(1)

\begin{align*} \zeta\left(s\right) & =\sum_{k=1}^{\infty}k^{-s}\\ & =1+\sum_{k=1}^{\infty}\left(k+1\right)^{-s}\\ & =1+\sum_{k=1}^{\infty}k^{-s}\left(1+\frac{1}{k}\right)^{-s}\\ & =1+\sum_{k=1}^{\infty}k^{-s}\sum_{j=0}^{\infty}C\left(-s,j\right)\left(\frac{1}{k}\right)^{j}\\ & =1+\sum_{j=0}^{\infty}C\left(-s,j\right)\sum_{k=1}^{\infty}k^{-\left(s+j\right)}\\ & =1+\sum_{j=0}^{\infty}C\left(-s,j\right)\zeta\left(s+j\right) \end{align*}

(2)

(1)より、
\begin{align*} \zeta\left(s\right) & =1+\sum_{j=0}^{\infty}C\left(-s,j\right)\zeta\left(s+j\right)\\ & =1+\sum_{j=0}^{\infty}\frac{P\left(-s,j\right)}{j!}\zeta\left(s+j\right)\\ & =1+\sum_{j=0}^{\infty}\left(-1\right)^{j}\frac{Q\left(s,j\right)}{j!}\zeta\left(s+j\right)\\ & =1+\zeta\left(s\right)-s\zeta\left(s+1\right)+\sum_{j=2}^{\infty}\left(-1\right)^{j}\frac{Q\left(s,j\right)}{j!}\zeta\left(s+j\right)\\ & =1+\LHS\left(s\right)-s\LHS\left(s\rightarrow s+1\right)+\sum_{j=2}^{\infty}\left(-1\right)^{j}\frac{Q\left(s,j\right)}{j!}\zeta\left(s+j\right)\\ & =\frac{1}{s-1}+\frac{1}{s-1}\sum_{j=2}^{\infty}\left(-1\right)^{j}\frac{Q\left(s-1,j\right)}{j!}\zeta\left(s+j-1\right)\\ & =\frac{1}{s-1}+\sum_{j=2}^{\infty}\left(-1\right)^{j}\frac{Q\left(s,j-1\right)}{j!}\zeta\left(s+j-1\right)\\ & =\frac{1}{s-1}+\sum_{j=1}^{\infty}\left(-1\right)^{j+1}\frac{Q\left(s,j\right)}{\left(j+1\right)!}\zeta\left(s+j\right) \end{align*}

(3)

(2)より、
\begin{align*} \zeta\left(s\right) & =\frac{1}{s-1}+\sum_{j=1}^{\infty}\left(-1\right)^{j+1}\frac{Q\left(s,j\right)}{\left(j+1\right)!}\zeta\left(s+j\right)\\ & =\frac{1}{s-1}-\sum_{j=1}^{\infty}\frac{P\left(-s,j\right)}{\left(j+1\right)!}\zeta\left(s+j\right)\\ & =\frac{1}{s-1}-\sum_{j=1}^{\infty}\frac{C\left(-s,j\right)}{j+1}\zeta\left(s+j\right)\\ & =\frac{1}{s-1}+\zeta\left(x\right)-\sum_{j=0}^{\infty}\frac{C\left(-s,j\right)}{j+1}\zeta\left(s+j\right) \end{align*} これより、
\[ \frac{1}{s-1}=\sum_{j=0}^{\infty}\frac{C\left(-s,j\right)}{j+1}\zeta\left(s+j\right) \] となるので与式は成り立つ。

(4)

(3)より\(s=n\in\mathbb{N}\)とすると、
\begin{align*} \frac{1}{-n-1} & =\sum_{j=0}^{\infty}\frac{C\left(n,j\right)}{j+1}\zeta\left(-n+j\right)\\ & =\zeta\left(-n\right)+\sum_{j=1}^{n}\frac{C\left(n,j\right)}{j+1}\zeta\left(-n+j\right)+\lim_{x\rightarrow0}\frac{C\left(n,n+1+x\right)}{n+2}\zeta\left(1+x\right)+\sum_{j=n+2}^{\infty}\frac{C\left(n,j\right)}{j+1}\zeta\left(-n+j\right)\\ & =\zeta\left(-n\right)+\sum_{j=1}^{n}\frac{C\left(n,j\right)}{j+1}\zeta\left(-n+j\right)+\lim_{x\rightarrow0}\frac{P\left(n,n\right)}{\left(n+2\right)\left(n+1\right)!}\left(-x\right)\zeta\left(1+x\right)\\ & =\zeta\left(-n\right)+\sum_{j=1}^{n}\frac{C\left(n,j\right)}{j+1}\zeta\left(-n+j\right)-\frac{n!}{\left(n+2\right)!}\\ & =\zeta\left(-n\right)+\sum_{j=1}^{n}\frac{C\left(n,j\right)}{j+1}\zeta\left(-n+j\right)-\frac{1}{\left(n+2\right)\left(n+1\right)}\\ & =\zeta\left(-n\right)+\sum_{j=1}^{n}\frac{C\left(n,j\right)}{j+1}\zeta\left(-n+j\right)-\frac{1}{\left(n+2\right)\left(n+1\right)}\\ & =\zeta\left(-n\right)+\frac{1}{n+1}\sum_{j=1}^{n}C\left(n+1,j+1\right)\zeta\left(-n+j\right)-\frac{1}{\left(n+2\right)\left(n+1\right)}\\ & =\zeta\left(-n\right)+\frac{1}{n+1}\sum_{j=0}^{n-1}C\left(n+1,j\right)\zeta\left(-j\right)-\frac{1}{\left(n+2\right)\left(n+1\right)}\cmt{n-j\rightarrow j}\\ & =\zeta\left(-n\right)+\frac{1}{n+1}\sum_{j=1}^{n}C\left(n+1,j-1\right)\zeta\left(-j+1\right)-\frac{1}{\left(n+2\right)\left(n+1\right)}\\ & =\zeta\left(-n\right)+\frac{1}{\left(n+1\right)\left(n+2\right)}\sum_{j=1}^{n}C\left(n+2,j\right)j\zeta\left(-j+1\right)-\frac{1}{\left(n+2\right)\left(n+1\right)}\\ & =\zeta\left(-n\right)+\frac{1}{\left(n+1\right)\left(n+2\right)}\left\{ \sum_{j=0}^{n}C\left(n+2,j\right)j\zeta\left(-j+1\right)-\lim_{j\rightarrow0}j\zeta\left(-j+1\right)\right\} -\frac{1}{\left(n+2\right)\left(n+1\right)}\\ & =\zeta\left(-n\right)+\frac{1}{\left(n+1\right)\left(n+2\right)}\left\{ \sum_{j=0}^{n}C\left(n+2,j\right)j\zeta\left(-j+1\right)+1\right\} -\frac{1}{\left(n+2\right)\left(n+1\right)}\\ & =\zeta\left(-n\right)+\frac{1}{\left(n+1\right)\left(n+2\right)}\sum_{j=0}^{n}C\left(n+2,j\right)j\zeta\left(-j+1\right)\\ & =\zeta\left(-n\right)+\frac{1}{\left(n+1\right)\left(n+2\right)}\sum_{j=0}^{n}C\left(n+2,j\right)\left(j\zeta\left(-j+1\right)+\delta_{1,j}\right)-\frac{1}{n+1}\left(1-\delta_{0,n}\right)\\ & =-\frac{1}{n+1}+\zeta\left(-n\right)+\delta_{0,n}+\frac{1}{\left(n+1\right)\left(n+2\right)}\sum_{j=0}^{n}C\left(n+2,j\right)\left(j\zeta\left(-j+1\right)+\delta_{1,j}\right) \end{align*} これより、
\[ \left(n+1\right)\left(\zeta\left(-n\right)+\delta_{0,n}\right)=-\frac{1}{n+2}\sum_{j=0}^{n}C\left(n+2,j\right)\left(j\left(\zeta\left(-j+1\right)+\delta_{1j}\right)\right) \] となる。
ここで\(B_{n+1}=\left(n+1\right)\left(\zeta\left(-n\right)+\delta_{0,n}\right)\)とおいて、\(n\rightarrow n-1\)とすると、
\[ B_{n+1}=-\frac{1}{n+1}\sum_{j=0}^{n-1}C\left(n+1,j\right)B_{j} \] これは\(n\in\mathbb{N}\)としたときのベルヌーイ数\(B_{n}\)の漸化式と同じで、初項は、
\begin{align*} B_{0} & =\lim_{n\rightarrow0}-n\left(\zeta\left(-n+1\right)+\delta_{1n}\right)\\ & =-\lim_{n\rightarrow0}n\zeta\left(-n+1\right)\\ & =\lim_{n\rightarrow1}\left(n-1\right)\zeta\left(n\right)\\ & =\lim_{n\rightarrow1}\left(n-1\right)\left(\frac{1}{n-1}+\sum_{k=0}^{\infty}\frac{\left(-1\right)^{k}}{k!}\gamma_{k}\left(n-1\right)^{k}\right)\\ & =1 \end{align*} となりベルヌーイ数と一致するので、
\[ B_{n}=-n\left(\zeta\left(-n+1\right)+\delta_{1n}\right) \] となる。
これをゼータ関数について解くと、
\[ \zeta\left(1-n\right)=-\left(\frac{B_{n}}{n}+\delta_{1n}\right) \] となるので与式が成り立つ。

(4)-2

途中まで示す。
\begin{align*} \zeta\left(-n\right) & =\frac{1}{-n-1}+\sum_{j=1}^{\infty}\left(-1\right)^{j+1}\frac{Q\left(-n,j\right)}{\left(j+1\right)!}\zeta\left(-n+j\right)\\ & =-\frac{1}{n+1}-\sum_{j=1}^{\infty}\frac{P\left(n,j\right)}{\left(j+1\right)!}\zeta\left(j-n\right)\\ & =-\frac{1}{n+1}-\sum_{j=1}^{n}\frac{P\left(n,j\right)}{\left(j+1\right)!}\zeta\left(j-n\right)-\lim_{x\rightarrow0}\frac{P\left(n,n\right)}{\left(n+2\right)!}x\zeta\left(1-x\right)-\sum_{j=n+2}^{\infty}\frac{P\left(n,j\right)}{\left(j+1\right)!}\zeta\left(j-n\right)\\ & =-\frac{1}{n+1}-\sum_{j=1}^{n}\frac{P\left(n,j\right)}{\left(j+1\right)!}\zeta\left(j-n\right)+\frac{1}{\left(n+1\right)\left(n+2\right)}-\sum_{j=0}^{\infty}\frac{P\left(n,j+n+2\right)}{\left(j+n+3\right)!}\zeta\left(j+2\right)\\ & =-\frac{1}{n+1}+\left(\frac{1}{n+1}-\frac{1}{n+2}\right)-\sum_{j=1}^{n}\frac{P\left(n,j\right)}{\left(j+1\right)!}\zeta\left(j-n\right)\\ & =-\frac{1}{n+2}-\sum_{j=1}^{n}\frac{P\left(n,j\right)}{\left(j+1\right)!}\zeta\left(j-n\right)\\ & =-\frac{1}{n+2}-\frac{1}{n+1}\sum_{j=1}^{n}\frac{P\left(n+1,j+1\right)}{\left(j+1\right)!}\zeta\left(j-n\right)\\ & =-\frac{1}{n+2}-\frac{1}{n+1}\sum_{j=1}^{n}C\left(n+1,j+1\right)\zeta\left(j-n\right)\\ & =-\frac{1}{n+2}-\frac{1}{n+1}\sum_{j=0}^{n-1}C\left(n+1,j\right)\zeta\left(-j\right)\tag{*}\\ & =-\frac{1}{n+2}-\frac{1}{n+1}\frac{1}{n+2}\sum_{j=0}^{n-1}C\left(n+2,j+1\right)\left(j+1\right)\zeta\left(-j\right)\\ & =-\frac{1}{n+2}-\frac{1}{n+1}\frac{1}{n+2}\sum_{j=1}^{n}C\left(n+2,j\right)j\zeta\left(-j+1\right)\\ & =-\frac{1}{n+2}-\left(\frac{1}{n+1}-\frac{1}{n+2}\right)-\frac{1}{n+1}\frac{1}{n+2}\sum_{j=0}^{n}C\left(n+2,j\right)j\zeta\left(-j+1\right)\\ & =-\frac{1}{n+1}-\frac{1}{n+1}\frac{1}{n+2}\sum_{j=0}^{n}C\left(n+2,j\right)j\zeta\left(-j+1\right)\\ & =-\frac{1}{n+1}-\delta_{n0}+\delta_{n0}-\frac{1}{n+1}\frac{1}{n+2}\sum_{j=0}^{n}C\left(n+2,j\right)j\zeta\left(-j+1\right)\\ & =-\delta_{n0}-\frac{1}{n+1}\frac{1}{n+2}\left(n+2\right)\left(1-\delta_{n0}\right)-\frac{1}{n+1}\frac{1}{n+2}\sum_{j=0}^{n}C\left(n+2,j\right)j\zeta\left(-j+1\right)\\ & =-\delta_{n0}-\frac{1}{n+1}\frac{1}{n+2}\sum_{j=0}^{n}C\left(n+2,j\right)\left(j\zeta\left(-j+1\right)+\left(n+2\right)\left(1-\delta_{n0}\right)\right)\\ & =-\delta_{n0}-\frac{1}{n+1}\frac{1}{n+2}\sum_{j=0}^{n}C\left(n+2,j\right)\left(j\zeta\left(-j+1\right)+\delta_{1j}\left(1-\delta_{n0}\right)\right)\\ & =-\delta_{n0}-\frac{1}{n+1}\frac{1}{n+2}\sum_{j=0}^{n}C\left(n+2,j\right)\left(j\zeta\left(-j+1\right)+\delta_{1j}\right) \end{align*} これより、
\[ \left(n+1\right)\left(\zeta\left(-n\right)+\delta_{n0}\right)=-\frac{1}{n+2}\sum_{j=0}^{n}C\left(n+2,j\right)\left(j\left(\zeta\left(-j+1\right)+\delta_{1j}\right)\right) \] となる。
以下同じ。

(5)

(4)より、\(n\in\mathbb{N}_{0}\)のとき、
\[ \zeta\left(-n\right)=\frac{B_{n+1}}{n+1}+\delta_{1,n+1}B_{0} \] となるので、
\begin{align*} \sum_{j=0}^{n}C\left(n+1,j\right)\zeta\left(-j\right) & =-\sum_{j=0}^{n}C\left(n+1,j\right)\left(\frac{B_{j+1}}{j+1}+\delta_{1,j+1}B_{0}\right)\\ & =-1-\sum_{j=0}^{n}C\left(n+1,j\right)\frac{B_{j+1}}{j+1}\\ & =-1-\frac{1}{n+2}\sum_{j=0}^{n}C\left(n+2,j+1\right)B_{j+1}\\ & =-1-\frac{1}{n+2}\sum_{j=1}^{n+1}C\left(n+2,j\right)B_{j}\\ & =-1+\frac{1}{n+2}-\frac{1}{n+2}\sum_{j=0}^{n+1}C\left(n+2,j\right)B_{j}\\ & =-\frac{n+1}{n+2}-\frac{1}{n+2}\delta_{0,n+1}\cmt{\because\delta_{0,n}=\sum_{k=0}^{n}C\left(n+1,k\right)B_{k}}\\ & =-\frac{n+1}{n+2} \end{align*} となる。
従って、
\[ \frac{1}{n+1}\sum_{j=0}^{n}C\left(n+1,j\right)\zeta\left(-j\right)=-\frac{1}{n+2} \] となり与式は成り立つ。

(6)

(5)より、
\begin{align*} -\frac{1}{n+2} & =\frac{1}{n+1}\sum_{j=0}^{n}C\left(n+1,j\right)\zeta\left(-j\right)\\ & =\frac{1}{n+1}\sum_{j=0}^{n}\frac{\left(n+1\right)}{n+1-j}C\left(n,j\right)\zeta\left(-j\right)\\ & =\sum_{j=0}^{n}C\left(n,j\right)\frac{\zeta\left(-j\right)}{n+1-j} \end{align*} となるので与式は成り立つ。