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.

$\ds H_n^{(m)}=\sum_{k=1}^n\frac1{k^m}$ とします.

$$\beq \sumn\frac{H_n^{(a)}}{n^b}&=&\sumn\frac1{n^b}\sum_{m=1}^n\frac1{m^a}\\ &=&\ds\sum_{m=1}^\infty\sum_{n=m^{_{}}}^\infty\frac{1}{m^an^b}\\ &=&\sum_{n=1}^\infty\frac1{n^{a+b}}+\sum_{0< m< n}\frac1{m^an^b}\\ &=&\zeta(a+b)+\zeta(a,b) \eeq$$
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$$\beq \sumn\bigg(\frac{H_n^{(a)}}{n^b}+\frac{H_n^{(b)}}{n^a}\bigg)&=&\ds\sum_{m=1}^\infty\sum_{n=m^{_{}}}^\infty\frac{1}{m^an^b}+\ds\sum_{n=1}^\infty\sum_{m=1}^n\frac{1}{n^am^b}\\ &=&\sum_{n=1}^\infty\frac1{n^{a+b}}+\sum_{0< m,n}\frac1{m^an^b}\\ &=&\zeta(a+b)+\zeta(a)\zeta(b) \eeq$$

特に$a=b$として,

$$\beq \sumn\frac{H_n^{(a)}}{n^a}&=&\frac12\big(\zeta(a)^2+\zeta(2a)\big) \eeq$$
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$$\beq \sum_{n=0}^\infty\big(\zeta(a)-H_n^{(a)}\big)&=&\sum_{n=0}^\infty\sum_{m=n+1}^\infty\frac1{m^a}\\ &=&\sum_{m=1}^\infty\frac1{m^a}\sum_{n=0}^{m-1}1\\ &=&\zeta(a-1) \eeq$$
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$$\beq \sum_{0< a_1,a_2,\ldots,a_n}\frac1{a_1\cdots a_n(a_1+\cdots+a_n)}&=&\sum_{0< a_1,a_2,\ldots,a_n}\frac{1}{a_1\cdots a_n}\int_0^1x^{a_1+\cdots+a_n}\frac{dx}{x}\\ &=&(-1)^n\int_0^1\frac{\log^n(1-x)}{x}\,dx\\ &=&(-1)^n\int_0^1\frac{\log^n(x)}{1-x}\,dx\\ &=&n!\zeta(n+1) \eeq$$
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