调和数

根据定义，调和数满足递推关系

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

它也满足恒等式

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

对于第n项调和数，有以下公式

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

设:${\displaystyle x=1-u\,\!}$ ，由此得到

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

对于调和数${\displaystyle H_{n}}$ ，当n不是太大时，可以直接计算。

当n特别大时，可以进行估算。

因为${\displaystyle \lim _{n\to \infty }\left(\sum _{k=1}^{n}{\frac {1}{k}}-\ln n\right)=\gamma }$ ，

其中${\displaystyle \gamma \approx 0.5772156649}$ 称为欧拉-马斯刻若尼常数，

由此得到

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

当n越大时，估算越精确。

更精确的估算是

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

其中${\displaystyle B_{k}}$ 是第k项伯努利数。

广义调和数

广义调和数满足

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

由此，我们得到

${\displaystyle H_{\frac {3}{4}}={\tfrac {4}{3}}-3\ln {2}+{\tfrac {\pi }{2}}}$ 

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

${\displaystyle H_{\frac {1}{2}}=2-2\ln {2}}$ 

${\displaystyle H_{\frac {1}{3}}=3-{\tfrac {\pi }{2{\sqrt {3}}}}-{\tfrac {3}{2}}\ln {3}}$ 

${\displaystyle H_{\frac {1}{4}}=4-{\tfrac {\pi }{2}}-3\ln {2}}$ 

${\displaystyle H_{\frac {1}{6}}=6-{\tfrac {\pi }{2}}{\sqrt {3}}-2\ln {2}-{\tfrac {3}{2}}\ln {3}}$ 

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

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

对于任意两个正整数p和q，并且p＜q，我们有

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

微积分

对于每一个大于0的x，有

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

由此，得

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

对于每一个n，有

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

其他数列

根据定义，其他类似于调和数的数列有以下计算方法：

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

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

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