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*}

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