(1)

\begin{align*} \frac{d^{n}}{dz^{n}}H_{z} & =\frac{d^{n}}{dz^{n}}\left(\psi\left(z+1\right)+\gamma\right)\\ & =\psi^{\left(n\right)}\left(z+1\right)\\ & =\left(-1\right)^{n+1}n!\left(\zeta\left(n+1\right)-H_{z,n+1}\right) \end{align*}

(1)-2

\begin{align*} \frac{d^{n}}{dz^{n}}H_{z} & =\frac{d^{n}}{dz^{n}}\left(z\sum_{k=1}^{\infty}\frac{1}{k\left(k+z\right)}\right)\\ & =\frac{d^{n}}{dz^{n}}\sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{\left(k+z\right)}\right)\\ & =-\sum_{k=1}^{\infty}\frac{d^{n}}{dz^{n}}\frac{1}{\left(k+z\right)}\\ & =\left(-1\right)^{n+1}n!\sum_{k=1}^{\infty}\frac{1}{\left(k+z\right)^{n+1}}\\ & =\left(-1\right)^{n+1}n!\zeta\left(n+1,1+z\right)\\ & =\left(-1\right)^{n+1}n!\left(\zeta\left(n+1\right)-H_{z,n+1}\right) \end{align*}

(2)

\begin{align*} \frac{d^{n}H_{z,\alpha}}{dz^{n}} & =\frac{d^{n}}{dz^{n}}H_{z,\alpha}\\ & =\frac{d^{n}}{dz^{n}}\left\{ \zeta\left(\alpha\right)-\sum_{k=1}^{\infty}\frac{1}{\left(k+z\right)^{\alpha}}\right\} \\ & =\zeta\left(\alpha\right)\delta_{0n}+\sum_{k=1}^{\infty}\frac{d^{n}}{dz^{n}}\frac{1}{\left(k+z\right)^{\alpha}}\\ & =\zeta\left(\alpha\right)\delta_{0n}+P\left(-\alpha,n\right)\sum_{k=1}^{\infty}\frac{1}{\left(k+z\right)^{\alpha+n}}\\ & =\zeta\left(\alpha\right)\delta_{0n}+\left(-1\right)^{n+1}Q\left(\alpha,n\right)\sum_{k=1}^{\infty}\frac{1}{\left(k+z\right)^{\alpha+n}}\\ & =\zeta\left(\alpha\right)\delta_{0n}+\left(-1\right)^{n+1}Q\left(\alpha,n\right)\left(\zeta\left(\alpha+n\right)-H_{z,\alpha+n}\right) \end{align*}

(2)-2

\(\alpha\)が1以外の自然数\(\alpha\in\mathbb{N}\setminus\left\{ 1\right\} \)のときの証明
\begin{align*} \psi^{\left(n\right)}\left(z\right) & =\left(-1\right)^{n+1}n!\left(\zeta\left(n+1\right)-H_{z-1,n+1}\right) \end{align*} なので、
\begin{align*} \frac{d^{n}H_{z,m}}{dz^{n}} & =\frac{d^{n}}{dz^{n}}\left\{ \zeta\left(m\right)-\frac{\left(-1\right)^{m}}{\left(m-1\right)!}\psi^{\left(m-1\right)}\left(z+1\right)\right\} \\ & =\zeta\left(m\right)\delta_{0n}-\frac{\left(-1\right)^{m}}{\left(m-1\right)!}\psi^{\left(m+n-1\right)}\left(z+1\right)\\ & =\zeta\left(m\right)\delta_{0n}-\frac{\left(-1\right)^{m}}{\left(m-1\right)!}\left(-1\right)^{m+n}\left(m+n-1\right)!\left(\zeta\left(m+n\right)-H_{z,m+n}\right)\\ & =\zeta\left(m\right)\delta_{0n}+\left(-1\right)^{n+1}Q\left(m,n\right)\left(\zeta\left(m+n\right)-H_{z,m+n}\right) \end{align*}

(3)

\(H_{z}\)のテイラー展開は、
\begin{align*} H_{z} & =H_{0}+\sum_{k=1}^{\infty}\left[\frac{d^{k}H_{x}}{dz^{k}}\right]_{z=0}\frac{z^{k}}{k!}\\ & =\sum_{k=1}^{\infty}\left[\left(-1\right)^{k+1}k!\left(\zeta\left(k+1\right)-H_{z,k+1}\right)\right]_{x=0}\frac{z^{k}}{k!}\\ & =\sum_{k=1}^{\infty}\left(-1\right)^{k+1}\zeta\left(k+1\right)z^{k} \end{align*} となる。
\(H_{z,\alpha}\)のテイラー展開は、\(\alpha\in\mathbb{C}\setminus\left\{ 1\right\} \)のとき、
\begin{align*} H_{z,\alpha} & =\sum_{k=0}^{\infty}\left[\frac{d^{k}H_{z,\alpha}}{dz^{k}}\right]_{z=0}\frac{z^{k}}{k!}\\ & =\sum_{k=0}^{\infty}\left[\zeta\left(\alpha\right)\delta_{0k}+\left(-1\right)^{k+1}Q\left(\alpha,k\right)\left(\zeta\left(\alpha+k\right)-H_{z,\alpha+k}\right)\right]_{z=0}\frac{z^{k}}{k!}\\ & =\sum_{k=0}^{\infty}\left\{ \zeta\left(\alpha\right)\delta_{0k}+\left(-1\right)^{k+1}Q\left(\alpha,k\right)\zeta\left(\alpha+k\right)\right\} \frac{z^{k}}{k!}\\ & =\sum_{k=1}^{\infty}\left(-1\right)^{k+1}\frac{Q\left(\alpha,k\right)}{k!}\zeta\left(\alpha+k\right)z^{k}\\ & =\sum_{k=1}^{\infty}\left(-1\right)^{k+1}C\left(\alpha+k-1,\alpha-1\right)\zeta\left(\alpha+k\right)z^{k} \end{align*} となる。
これに\(\alpha=1\)を代入すると、
\begin{align*} H_{z,1} & =\sum_{k=1}^{\infty}\left(-1\right)^{k+1}C\left(1+k-1,1-1\right)\zeta\left(1+k\right)z^{k}\\ & =\sum_{k=1}^{\infty}\left(-1\right)^{k+1}\zeta\left(1+k\right)z^{k}\\ & =H_{z} \end{align*} となるので任意の\(\alpha\in\mathbb{C}\)に対し、
\[ H_{z,\alpha}=\sum_{k=1}^{\infty}\left(-1\right)^{k+1}C\left(\alpha+k-1,\alpha-1\right)\zeta\left(\alpha+k\right)z^{k} \] が成り立つ。
これより与式は成り立つ。