FANDOM


$ U_0>0,U_1>0,U_{n+2}=U_{n+1}+U_n $

より、帰納的に

$ 0<U_1<U_2<\cdots<U_n<U_{n+1}<U_{n+2}<U_{n+3}<\cdots,0<\frac{U_{n+1}}{U_{n+2}}<1,0<\frac{U_n}{U_{n+3}}<1 $

これより、

$ \sum_{k=0}^\infty\frac{(-1)^k}{2k+1}((\frac{U_{n+1}}{U_{n+2}})^{2k+1}+(\frac{U_n}{U_{n+3}})^{2k+1})=\arctan\frac{U_{n+1}}{U_{n+2}}+\arctan\frac{U_n}{U_{n+3}} $

かつ

$ 0<\arctan\frac{U_{n+1}}{U_{n+2}}<\frac{\pi}{4},0<\arctan\frac{U_{n+1}}{U_{n+2}}<\frac{\pi}{4} $

ここで

$ \begin{align} &\tan(\arctan\frac{U_{n+1}}{U_{n+2}}+\arctan\frac{U_n}{U_{n+3}})=\frac{\frac{U_{n+1}}{U_{n+2}}+\frac{U_n}{U_{n+3}}}{1-\frac{U_{n+1}}{U_{n+2}}\frac{U_n}{U_{n+3}}}=\frac{U_{n+1}U_{n+3}+U_nU_{n+2}}{U_{n+2}U_{n+3}-U_nU_{n+1}} \\ &=\frac{U_{n+1}(2U_{n+1}+U_n)+U_n(U_{n+1}+U_n)}{(2U_{n+1}+U_n)(U_{n+1}+U_n)-U_nU_{n+1}}=\frac{2U_{n+1}^2+2U_nU_{n+1}+U_n^2}{2U_{n+1}^2+2U_nU_{n+1}+U_n^2}=1\\ \end{align} $

$ 0<\arctan\frac{U_{n+1}}{U_{n+2}}+\arctan\frac{U_n}{U_{n+3}}<\frac{\pi}{2} $

より

$ \frac{\pi}{4}=\arctan \frac{U_{n+1}}{U_{n+2}}+\arctan \frac{U_n}{U_{n+3}} $

$ \pi = 4\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}((\frac{U_{n+1}}{U_{n+2}})^{2k+1}+(\frac{U_n}{U_{n+3}})^{2k+1}) $

$ \begin{align}\lim_{z\to0}\frac{d}{dz}\pi\cot\pi{z} &=-\frac{\pi^2}{\sin^2\pi{z}}\\ &=-\frac{\pi^2}{(\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}(\pi z)^{2n+1})^2}\\ &=-\frac{\pi^2}{\left(\pi{z}-\frac{1}{6}(\pi{z})^3+O(z^5)\right)^2}\\ &=-\frac{\pi^2}{(\pi{z})^2-\frac{1}{3}(\pi{z})^4+O(z^6)}\\ &=-\frac{1}{z^2}\frac{1}{1-\frac{1}{3}(\pi{z})^2+O(z^4)}\\ &=-\frac{1}{z^2}\sum_{n=0}^\infty(\frac{1}{3}(\pi{z})^2+O(z^4))^n\\ &=-\frac{1}{z^2}(1+\frac{1}{3}(\pi{z})^2+O(z^4))\\ &=-\frac{1}{z^2}-\frac{1}{3}\pi^2+O(z^2)\\ \end{align} $ $ \begin{align}\lim_{z\to0}\frac{d}{dz}\left(\frac{1}{z}+\sum_{n=1}^{\infty}\frac{2z}{z^2-n^2}\right) &=-\frac{1}{z^2}+\sum_{n=1}^{\infty}\frac{2}{z^2-n^2}-\sum_{n=1}^{\infty}\frac{4z^2}{(z^2-n^2)^2}\\ &=-\frac{1}{z^2}-\sum_{n=1}^{\infty}\frac{2}{n^2}+O(z^2)\\ \end{align} $