \rightarrow \infty}\vert\frac{a_{n}}{a_{n+1}}\vert = \lim_{n \rightarrow \infty}\vert\frac{n/3^{n}}{(n+1)/3^{n+1}}\vert$  
  $\displaystyle =$ $\displaystyle \lim_{n \rightarrow \infty}\vert\frac{3n}{n+1}\vert = 3. \ \ \framebox{終}$  

(b)


$\displaystyle \rho$ $\displaystyle =$ $\displaystyle \lim_{n \rightarrow \infty}\vert\frac{a_{n}}{a_{n+1}}\vert = \lim_{n \rightarrow \infty}\vert\frac{n^n/n!}{(n+1)^{n+1}/(n+1)!}\vert$  
  $\displaystyle =$ $\displaystyle \lim_{n \rightarrow \infty}\vert(\frac{n}{n+1})^{n}\vert = \lim_{...
...tarrow \infty}\vert(\frac{1}{1 + 1/n})^{n}\vert = \frac{1}{e}. \ \ \framebox{終}$  

2.

(a) $S_{n} = \sum_{k=0}^{n}x^{k}$とおくと

$\displaystyle S_{n}-xS_{n} = \sum_{k=0}^{n}x^{k} - \sum_{k=0}^{n}x^{k+1} = 1 - x^{n+1}$

よって $S_{n} = \frac{1 - x^{n+1}}{1 - x}$ .ここで

$\displaystyle \lim_{n \rightarrow \infty}x^{n} = \left\{\begin{array}{cl}
\infty & x > 1\\
0 & \vert x\vert < 1\\
振動 & x < -1
\end{array} \right . $

に注意すると

$\displaystyle S = \lim S_{n} = \frac{1}{1-x}, \vert x\vert < 1. \ \framebox{終}$

(b)


$\displaystyle \log{(1+x)}$ $\displaystyle =$ $\displaystyle \int_{0}^{x}\frac{1}{1+t}dt = \int_{0}^{x}\sum_{n=0}^{\infty}(-1)^{n}t^{n}dt \ (項別積分可能)$  
  $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}\int_{0}^{x}(-1)^{n}t^{n}dt = \sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n+1}}{n+1}, \ \vert x\vert < 1$  

$x=1$のとき $\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n+1}}{n+1}$は収束するので,

$\displaystyle \log{(1+x)} =\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n+1}}{n+1}, \ -1 < x \leq 1 \ \framebox{終}$

3.

(a) $\vert\frac{\sin{nx}}{n^3}\vert \leq \frac{1}{n^3} = M_{n}$とおくと,

$\displaystyle \sum M_{n} = \sum \frac{1}{n^3} < \infty $

よってWeierstrassのM-testにより一様収束. \framebox{終}
(b) $f(x) =\sum_{n=1}^{\infty} \frac{\sin{nx}}{n^3}$とおくと

$\displaystyle \ f^{\prime}(x) = (\sum_{n=1}^{\infty}\frac{\sin{nx}}{n^3})^{\prime} $

ここで(a)より$f(x)$は一様収束するので項別微分が可能.よって

$\displaystyle f^{\prime}(x) = \sum_{n=1}^{\infty}(\frac{\sin{nx}}{n^3})^{\prime} = \sum_{n=1}^{\infty}\frac{\cos{nx}}{n^2} \ \framebox{終}$

(c) $f(x) =\sum_{n=1}^{\infty} \frac{\sin{nx}}{n^3}$とおくと

$\displaystyle \int_{0}^{\pi}f(x)dx = \int_{0}^{\pi}\sum_{n=1}^{\infty}\frac{\sin{nx}}{n^3} $

ここで(a)より$f(x)$は一様収束するので項別積分が可能.よって

$\displaystyle \sum_{n=1}^{\infty}\int_{0}^{\pi}\frac{\sin{nx}}{n^3}dx = \sum_{n=1}^{\infty}\frac{2}{(2n-1)^{4}} \ \framebox{終} $