# 调和级数

${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k}}=1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+\cdots \,\!}$

${\displaystyle \zeta (s)=\sum _{k=1}^{\infty }{\frac {1}{k^{s}}}}$

## 佯谬

${\displaystyle {\frac {1}{100}}\sum _{k=1}^{n}{\frac {1}{k}}}$

${\displaystyle d_{n+1}\,=\,{\frac {(d_{n}+l_{n})n+{\frac {l_{0}}{2}}}{n+1}}\,=\,{\frac {l_{0}\cdot n+{\frac {l_{0}}{2}}}{n+1}}\,=\,{\frac {l_{0}\cdot (n+1)-{\frac {l_{0}}{2}}}{n+1}}\,=\,l_{0}-{\frac {\frac {l_{0}}{2}}{n+1}}}$

${\displaystyle l_{n+1}=l_{0}-d_{n+1}={\frac {\frac {l_{0}}{2}}{n+1}}}$，即${\displaystyle l_{n}={\frac {l_{0}}{2}}\cdot {\frac {1}{n}}}$

${\displaystyle l_{\mathrm {total} }={\frac {l_{0}}{2}}\cdot \sum _{k=1}^{n}{\frac {1}{k}}}$

## 发散

### 比较审敛法

${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k}}=1+\left[{\frac {1}{2}}\right]+\left[{\frac {1}{3}}+{\frac {1}{4}}\right]+\left[{\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+{\frac {1}{8}}\right]+\left[{\frac {1}{9}}+\cdots \right.}$
${\displaystyle \geq \sum _{k=1}^{\infty }2^{-\lceil \log _{2}k\rceil }\,\!}$
${\displaystyle =1+\left[{\frac {1}{2}}\right]+\left[{\frac {1}{4}}+{\frac {1}{4}}\right]+\left[{\frac {1}{8}}+{\frac {1}{8}}+{\frac {1}{8}}+{\frac {1}{8}}\right]+\left[{\frac {1}{16}}+\cdots \right.\,\!}$
${\displaystyle =1+\ {\frac {1}{2}}\ +\qquad {\frac {1}{2}}\ \quad +\ \qquad \quad {\frac {1}{2}}\qquad \ \quad \ +\ \quad \ \cdots \,\!\;=\;\;\infty .}$

### 积分判别法（integral test）

${\displaystyle =1\,+\,{\frac {1}{2}}\,+\,{\frac {1}{3}}\,+\,{\frac {1}{4}}\,+\,{\frac {1}{5}}\,+\,\cdots }$

${\displaystyle =\int _{1}^{\infty }{\frac {1}{x}}\,dx\;=\;\infty }$。这部分面积真包含于（换言之，小于）长方形总面积，长方形的总面积也必定趋于无穷。更准确说，这证明了

${\displaystyle \sum _{n=1}^{k}\,{\frac {1}{n}}\;>\;\int _{1}^{k+1}{\frac {1}{x}}\,dx\;=\;\ln(k+1)}$

## 发散率

${\displaystyle \sum _{n=1}^{k}\,{\frac {1}{n}}\;=\;\ln k+\gamma +\varepsilon _{k}}$

## 部分和

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

## 相关级数

### 交错调和级数

${\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}\;=\;1\,-\,{\frac {1}{2}}\,+\,{\frac {1}{3}}\,-\,{\frac {1}{4}}\,+\,{\frac {1}{5}}\,-\,\cdots }$

${\displaystyle 1\,-\,{\frac {1}{2}}\,+\,{\frac {1}{3}}\,-\,{\frac {1}{4}}\,+\,{\frac {1}{5}}\,-\,\cdots \;=\;\ln 2}$

${\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2n+1}}\;\;=\;\;1\,-\,{\frac {1}{3}}\,+\,{\frac {1}{5}}\,-\,{\frac {1}{7}}\,+\,\cdots \;\;=\;\;{\frac {\pi }{4}}}$

### 广义调和级数

${\displaystyle \sum _{n=0}^{\infty }{\frac {1}{an+b}}}$

### ${\displaystyle p}$-级数

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{p}}}}$

P是任何正实数。当${\displaystyle p=1}$${\displaystyle p}$-级数即调和级数。由积分判别法柯西稠密判定法可知${\displaystyle p}$-级数在${\displaystyle p>1}$时收敛（此时级数又叫过调和级数（over-harmonic series）），而在${\displaystyle p\leq 1}$时发散。当${\displaystyle p>1}$时，${\displaystyle p}$-级数的和即${\displaystyle \zeta (p)}$，也就是黎曼ζ函数${\displaystyle p}$的值。

### ${\displaystyle \varphi }$-级数

${\displaystyle \limsup _{u\to 0^{+}}{\frac {\varphi ({\frac {u}{2}})}{\varphi (u)}}<{\frac {1}{2}}}$

### 随机调和级数

${\displaystyle \sum _{n=1}^{\infty }{\frac {s_{n}}{n}}}$

## 注釋

1. ^ 泛音列与调和级数英文同为harmonic series
2. ^

## 参考

1. ^ George L. Hersey, Architecture and Geometry in the Age of the Baroque, p 11-12 and p37-51.
2. Graham, Ronald; Knuth, Donald E.; Patashnik, Oren, Concrete Mathematics 2nd, Addison-Wesley: 258–264, 1989, ISBN 978-0-201-55802-9
3. ^ Sharp, R.T., Problem 52: Overhanging dominoes, Pi Mu Epsilon Journal, 1954: 411–412
4. ^ Sequence A082912 in the On-Line Encyclopedia of Integer Sequences
5. ^ Weisstein, Eric W. (编). Harmonic Number. at MathWorld--A Wolfram Web Resource. Wolfram Research, Inc. [2011-01-16]. （原始内容存档于2013-05-16） （英语）.
6. ^ Art of Problem Solving: "General Harmonic Series"页面存档备份，存于互联网档案馆
7. ^ "Random Harmonic Series", American Mathematical Monthly 110, 407-416, May 2003
8. ^ Schmuland's preprint of Random Harmonic Series (PDF). [2011-01-16]. （原始内容 (PDF)存档于2011-06-08）.
9. ^ Weisstein, Eric W. (编). Infinite Cosine Product Integral. at MathWorld--A Wolfram Web Resource. Wolfram Research, Inc. [2010-11-14]. （原始内容存档于2011-12-28） （英语）.
10. ^ Nick's Mathematical Puzzles: Solution 72. [2011-01-16]. （原始内容存档于2010-09-28）.