級数解説14

2020/12/31に PCT 君がツイートした級数です。

https://twitter.com/pctprobability/status/1344534262121316352?s=21

[解説]

$\begin{array}{rcl}& & \sum _{k=0}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{4(2k+1{)}^2}\right)}^n\\ & =& \sum _{k=0}^{\mathrm{\infty }}\frac{1}{4(2k+1{)}^2-1}\\ & =& \frac{1}{2}\sum _{k=0}^{\mathrm{\infty }}(\frac{1}{4k+1}-\frac{1}{4k+3})\\ & =& \frac{1}{2}\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k}{2k+1}\\ & =& \frac{1}{2}\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k{\int }_0^1{x}^{2k}dx\\ & =& \frac{1}{2}{\int }_0^1\frac{1}{1+x^2}dx\\ & =& \frac{\pi }{8}◻\end{array}$

また、

$\begin{array}{rcl}& & \sum _{n=0}^{\mathrm{\infty }}(64n^2+64n+19)\sum _{m=1}^{\mathrm{\infty }}{\left(\frac{1}{4096(n^2+n{)}^2+1048(n^2+n)+106}\right)}^m\\ & =& \sum _{n=0}^{\mathrm{\infty }}\frac{64n^2+64n+19}{4096(n^2+n{)}^2+1048(n^2+n)+105}\\ & =& \frac{1}{4}\sum _{n=0}^{\mathrm{\infty }}(\frac{1}{8n+1}-\frac{1}{8n+3}+\frac{1}{8n+5}-\frac{1}{8n+7})\\ & =& \frac{1}{4}\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{2n+1}\\ & =& \frac{\pi }{16}◻\end{array}$

以上より、この2つの級数が証明されました。