查看“︁級數列表”︁的源代码


'''本表''' 列出基本或常見的有限級數與無限級數的計算公式。
*規定<math>0^0=1</math>；
*<math>B_n(x)</math>表示[[伯努利多項式|Bernoulli多項式]]；
*<math>B_n</math>表示[[Bernoulli數]]，其中，<math>B_1=-\frac{1}{2}</math>；
*<math>E_n</math>表示[[歐拉數|Euler數]]；
*<math>\zeta(s)</math>表示[[黎曼ζ函數]]；
*<math>\Gamma(z)</math>表示[[Γ函數]]；
*<math>\psi_n(z)</math>表示[[多伽瑪函數]]；
*<math>\operatorname{Li}_s(z)</math>表示[[多重對數函數]]。

== 冪求和 ==
:參見[[等冪求和]]。
*<math>\sum_{k=0}^m k^{n-1}=\frac{B_n(m+1)-B_n}{n}</math>
其中，一次方項、[[平方]]項及[[立方]]項之和的公式如下︰
*<math>\sum_{k=1}^m k=\frac{m(m+1)}{2}</math>
*<math>\sum_{k=1}^m k^2=\frac{m(m+1)(2m+1)}{6}=\frac{m^3}{3}+\frac{m^2}{2}+\frac{m}{6}</math>
*<math>\sum_{k=1}^m k^3=\left[\frac{m(m+1)}{2}\right]^2=\frac{m^4}{4}+\frac{m^3}{2}+\frac{m^2}{4}</math>

:參見[[ζ常數]].
*<math>\zeta(2n)=\sum^{\infty}_{k=1} \frac{1}{k^{2n}}=(-1)^{n+1} \frac{B_{2n} (2\pi)^{2n}}{2(2n)!} </math>
其中，前幾項為︰
*<math>\zeta(2)=\sum^{\infty}_{k=1} \frac{1}{k^2}=\frac{\pi^2}{6}</math> （[[巴塞爾問題]]）
*<math>\zeta(4)=\sum^{\infty}_{k=1} \frac{1}{k^4}=\frac{\pi^4}{90}</math>
*<math>\zeta(6)=\sum^{\infty}_{k=1} \frac{1}{k^6}=\frac{\pi^6}{945}</math>

== 冪級數 ==

=== 低次數多重對數函數 ===
有限項求和︰
*<math>\sum_{k=0}^{n} z^k = \frac{1-z^{n+1}}{1-z}</math>, ([[等比數列]])
*<math>\sum_{k=1}^n k z^k = z\frac{1-(n+1)z^n+nz^{n+1}}{(1-z)^2}</math>
*<math>\sum_{k=1}^n k^2 z^k = z\frac{1+z-(n+1)^2z^n+(2n^2+2n-1)z^{n+1}-n^2z^{n+2}}{(1-z)^3} </math>
*<math>\sum_{k=1}^n k^m z^k = \left(z \frac{d}{dz}\right)^m \frac{1-z^{n+1}}{1-z}</math>

無限項求和，其中<math>|z|<1</math>（參見[[多重對數函數]]）︰
*<math>\operatorname{Li}_n(z)=\sum_{k=1}^{\infty} \frac{z^k}{k^n}</math>
以下是[[遞歸]]計算低整數次冪的多重對數函數以得出[[解析解]]時所用到的一個性質︰
*<math>\frac{d}{dz}\operatorname{Li}_n(z)=\frac{\operatorname{Li}_{n-1}(z)}{z}</math>
前幾項分別為︰
*<math>\operatorname{Li}_{1}(z)=\sum_{k=1}^\infty \frac{z^k}{k}=-\ln(1-z)</math>
*<math>\operatorname{Li}_{0}(z)=\sum_{k=1}^\infty z^k=\frac{z}{1-z}</math>
*<math>\operatorname{Li}_{-1}(z)=\sum_{k=1}^\infty k z^k=\frac{z}{(1-z)^2}</math>
*<math>\operatorname{Li}_{-2}(z)=\sum_{k=1}^\infty k^2 z^k=\frac{z(1+z)}{(1-z)^3}</math>
*<math>\operatorname{Li}_{-3}(z)=\sum_{k=1}^\infty k^3 z^k =\frac{z(1+4z+z^2)}{(1-z)^4}</math>
*<math>\operatorname{Li}_{-4}(z)=\sum_{k=1}^\infty k^4 z^k =\frac{z(1+z)(1+10z+z^2)}{(1-z)^5}</math>

=== 指數函數 ===
*<math>\sum_{k=0}^\infty \frac{z^k}{k!} = e^z</math>
*<math>\sum_{k=0}^\infty k\frac{z^k}{k!} = z e^z</math> （參見[[Poisson分佈]]的[[數學期望]]）
*<math>\sum_{k=0}^\infty k^2 \frac{z^k}{k!} = (z + z^2) e^z</math> （參見[[Poisson分佈]]的[[矩|二階矩]]）
*<math>\sum_{k=0}^\infty k^3 \frac{z^k}{k!} = (z + 3z^2 + z^3) e^z</math>
*<math>\sum_{k=0}^\infty k^4 \frac{z^k}{k!} = (z + 7z^2 + 6z^3 + z^4) e^z</math>
*<math>\sum_{k=0}^\infty k^n \frac{z^k}{k!} = z \frac{d}{dz} \sum_{k=0}^\infty k^{n-1} \frac{z^k}{k!}\,\! = e^z T_{n}(z) </math>

其中，<math>T_{n}(z)</math>表示[[圖沙德多項式]]。

=== [[三角函數]]、[[反三角函數]]、[[雙曲函數]]及[[反雙曲函數]] ===

*<math>\sum_{k=0}^\infty \frac{(-1)^k z^{2k+1}}{(2k+1)!}=\sin z</math>
*<math>\sum_{k=0}^\infty \frac{z^{2k+1}}{(2k+1)!}=\sinh z</math>
*<math>\sum_{k=0}^\infty \frac{(-1)^k z^{2k}}{(2k)!}=\cos z</math>
*<math>\sum_{k=0}^\infty \frac{z^{2k}}{(2k)!}=\cosh z</math>
*<math>\sum_{k=1}^\infty \frac{(-1)^{k-1}(2^{2k}-1)2^{2k}B_{2k}z^{2k-1}}{(2k)!}=\tan z, |z|<\frac{\pi}{2}</math>
*<math>\sum_{k=1}^\infty \frac{(2^{2k}-1)2^{2k}B_{2k}z^{2k-1}}{(2k)!}=\tanh z, |z|<\frac{\pi}{2}</math>
*<math>\sum_{k=0}^\infty \frac{(-1)^k2^{2k}B_{2k}z^{2k-1}}{(2k)!}=\cot z, |z|<\pi</math>
*<math>\sum_{k=0}^\infty \frac{2^{2k}B_{2k}z^{2k-1}}{(2k)!}=\coth z, |z|<\pi</math>
*<math>\sum_{k=0}^\infty \frac{(-1)^{k-1}(2^{2k}-2)B_{2k}z^{2k-1}}{(2k)!}=\csc z, |z|<\pi</math>
*<math>\sum_{k=0}^\infty \frac{-(2^{2k}-2)B_{2k}z^{2k-1}}{(2k)!}=\operatorname{csch} z, |z|<\pi</math>
*<math>\sum_{k=0}^\infty \frac{(-1)^kE_{2k}z^{2k}}{(2k)!}=\sec z, |z|<\frac{\pi}{2}</math>
*<math>\sum_{k=0}^\infty \frac{E_{2k}z^{2k}}{(2k)!}=\operatorname{sech} z, |z|<\frac{\pi}{2}</math>
*<math>\sum_{k=1}^\infty \frac{(-1)^{k-1} z^{2k}}{(2k)!}=\operatorname{ver}z</math>（[[正矢]]）
*<math>\sum_{k=1}^\infty \frac{(-1)^{k-1} z^{2k}}{2(2k)!}=\operatorname{hav}z</math><ref name="Weisstein_hav">{{cite web | first=Eric W. | last=Weisstein | author-link=Eric W. Weisstein | title=Haversine | work=[[MathWorld]] | publisher=[[Wolfram Research, Inc.]] | url=http://mathworld.wolfram.com/Haversine.html | access-date=2015-11-06 |dead-url=no |archive-url=https://web.archive.org/web/20050310194740/http://mathworld.wolfram.com/Haversine.html |archive-date=2005-03-10}}</ref>（[[半正矢]]）
*<math>\sum_{k=0}^\infty \frac{(2k)!z^{2k+1}}{2^{2k}(k!)^2(2k+1)}=\arcsin z, |z|\le1</math>
*<math>\sum_{k=0}^\infty \frac{(-1)^k(2k)!z^{2k+1}}{2^{2k}(k!)^2(2k+1)}=\operatorname{arcsinh} {z}, |z|\le1</math>
*<math>\sum_{k=0}^\infty \frac{(-1)^kz^{2k+1}}{2k+1}=\arctan z, |z|<1</math>
*<math>\sum_{k=0}^\infty \frac{z^{2k+1}}{2k+1}=\operatorname{arctanh} z, |z|<1</math>
*<math>\ln2+\sum_{k=1}^\infty \frac{(-1)^{k-1}(2k)!z^{2k}}{2^{2k+1}k(k!)^2}=\ln\left(1+\sqrt{1+z^2}\right), |z|\le1</math>

=== 修正的[[分母]][[階乘]] ===
*<math>\sum^{\infty}_{k=0} \frac{(4k)!}{2^{4k} \sqrt{2} (2k)! (2k+1)!} z^k = \sqrt{\frac{1-\sqrt{1-z}}{z}}, |z|<1</math><ref name="gfo">{{Cite web |url=http://www.math.upenn.edu/~wilf/gfologyLinked2.pdf |title=generatingfunctionology |access-date=2017-08-08 |archive-date=2021-04-27 |archive-url=https://web.archive.org/web/20210427234643/https://www2.math.upenn.edu/~wilf/gfologyLinked2.pdf |dead-url=no }}</ref>
*<math>\sum^{\infty}_{k=0} \frac{2^{2k} (k!)^2}{(k+1) (2k+1)!} z^{2k+2} = \left(\arcsin{z}\right)^2, |z|\le1 </math><ref name="gfo"/>
*<math>\sum^{\infty}_{n=0} \frac{\prod_{k=0}^{n-1}(4k^2+\alpha^2)}{(2n)!} z^{2n} + \sum^{\infty}_{n=0} \frac{\alpha \prod_{k=0}^{n-1}[(2k+1)^2+\alpha^2]}{(2n+1)!}  z^{2n+1} = e^{\alpha \arcsin{z}}, |z|\le1 </math>

=== [[二項式係數]] ===
*<math>(1+z)^\alpha = \sum_{k=0}^\infty {\alpha \choose k} z^k, |z|<1</math> （參見[[二項式定理]]）
*<ref name="ctcs">{{Cite web |url=http://www.tug.org/texshowcase/cheat.pdf |title=Theoretical computer science cheat sheet |accessdate=2017-08-08 |archive-date=2018-07-30 |archive-url=https://web.archive.org/web/20180730203751/http://www.tug.org/texshowcase/cheat.pdf |dead-url=no }}</ref> <math>\sum_{k=0}^\infty {{\alpha+k-1} \choose k} z^k = \frac{1}{(1-z)^\alpha}, |z|<1</math>
*<ref name="ctcs"/> <math>\sum_{k=0}^\infty \frac{1}{k+1}{2k \choose k} z^k = \frac{1-\sqrt{1-4z}}{2z}, |z|\leq\frac{1}{4}</math>（[[卡塔蘭數]]的[[母函數]]）
*<ref name="ctcs"/> <math>\sum_{k=0}^\infty {2k \choose k} z^k = \frac{1}{\sqrt{1-4z}}, |z|<\frac{1}{4}</math>（[[中心二項式係數]]的[[母函數]]）
*<ref name="ctcs"/> <math>\sum_{k=0}^\infty {2k + \alpha \choose k} z^k = \frac{1}{\sqrt{1-4z}}\left(\frac{1-\sqrt{1-4z}}{2z}\right)^\alpha, |z|<\frac{1}{4}</math>

=== [[調和數]] ===
*<math> \sum_{k=1}^\infty H_k z^k = \frac{-\ln(1-z)}{1-z}, |z|<1</math>
*<math> \sum_{k=1}^\infty \frac{H_k}{k+1} z^{k+1} = \frac{1}{2}\left[\ln(1-z)\right]^2, \qquad |z|<1</math>
*<math> \sum_{k=1}^\infty \frac{(-1)^{k-1} H_{2k}}{2k+1} z^{2k+1} = \frac{1}{2} \arctan{z} \log{(1+z^2)}, \qquad |z|<1 </math><ref name="gfo"/>
*<math> \sum_{n=0}^\infty \sum_{k=0}^{2n} \frac{(-1)^k}{2k+1} \frac{z^{4n+2}}{4n+2} = \frac{1}{4} \arctan{z} \log{\frac{1+z}{1-z}},\qquad  |z|<1 </math><ref name="gfo"/>

== [[二項式係數]] ==
*<math>\sum_{k=0}^n {n \choose k} = 2^n</math>
*<math>\sum_{k=0}^n (-1)^k {n \choose k} = 0, \text{ 其中 }n>0</math>
*<math>\sum_{k=0}^n {k \choose m} = { n+1 \choose m+1 }</math>
*<math>\sum_{k=0}^n {m+k-1 \choose k} = { n+m \choose n }</math> （參見[[多重集]]）
*<math>\sum_{k=0}^n {\alpha \choose k}{\beta \choose n-k} = {\alpha+\beta \choose n}</math>（參見[[范德蒙恆等式|Vandermonde恆等式]]）

== [[三角函數]] ==
:[[正弦]]及[[餘弦]]的求和詳見[[傅立葉級數|Fourier級數]]。

*<math>\sum_{k=1}^\infty \frac{\sin(k\theta)}{k}=\frac{\pi-\theta}{2}, 0<\theta<2\pi</math>
*<math>\sum_{k=1}^\infty \frac{\cos(k\theta)}{k}=-\frac{1}{2}\ln(2-2\cos\theta), \theta\in\mathbb{R}</math>
*<math>\sum_{k=0}^\infty \frac{\sin[(2k+1)\theta]}{2k+1}=\frac{\pi}{4}, 0<\theta<\pi</math>
*<math>B_n(x)=-\frac{n!}{2^{n-1}\pi^n}\sum_{k=1}^\infty \frac{1}{k^n}\cos\left(2\pi kx-\frac{\pi n}{2}\right), 0<x<1</math><ref>{{cite web|url=http://functions.wolfram.com/Polynomials/BernoulliB2/06/02/|title=Bernoulli polynomials: Series representations (subsection 06/02)|accessdate=2011-06-02|archive-date=2020-07-10|archive-url=https://web.archive.org/web/20200710172826/https://functions.wolfram.com/Polynomials/BernoulliB2/06/02/|dead-url=no}}</ref>
*<math>\sum_{k=0}^n \sin(\theta+k\alpha)=\frac{\sin\frac{(n+1)\alpha}{2}\sin(\theta+\frac{n\alpha}{2})}{\sin\frac{\alpha}{2}}</math>
*<math>\sum_{k=1}^{n-1} \sin\frac{\pi k}{n}=\cot\frac{\pi}{2n}</math>
*<math>\sum_{k=1}^{n-1} \sin\frac{2\pi k}{n}=0</math>
*<math>\sum_{k=0}^{n-1} \csc^2\left(\theta+\frac{\pi k}{n}\right)=n^2\csc^2(n\theta)</math><ref>{{cite web|last=Hofbauer|first=Josef|title=A simple proof of 1+1/2^2+1/3^2+...=PI^2/6 and related identities|url=http://homepage.univie.ac.at/josef.hofbauer/02amm.pdf|accessdate=2011-06-02|archive-date=2021-01-23|archive-url=https://web.archive.org/web/20210123060039/https://homepage.univie.ac.at/josef.hofbauer/02amm.pdf|dead-url=no}}</ref>
*<math>\sum_{k=1}^{n-1} \csc^2\frac{\pi k}{n}=\frac{n^2-1}{3}</math>
*<math>\sum_{k=1}^{n-1} \csc^4\frac{\pi k}{n}=\frac{n^4+10n^2-11}{45}</math>

== [[有理函數]] ==
*<math>\sum_{n=a+1}^{\infty} \frac{a}{n^2 - a^2} = \frac{1}{2} H_{2a}</math><ref>[http://mathworld.wolfram.com/RiemannZetaFunction.html Riemann Zeta Function] {{Wayback|url=http://mathworld.wolfram.com/RiemannZetaFunction.html |date=20100209094215 }}" from [[MathWorld]], equation 52</ref>
*<math>\sum_{n=0}^\infty\frac{1}{n^2+a^2}=\frac{1+a\pi\coth (a\pi)}{2a^2}</math>
*<math>\displaystyle \sum_{n=0}^\infty \frac {1}{n^4+4a^4} = \dfrac{1 + a \pi \coth (a \pi)}{8a^4}</math>


使用[[部分分式分解]]方法，能夠將任何關於<math>n</math>的[[有理函數]]的無限項級數都被化簡為一個[[多伽瑪函數]]的有限項級數。<ref>{{Cite web |url=http://people.math.sfu.ca/~cbm/aands/ |title=Abramowitz and Stegun |accessdate=2017-08-08 |archive-date=2014-05-09 |archive-url=https://web.archive.org/web/20140509012019/http://people.math.sfu.ca/~cbm/aands/ |dead-url=no }}</ref>這個方法也被應用於[[有理函數]]的有限項級數中，使得即便所求[[級數]]的項數極多，其計算也能在[[常數時間]]內完成。

== 參閱 ==
{{Div col|cols=3}}
* [[級數]]
* [[積分表]]
* [[求和符號]]
* [[泰勒級數]]
* [[二項式定理]]
* [[Gregory級數]]
* [[整數數列線上大全]]
{{Div col end}}

== 註釋 ==
{{Reflist|30em}}

[[Category:級數]]
[[Category:數學列表]]
[[Category:數學用表]]