級数解説10

級数botで投下された級数です。

https://twitter.com/infseriesbot/status/1342327861105987584?s=21

[解説]

まず、${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\zeta (2n)x^{2n-1}}$を考えます。

$\begin{array}{rcl}& & \sum _{n=1}^{\mathrm{\infty }}\zeta (2n)x^{2n-1}\\ & =& \frac{1}{x}\sum _{n=1}^{\mathrm{\infty }}{x}^{2n}\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^{2n}}\\ & =& \frac{1}{x}\sum _{k=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{x^2}{k^2}\right)}^n\\ & =& \frac{1}{x}\sum _{k=1}^{\mathrm{\infty }}\frac{\frac{x^2}{k^2}}{1-\frac{x^2}{k^2}}\\ & =& -\frac{1}{2}\sum _{k=1}^{\mathrm{\infty }}\frac{2x}{x^2-k^2}\\ & =& -\frac{1}{2}(\pi \cot \pi x-\frac{1}{x})\end{array}$

両辺を1回積分し、

$\begin{array}{rcl}& & \sum _{n=1}^{\mathrm{\infty }}\frac{\zeta (2n)x^{2n}}{2n}\\ & =& -\frac{1}{2}\int (\pi \cot \pi x-\frac{1}{x})dx\\ & =& \frac{1}{2}(\log x-\log \sin \pi x)+C_1\end{array}$

$x\to 0$の極限を取ると、

$\begin{array}{rcl}& & \underset{x\to 0}{lim}\frac{1}{2}(\log x-\log \sin \pi x)+C_1\\ & =& \underset{x\to 0}{lim}\frac{1}{2}(\log \frac{x}{\sin \pi x})+C_1\\ & =& -\frac{1}{2}\log \pi +C_1\\ & =& 0\end{array}$

より、${\displaystyle C_1=\frac{1}{2}\log \pi }$であるから、

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{\zeta (2n)x^{2n}}{2n}=\frac{1}{2}(\log \pi x-\log \sin \pi x)}$

両辺を${\displaystyle [0,1]}$で積分して、

$\begin{array}{rcl}& & \sum _{n=1}^{\mathrm{\infty }}\frac{\zeta (2n)}{2n(2n+1)}\\ & =& \frac{1}{2}{\int }_0^1(\log \pi x-\log \sin \pi x)dx\\ & =& \frac{1}{2}{\int }_0^1(\log 2\pi x-\log 2\sin \pi x)dx\\ & =& \frac{1}{2}{[x\log 2\pi x-x]}_0^1\\ & =& \frac{1}{2}\log \frac{2\pi }{e}\end{array}$

であるから、

${\displaystyle \exp \left(\sum _{n=1}^{\mathrm{\infty }}\frac{\zeta (2n)}{2n(2n+1)}\right)=\exp (\frac{1}{2}\log \frac{2\pi }{e})}$

$\begin{array}{rcl}& & \exp \left(\sum _{n=1}^{\mathrm{\infty }}\frac{\zeta (2n)}{2n(2n+1)}\right)\\ & =& \exp (\sum _{n=1}^{\mathrm{\infty }}\zeta (2n)(\frac{1}{2n}-\frac{1}{2n+1}))\\ & =& \exp (\frac{\zeta (2)}{2}-\frac{\zeta (2)}{3}+\frac{\zeta (4)}{4}-\frac{\zeta (4)}{5}+\dots )\end{array}$

$\begin{array}{rcl}& & \exp (\frac{1}{2}\log \frac{2\pi }{e})\\ & =& \sqrt{e^{\log \frac{2\pi }{e}}}\\ & =& \sqrt{\frac{2\pi }{e}}\end{array}$

以上より、

${\displaystyle \exp (\frac{\zeta (2)}{2}-\frac{\zeta (2)}{3}+\frac{\zeta (4)}{4}-\frac{\zeta (4)}{5}+\dots )=\sqrt{\frac{2\pi }{e}}}$

が証明されました。