調和數

${\displaystyle H_{n}=1+{\frac {1}{2}}+{\frac {1}{3}}+\cdots +{\frac {1}{n}}=\sum _{k=1}^{n}{\frac {1}{k}}}$

調和級數的性質

${\displaystyle H_{n+1}=H_{n}+{\frac {1}{n+1}}}$

${\displaystyle \sum _{k=1}^{n}H_{k}=(n+1)H_{n}-n}$

計算

${\displaystyle H_{n}=\int _{0}^{1}{\frac {1-x^{n}}{1-x}}\,dx.}$

{\displaystyle {\begin{aligned}H_{n}&=\int _{0}^{1}{\frac {1-x^{n}}{1-x}}\,dx\\&=-\int _{1}^{0}{\frac {1-(1-u)^{n}}{u}}\,du\\&=\int _{0}^{1}{\frac {1-(1-u)^{n}}{u}}\,du\\&=\int _{0}^{1}\left[\sum _{k=1}^{n}(-1)^{k-1}{\binom {n}{k}}u^{k-1}\right]\,du\\&=\sum _{k=1}^{n}(-1)^{k-1}{\binom {n}{k}}\int _{0}^{1}u^{k-1}\,du\\&=\sum _{k=1}^{n}(-1)^{k-1}{\frac {1}{k}}{\binom {n}{k}}.\end{aligned}}}

${\displaystyle H_{n}\sim \ln {n}+\gamma }$

${\displaystyle H_{n}\sim \ln {n}+\gamma +{\frac {1}{2n}}-\sum _{k=1}^{\infty }{\frac {B_{2k}}{2kn^{2k}}}=\ln {n}+\gamma +{\frac {1}{2n}}-{\frac {1}{12n^{2}}}+{\frac {1}{120n^{4}}}-\cdots ,}$

廣義調和數

${\displaystyle H_{\alpha }=\int _{0}^{1}{\frac {1-x^{\alpha }}{1-x}}\,dx\,.}$

${\displaystyle H_{\frac {3}{4}}={\tfrac {4}{3}}-3\ln {2}+{\tfrac {\pi }{2}}}$
${\displaystyle H_{\frac {2}{3}}={\tfrac {3}{2}}(1-\ln {3})+{\sqrt {3}}{\tfrac {\pi }{6}}}$
${\displaystyle H_{\frac {1}{2}}=2-2\ln {2}}$
${\displaystyle H_{\frac {1}{3}}=3-{\tfrac {\pi }{2{\sqrt {3}}}}-{\tfrac {3}{2}}\ln {3}}$
${\displaystyle H_{\frac {1}{4}}=4-{\tfrac {\pi }{2}}-3\ln {2}}$
${\displaystyle H_{\frac {1}{6}}=6-{\tfrac {\pi }{2}}{\sqrt {3}}-2\ln {2}-{\tfrac {3}{2}}\ln {3}}$
${\displaystyle H_{\frac {1}{8}}=8-{\tfrac {\pi }{2}}-4\ln {2}-{\tfrac {1}{\sqrt {2}}}\left\{\pi +\ln \left(2+{\sqrt {2}}\right)-\ln \left(2-{\sqrt {2}}\right)\right\}}$
${\displaystyle H_{\frac {1}{12}}=12-3\left(\ln {2}+{\tfrac {\ln {3}}{2}}\right)-\pi \left(1+{\tfrac {\sqrt {3}}{2}}\right)+2{\sqrt {3}}\ln \left({\sqrt {2-{\sqrt {3}}}}\right)}$

${\displaystyle H_{\frac {p}{q}}={\frac {q}{p}}+2\sum _{k=1}^{\lfloor {\frac {q-1}{2}}\rfloor }\cos({\frac {2\pi pk}{q}})ln({\sin({\frac {\pi k}{q}})})-{\frac {\pi }{2}}cot({\frac {\pi p}{q}})-ln({2q})}$

微積分

${\displaystyle H_{x}=x\sum _{k=1}^{\infty }{\frac {1}{k(x+k)}}\,.}$

${\displaystyle \int _{0}^{1}H_{x}\,dx=\gamma \,,}$

${\displaystyle \int _{0}^{n}H_{x}\,dx=\ln {(n!)}+n\gamma \,.}$

其他數列

${\displaystyle \sum _{k=1}^{n}{\frac {1}{k}}=\psi (n-1)+\gamma }$

${\displaystyle \sum _{k=0}^{n}{\frac {1}{2k+1}}={\frac {1}{2}}\left[\psi \left(n+{\frac {3}{2}}\right)+\gamma \right]+\ln {2}}$

${\displaystyle \sum _{k=1}^{n}{\frac {1}{2k}}={\frac {H_{n}}{2}}}$