라이프니츠 급수

$$\begin{array}{rl}& 1+r+r^2+\cdots +r^n=\frac{1-r^{n+1}}{1-r}(r\ne 1)\\ & 1+(-x^2)+(-x^2{)}^2+\cdots +(-x^2{)}^n=\frac{1+x^{2n+2}}{1+x^2}\\ & \frac{1}{1+x^2}+\frac{x^{2n+2}}{1+x^2}=1-x^2+x^4+\cdots x^{2n}\\ & \frac{1}{1+x^2}=1-x^2+x^4+\cdots x^{2n}-\frac{x^{2n+2}}{1+x^2}\end{array}$$

이를 $[0,x]$에 대해 적분하면 (단, $0\le x\le 1$)

$${\int }_0^x\frac{1}{1+t^2}dt=x-\frac{1}{3}{x}^3+\frac{1}{5}{x}^5\cdots +\frac{1}{2n+1}{x}^{2n+1}-{\int }_0^x\frac{t^{2n+2}}{1+t^2}dt$$

이때

$$0\le {\int }_0^a\frac{t^{2n+2}}{1+t^2}dx\le {\int }_0^a{t}^{2n+2}dx\le \frac{1}{2n+3}$$

이므로

$$\begin{array}{rl}{\int }_0^x\frac{1}{1+t^2}dt& =\underset{n\to \infty }{\lim }(x-\frac{1}{3}{x}^3+\frac{1}{5}{x}^5\cdots +\frac{1}{2n+1}{x}^{2n+1})\\ {\tan }^{-1}x& =\underset{n\to \infty }{\lim }(x-\frac{1}{3}{x}^3+\frac{1}{5}{x}^5\cdots +\frac{1}{2n+1}{x}^{2n+1})\\ \therefore \frac{\pi }{4}& =1-{\frac{1}{3}}^3+{\frac{1}{5}}^5\cdots +\frac{1}{2n+1}(x=1)\end{array}$$ $$L:=\sum _{k=1}^{\infty }\frac{(-1{)}^{k+1}}{2k-1}=\frac{\pi }{4}$$