調和数つき級数について少し

調和数つき級数について, 簡単なことですが, メモ程度に少し書かせていただきます.

${\displaystyle H_n^{(m)}=\sum _{k=1}^n\frac{1}{k^m}}$ とします.

$$\begin{array}{rcl}\sum _{n=1}^{\mathrm{\infty }}\frac{H_n^{(a)}}{n^b}& =& \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^b}\sum _{m=1}^n\frac{1}{m^a}\\ & =& {\displaystyle \sum _{m=1}^{\mathrm{\infty }}\sum _{n=m^{{}_{}}}^{\mathrm{\infty }}\frac{1}{m^a{n}^b}}\\ & =& \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^{a+b}}+\sum _{0<m<n}\frac{1}{m^a{n}^b}\\ & =& \zeta (a+b)+\zeta (a,b)\end{array}$$
$$

$$\begin{array}{rcl}\sum _{n=1}^{\mathrm{\infty }}(\frac{H_n^{(a)}}{n^b}+\frac{H_n^{(b)}}{n^a})& =& {\displaystyle \sum _{m=1}^{\mathrm{\infty }}\sum _{n=m^{{}_{}}}^{\mathrm{\infty }}\frac{1}{m^a{n}^b}+{\displaystyle \sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^n\frac{1}{n^a{m}^b}}}\\ & =& \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^{a+b}}+\sum _{0<m,n}\frac{1}{m^a{n}^b}\\ & =& \zeta (a+b)+\zeta (a)\zeta (b)\end{array}$$

特に$a=b$として,

$$\begin{array}{rcl}\sum _{n=1}^{\mathrm{\infty }}\frac{H_n^{(a)}}{n^a}& =& \frac{1}{2}(\zeta (a{)}^2+\zeta (2a))\end{array}$$
$$

$$\begin{array}{rcl}\sum _{n=0}^{\mathrm{\infty }}(\zeta (a)-H_n^{(a)})& =& \sum _{n=0}^{\mathrm{\infty }}\sum _{m=n+1}^{\mathrm{\infty }}\frac{1}{m^a}\\ & =& \sum _{m=1}^{\mathrm{\infty }}\frac{1}{m^a}\sum _{n=0}^{m-1}1\\ & =& \zeta (a-1)\end{array}$$
$$

$$\begin{array}{rcl}\sum _{0<a_1,a_2,\dots ,a_n}\frac{1}{a_1\cdots a_n(a_1+\cdots +a_n)}& =& \sum _{0<a_1,a_2,\dots ,a_n}\frac{1}{a_1\cdots a_n}{\int }_0^1{x}^{a_1+\cdots +a_n}\frac{dx}{x}\\ & =& (-1{)}^n{\int }_0^1\frac{{\log }^n(1-x)}{x}dx\\ & =& (-1{)}^n{\int }_0^1\frac{{\log }^n(x)}{1-x}dx\\ & =& n!\zeta (n+1)\end{array}$$
$$