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
;
表示Bernoulli多項式;
表示Bernoulli數,其中,
;
表示Euler數;
表示黎曼ζ函數;
表示Γ函數;
表示多伽瑪函數;
表示多重對數函數。
冪求和[編輯]
- 參見等冪求和。
![{\displaystyle \sum _{k=0}^{m}k^{n-1}={\frac {B_{n}(m+1)-B_{n}}{n}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e82797674c101a71a773fa28db688ccaba2e827)
其中,一次方項、平方項及立方項之和的公式如下︰
![{\displaystyle \sum _{k=1}^{m}k={\frac {m(m+1)}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/615f66562931b8bfd0238dc8ccc87b7a6e83d9e8)
![{\displaystyle \sum _{k=1}^{m}k^{2}={\frac {m(m+1)(2m+1)}{6}}={\frac {m^{3}}{3}}+{\frac {m^{2}}{2}}+{\frac {m}{6}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/590a25a336ef2d10df6962aee36d70dc8c623a5f)
![{\displaystyle \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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/83655857c974dd27c9b29de8cda04d7c65d334e3)
- 參見ζ常數.
![{\displaystyle \zeta (2n)=\sum _{k=1}^{\infty }{\frac {1}{k^{2n}}}=(-1)^{n+1}{\frac {B_{2n}(2\pi )^{2n}}{2(2n)!}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/39c16e56068bfb1b7c7a16876faecbd23cae1fb9)
其中,前幾項為︰
(巴塞爾問題)
![{\displaystyle \zeta (4)=\sum _{k=1}^{\infty }{\frac {1}{k^{4}}}={\frac {\pi ^{4}}{90}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/57d340ce3e07c8d682543de1ee543ddb28dbf071)
![{\displaystyle \zeta (6)=\sum _{k=1}^{\infty }{\frac {1}{k^{6}}}={\frac {\pi ^{6}}{945}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c150edab196b63b262f0bcbb971ee895456f8e4)
冪級數[編輯]
低次數多重對數函數[編輯]
有限項求和︰
, (等比數列)
![{\displaystyle \sum _{k=1}^{n}kz^{k}=z{\frac {1-(n+1)z^{n}+nz^{n+1}}{(1-z)^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5ba5195ab25644b0202fb60e7c30b94d044ea38d)
![{\displaystyle \sum _{k=1}^{n}k^{2}z^{k}=z{\frac {1+z-(n+1)^{2}z^{n}+(2n^{2}+2n-1)z^{n+1}-n^{2}z^{n+2}}{(1-z)^{3}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5274ec4b72fcd2bb8ed27ddf604ed21d8dd126f2)
![{\displaystyle \sum _{k=1}^{n}k^{m}z^{k}=\left(z{\frac {d}{dz}}\right)^{m}{\frac {1-z^{n+1}}{1-z}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c7a59ad2bafdc84f1a2ed59d06acdf45a9cb4789)
無限項求和,其中
(參見多重對數函數)︰
![{\displaystyle \operatorname {Li} _{n}(z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k^{n}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/269bc4ebc751699b90632451c1506b0d12aef7a9)
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︰
![{\displaystyle {\frac {d}{dz}}\operatorname {Li} _{n}(z)={\frac {\operatorname {Li} _{n-1}(z)}{z}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9fc806c7174eab906a2477189cebb285c41cb0ed)
前幾項分別為︰
![{\displaystyle \operatorname {Li} _{1}(z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k}}=-\ln(1-z)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/78c0907fa4e026586a3dec2121860a12c13a62c5)
![{\displaystyle \operatorname {Li} _{0}(z)=\sum _{k=1}^{\infty }z^{k}={\frac {z}{1-z}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/df5a61f7feaffd247a5450eba4968debd0f9bf6e)
![{\displaystyle \operatorname {Li} _{-1}(z)=\sum _{k=1}^{\infty }kz^{k}={\frac {z}{(1-z)^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2505cfc24d99fe2c95e297738310c1347577f017)
![{\displaystyle \operatorname {Li} _{-2}(z)=\sum _{k=1}^{\infty }k^{2}z^{k}={\frac {z(1+z)}{(1-z)^{3}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d703061c9125105bede161bf3adc41091b2fb830)
![{\displaystyle \operatorname {Li} _{-3}(z)=\sum _{k=1}^{\infty }k^{3}z^{k}={\frac {z(1+4z+z^{2})}{(1-z)^{4}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c15985776b2b6a3638ec04c0bf292b81cd6b72a)
![{\displaystyle \operatorname {Li} _{-4}(z)=\sum _{k=1}^{\infty }k^{4}z^{k}={\frac {z(1+z)(1+10z+z^{2})}{(1-z)^{5}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f08ae7cc5ef199773da7054d9ba3b27aec21012d)
指數函數[編輯]
![{\displaystyle \sum _{k=0}^{\infty }{\frac {z^{k}}{k!}}=e^{z}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1d3c8535bc3feb0e123e11fe343171dd9d4776da)
(參見Poisson分佈的數學期望)
(參見Poisson分佈的二階矩)
![{\displaystyle \sum _{k=0}^{\infty }k^{3}{\frac {z^{k}}{k!}}=(z+3z^{2}+z^{3})e^{z}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/62129fb023e2b6de038703c670c0394abdb87315)
![{\displaystyle \sum _{k=0}^{\infty }k^{4}{\frac {z^{k}}{k!}}=(z+7z^{2}+6z^{3}+z^{4})e^{z}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/738269671a82829e80dca30df6a8c4aa93c98653)
![{\displaystyle \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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8ff42a20c13815fd8611f979983110d5f8d9b3a6)
其中,
表示圖沙德多項式。
![{\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k+1}}{(2k+1)!}}=\sin z}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0eeb6209d2ef99d44eb022f43b79787eade4c648)
![{\displaystyle \sum _{k=0}^{\infty }{\frac {z^{2k+1}}{(2k+1)!}}=\sinh z}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5eed9faf752bff168c51a2901e44421778e377b6)
![{\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k}}{(2k)!}}=\cos z}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9386a3bfce6368adbad6c7962f37b18b9b995012)
![{\displaystyle \sum _{k=0}^{\infty }{\frac {z^{2k}}{(2k)!}}=\cosh z}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e495ed1e2d351c9644a9b2b9b62814f0255d911)
![{\displaystyle \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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2256f274843b5a8dd7338fcd46d89457f27d39b8)
![{\displaystyle \sum _{k=1}^{\infty }{\frac {(2^{2k}-1)2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\tanh z,|z|<{\frac {\pi }{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b10f67088d6d4a62eee48692deda3065a9ef72f8)
![{\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\cot z,|z|<\pi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/462f64ebe4b22d9eb36d69972a2c16259d72ea16)
![{\displaystyle \sum _{k=0}^{\infty }{\frac {2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\coth z,|z|<\pi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/00bfdc23630f34df2a588dcd3f1d5c7b3c9fc6f5)
![{\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k-1}(2^{2k}-2)B_{2k}z^{2k-1}}{(2k)!}}=\csc z,|z|<\pi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/d223384181921eadadcc9acb38bbbd886d85c7ee)
![{\displaystyle \sum _{k=0}^{\infty }{\frac {-(2^{2k}-2)B_{2k}z^{2k-1}}{(2k)!}}=\operatorname {csch} z,|z|<\pi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/b7564ad5932fa5f7084599d879730a4935370aab)
![{\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}E_{2k}z^{2k}}{(2k)!}}=\sec z,|z|<{\frac {\pi }{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d7ad03010883fd18659a8ea0b6759f1fa67eb04b)
![{\displaystyle \sum _{k=0}^{\infty }{\frac {E_{2k}z^{2k}}{(2k)!}}=\operatorname {sech} z,|z|<{\frac {\pi }{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d7071105db1a7775c950f733aae6f3fd9707b177)
(正矢)
[1](半正矢)
![{\displaystyle \sum _{k=0}^{\infty }{\frac {(2k)!z^{2k+1}}{2^{2k}(k!)^{2}(2k+1)}}=\arcsin z,|z|\leq 1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4fc3700c4addbf8311c6ff90b93ac759a750d6d8)
![{\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}(2k)!z^{2k+1}}{2^{2k}(k!)^{2}(2k+1)}}=\operatorname {arcsinh} {z},|z|\leq 1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e915cadf00a2f6f95ccc6ae99dbf5c5b574a820b)
![{\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k+1}}{2k+1}}=\arctan z,|z|<1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bde385b223a3706eb46a282d932a6dc758bbd8fa)
![{\displaystyle \sum _{k=0}^{\infty }{\frac {z^{2k+1}}{2k+1}}=\operatorname {arctanh} z,|z|<1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/33cab9855e7ab0d8b6e59cdfe1e8e99cef53d093)
![{\displaystyle \ln 2+\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|\leq 1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ea418d43688db9537a8b965838306a48a90840a7)
[2]
[2]
![{\displaystyle \sum _{n=0}^{\infty }{\frac {\prod _{k=0}^{n-1}(4k^{2}+\alpha ^{2})}{(2n)!}}z^{2n}+\sum _{n=0}^{\infty }{\frac {\alpha \prod _{k=0}^{n-1}[(2k+1)^{2}+\alpha ^{2}]}{(2n+1)!}}z^{2n+1}=e^{\alpha \arcsin {z}},|z|\leq 1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7690094e2c29c30c517059014511d42f93f0912a)
(參見二項式定理)
- [3]
![{\displaystyle \sum _{k=0}^{\infty }{{\alpha +k-1} \choose k}z^{k}={\frac {1}{(1-z)^{\alpha }}},|z|<1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d69e6455c13c71f8e74ce0760ccc2f9fc11ac70d)
- [3]
(卡塔蘭數的母函數)
- [3]
(中心二項式係數的母函數)
- [3]
![{\displaystyle \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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/10c3c2d66060add977823b4848d7212af4b4b68f)
![{\displaystyle \sum _{k=1}^{\infty }H_{k}z^{k}={\frac {-\ln(1-z)}{1-z}},|z|<1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/890b6859948e31ec717858a6a6b1582db3673345)
![{\displaystyle \sum _{k=1}^{\infty }{\frac {H_{k}}{k+1}}z^{k+1}={\frac {1}{2}}\left[\ln(1-z)\right]^{2},\qquad |z|<1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a1c2c3f140738f0c5c61f88f041f311fbda3a340)
[2]
[2]
![{\displaystyle \sum _{k=0}^{n}{n \choose k}=2^{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b30fdd28895f157a1d1f254f931879606064ce1c)
![{\displaystyle \sum _{k=0}^{n}(-1)^{k}{n \choose k}=0,{\text{ 其中 }}n>0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ce0294bc231afa2c42b5c2377a047d36b1b55b9f)
![{\displaystyle \sum _{k=0}^{n}{k \choose m}={n+1 \choose m+1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fad96c9dbb6c1228a1f7264d6feea813478e34ea)
(參見多重集)
(參見Vandermonde恆等式)
- 正弦及餘弦的求和詳見Fourier級數。
![{\displaystyle \sum _{k=1}^{\infty }{\frac {\sin(k\theta )}{k}}={\frac {\pi -\theta }{2}},0<\theta <2\pi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/e191794b1821b1f4608a4d21721396e2a705050b)
![{\displaystyle \sum _{k=1}^{\infty }{\frac {\cos(k\theta )}{k}}=-{\frac {1}{2}}\ln(2-2\cos \theta ),\theta \in \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/b7351fa56f21f8f5e5934934d36e7d98abb9176c)
![{\displaystyle \sum _{k=0}^{\infty }{\frac {\sin[(2k+1)\theta ]}{2k+1}}={\frac {\pi }{4}},0<\theta <\pi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/faf41e942b227c04e70af874e45bddb621602e58)
[4]
![{\displaystyle \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}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6c9a71d157f3e6aecf7c679c9d826cf2ed78772)
![{\displaystyle \sum _{k=1}^{n-1}\sin {\frac {\pi k}{n}}=\cot {\frac {\pi }{2n}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cd1e592cdc3214ad2a61e0a4d6c8c171b9bbc237)
![{\displaystyle \sum _{k=1}^{n-1}\sin {\frac {2\pi k}{n}}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/538dd88d3f15d24a398e3f106d0a6092725fbeca)
[5]
![{\displaystyle \sum _{k=1}^{n-1}\csc ^{2}{\frac {\pi k}{n}}={\frac {n^{2}-1}{3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/036c3d6e188cf05baf35356bf314e236fb5a45ed)
![{\displaystyle \sum _{k=1}^{n-1}\csc ^{4}{\frac {\pi k}{n}}={\frac {n^{4}+10n^{2}-11}{45}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e969e8c1e28c457892ad6902866438f84193c32)
[6]
![{\displaystyle \sum _{n=0}^{\infty }{\frac {1}{n^{2}+a^{2}}}={\frac {1+a\pi \coth(a\pi )}{2a^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ca1fc8f8afa2921f121e9d5b13b9c03a3b9f7dac)
![{\displaystyle \displaystyle \sum _{n=0}^{\infty }{\frac {1}{n^{4}+4a^{4}}}={\dfrac {1+a\pi \coth(a\pi )}{8a^{4}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3d22f3121e6ee4f65f0d91e632f866b937cb0ddf)
使用部分分式分解方法,能夠將任何關於
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。[7]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。