級数を解く

143

級数を解く

どうも、らららです。

(今回の記事ほとんどメモ用みたいな感じです、なので雑です)

$\sum_{n = 0}^{\infty} \frac{(\binom{2 n}{n})}{(2 n + 1)^{2} 2^{2 n}} = \frac{π}{2} \log 2$

$\begin{aligned} f (x) & = \sum_{n = 0}^{\infty} \frac{x^{2 n + 1}}{(2 n + 1)^{2}} \frac{(\binom{2 n}{n})}{2^{2 n}} \\ f^{'} (x) & = \sum_{n = 0}^{\infty} \frac{x^{2 n}}{2 n + 1} \frac{(\binom{2 n}{n})}{2^{2 n}} \\ = \frac{\arcsin x}{x} \end{aligned}$

求める値は $f (1)$

$f (0) = 0$ であることに注意して

$\begin{aligned} f (1) & = f (1) - 0 \\ = f (1) - f (0) \\ = \int_{0}^{1} f^{'} (x) d x \\ = \int_{0}^{1} \frac{\arcsin x}{x} d x \\ = \int_{0}^{\frac{π}{2}} \frac{x}{\tan x} d x \\ = [x \log \sin x]_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} \log \sin x d x \\ = - \int_{0}^{\frac{π}{2}} \log \sin x d x \\ = \frac{π}{2} \log 2 \end{aligned}$

$\begin{aligned} f (x) & = \sum_{n = 0}^{\infty} \frac{x^{2 n + 1}}{(2 n + 1)^{3}} \frac{(\binom{2 n}{n})}{2^{2 n}} \\ x f^{'} (x) & = \sum_{n = 0}^{\infty} \frac{x^{2 n + 1}}{(2 n + 1)^{2}} \frac{(\binom{2 n}{n})}{2^{2 n}} \\ f^{'} (x) + x f^{″} (x) & = \sum_{n = 0}^{\infty} \frac{x^{2 n}}{2 n + 1} \frac{(\binom{2 n}{n})}{2^{2 n}} \\ = \frac{\arcsin x}{x} \end{aligned}$

$\begin{aligned} f (1) & = \int_{0}^{1} f^{'} (x) d x \\ = \int_{0}^{1} \frac{\arcsin x}{x} d x - \int_{0}^{1} x f^{″} (x) d x \\ = \frac{π}{2} \log 2 - [x f^{'} (x)]_{0}^{1} + \int_{0}^{1} f^{'} (x) d x \\ = \frac{π}{2} \log 2 - f^{'} (1) + f (1) \end{aligned}$