偶数ゼータ・奇数ゼータ・ゼータの総和
偶数ゼータ・奇数ゼータ・ゼータの総和
(1)偶数ゼータの総和
\[ \sum_{k=1}^{\infty}\left(\zeta(2k)-1\right)=\frac{3}{4} \](2)奇数ゼータの総和
\[ \sum_{k=1}^{\infty}\left(\zeta\left(2k+1\right)-1\right)=\frac{1}{4} \](3)ゼータの総和
\[ \sum_{k=2}^{\infty}\left(\zeta\left(k\right)-1\right)=1 \](1)
\begin{align*} \sum_{k=1}^{\infty}\left(\zeta(2k)-1\right) & =\lim_{x\rightarrow1}\sum_{k=1}^{\infty}\left(\zeta(2k)-1\right)x^{2k}\\ & =\lim_{x\rightarrow1}\left(\frac{1}{2}\left(1-\pi x\tan^{-1}\left(\pi x\right)\right)-\frac{x^{2}}{1-x^{2}}\right)\cmt{\sum_{k=1}^{\infty}\zeta(2k)x^{2k}=\frac{1}{2}\left(1-\pi x\tan^{-1}\left(\pi x\right)\right)}\\ & =\frac{1}{2}-\lim_{x\rightarrow1}\left(\pi x\frac{\cos\left(\pi x\right)}{2\sin\left(\pi x\right)}+\frac{x^{2}}{1-x^{2}}\right)\\ & =\frac{1}{2}-\lim_{x\rightarrow1}\left(\frac{x}{2}\frac{\pi\cos\left(\pi x\right)\left(1-x^{2}\right)+2\sin\left(\pi x\right)x}{\sin\left(\pi x\right)\left(1-x^{2}\right)}\right)\\ & =\frac{1}{2}-\lim_{x\rightarrow1}\left(\frac{x}{2}\frac{\left\{ 2-\pi^{2}\left(1-x^{2}\right)\right\} \sin\left(\pi x\right)}{\pi\left(1-x^{2}\right)\cos\left(\pi x\right)-2x\sin\left(\pi x\right)}\right)\\ & =\frac{1}{2}-\lim_{x\rightarrow1}\left(\frac{x}{2}\frac{2\pi^{2}x\sin\left(\pi x\right)+\pi\left\{ 2-\pi^{2}\left(1-x^{2}\right)\right\} \cos\left(\pi x\right)}{\left(-2-\pi^{2}\left(1-x^{2}\right)\right)\sin\left(\pi x\right)-4\pi x\cos\left(\pi x\right)}\right)\\ & =\frac{1}{2}+\frac{1}{4}\\ & =\frac{3}{4} \end{align*}(2)
\begin{align*} \sum_{k=1}^{\infty}\left(\zeta\left(2k+1\right)-1\right) & =-\sum_{k=1}^{\infty}\left\{ \left(\zeta\left(2k\right)-1\right)-\left(\zeta\left(2k+1\right)-1\right)\right\} +\sum_{k=1}^{\infty}\left(\zeta\left(2k\right)-1\right)\\ & =-\sum_{k=1}^{\infty}\left\{ \zeta\left(2k\right)-\zeta\left(2k+1\right)\right\} +\sum_{k=1}^{\infty}\left(\zeta\left(2k\right)-1\right)\\ & =-\frac{1}{2}+\frac{3}{4}\\ & =\frac{1}{4} \end{align*}(3)
\begin{align*} \sum_{k=2}^{\infty}\left(\zeta\left(k\right)-1\right) & =\sum_{k=1}^{\infty}\left(\zeta\left(2k\right)-1\right)+\sum_{k=1}^{\infty}\left(\zeta\left(2k+1\right)-1\right)\\ & =\frac{3}{4}+\frac{1}{4}\\ & =1 \end{align*}ページ情報
タイトル | 偶数ゼータ・奇数ゼータ・ゼータの総和 |
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完備リーマンゼータ関数の関数等式
\[
\xi(s)=\xi(1-s)
\]
ゼータ関数の絶対収束条件
ゼータ関数$\zeta\left(s\right)$は$\Re\left(s\right)>1$で絶対収束
リーマン・ゼータ関数とフルヴィッツ・ゼータ関数の関係
\[
\zeta\left(s,1\right)=\zeta\left(s\right)
\]
リーマンゼータ関数の関数等式
\[
\pi^{-\frac{s}{2}}\Gamma\left(\frac{s}{2}\right)\zeta(s)=\pi^{-\frac{1-s}{2}}\Gamma\left(\frac{1-s}{2}\right)\zeta(1-s)
\]