級数(ウォリス積分の応用)

記事投稿の練習も兼ねて、試しに投稿してみました。

目標$$\sum _{n=0}^{\mathrm{\infty }}\frac{(2n)!}{8^n(n!{)}^2}$$の値を求める

上を使い求める。
$$\begin{array}{rcl}\sum _{n=0}^{\mathrm{\infty }}\frac{(2n)!}{8^n(n!{)}^2}& =& \sum _{n=0}^{\mathrm{\infty }}\frac{1}{2^n}\cdot \frac{(2n)!}{4^n(n!{)}^2}\\ & =& \frac{2}{\pi }\sum _{n=0}^{\mathrm{\infty }}\frac{1}{2^n}{\int }_0^{\frac{\pi }{2}}{\sin }^{2n}xdx\\ & =& \frac{2}{\pi }{\int }_0^{\frac{\pi }{2}}\sum _n^{\mathrm{\infty }}{\left(\frac{{\sin }^{2}x}{2}\right)}^{n}dx\\ & =& \frac{2}{\pi }{\int }_0^{\frac{\pi }{2}}\frac{1}{1-\frac{{\sin }^{2}x}{2}}dx\\ & =& \frac{4}{\pi }{\int }_0^{\frac{\pi }{2}}\frac{1}{1+{\cos }^{2}x}dx\\ & =& \frac{4}{\pi }{\int }_0^{\frac{\pi }{2}}\frac{1}{{\cos }^{2}x}\cdot \frac{1}{{\tan }^{2}x+2}dx\\ & =& \frac{4}{\pi }{\int }_0^{\mathrm{\infty }}\frac{1}{t^2+2}dt(t=\tan x\text{で置換})\\ & =& \frac{2\sqrt{2}}{\pi }{\int }_0^{\mathrm{\infty }}\frac{1}{u^2+1}du(u=\frac{t}{\sqrt{2}}\text{で置換})\\ & =& \frac{2\sqrt{2}}{\pi }{[\arctan u]}_0^{\mathrm{\infty }}\\ & =& \frac{2\sqrt{2}}{\pi }\cdot \frac{\pi }{2}\\ & =& \sqrt{2}\end{array}$$

よって$$\sum _{n=0}^{\mathrm{\infty }}\frac{(2n)!}{4^n(n!{)}^2}=\sqrt{2}$$