Remove ads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 0 0 = 1 {\displaystyle 0^{0}=1} ； B n ( x ) {\displaystyle B_{n}(x)} 表示Bernoulli多項式； B n {\displaystyle B_{n}} 表示Bernoulli數，其中， B 1 = − 1 2 {\displaystyle B_{1}=-{\frac {1}{2}}} ； E n {\displaystyle E_{n}} 表示Euler數； ζ ( s ) {\displaystyle \zeta (s)} 表示黎曼ζ函數； Γ ( z ) {\displaystyle \Gamma (z)} 表示Γ函數； ψ n ( z ) {\displaystyle \psi _{n}(z)} 表示多伽瑪函數； Li s ( z ) {\displaystyle \operatorname {Li} _{s}(z)} 表示多重對數函數。 Remove ads冪求和 參見等冪求和。 ∑ k = 0 m k n − 1 = B n ( m + 1 ) − B n n {\displaystyle \sum _{k=0}^{m}k^{n-1}={\frac {B_{n}(m+1)-B_{n}}{n}}} 其中，一次方項、平方項及立方項之和的公式如下︰ ∑ k = 1 m k = m ( m + 1 ) 2 {\displaystyle \sum _{k=1}^{m}k={\frac {m(m+1)}{2}}} ∑ k = 1 m k 2 = m ( m + 1 ) ( 2 m + 1 ) 6 = m 3 3 + m 2 2 + m 6 {\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}}} ∑ k = 1 m k 3 = [ m ( m + 1 ) 2 ] 2 = m 4 4 + m 3 2 + m 2 4 {\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}}} 參見ζ常數. ζ ( 2 n ) = ∑ k = 1 ∞ 1 k 2 n = ( − 1 ) n + 1 B 2 n ( 2 π ) 2 n 2 ( 2 n ) ! {\displaystyle \zeta (2n)=\sum _{k=1}^{\infty }{\frac {1}{k^{2n}}}=(-1)^{n+1}{\frac {B_{2n}(2\pi )^{2n}}{2(2n)!}}} 其中，前幾項為︰ ζ ( 2 ) = ∑ k = 1 ∞ 1 k 2 = π 2 6 {\displaystyle \zeta (2)=\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}={\frac {\pi ^{2}}{6}}} （巴塞爾問題） ζ ( 4 ) = ∑ k = 1 ∞ 1 k 4 = π 4 90 {\displaystyle \zeta (4)=\sum _{k=1}^{\infty }{\frac {1}{k^{4}}}={\frac {\pi ^{4}}{90}}} ζ ( 6 ) = ∑ k = 1 ∞ 1 k 6 = π 6 945 {\displaystyle \zeta (6)=\sum _{k=1}^{\infty }{\frac {1}{k^{6}}}={\frac {\pi ^{6}}{945}}} Remove ads冪級數 低次數多重對數函數 有限項求和︰ ∑ k = 0 n z k = 1 − z n + 1 1 − z {\displaystyle \sum _{k=0}^{n}z^{k}={\frac {1-z^{n+1}}{1-z}}} , (等比數列) ∑ k = 1 n k z k = z 1 − ( n + 1 ) z n + n z n + 1 ( 1 − z ) 2 {\displaystyle \sum _{k=1}^{n}kz^{k}=z{\frac {1-(n+1)z^{n}+nz^{n+1}}{(1-z)^{2}}}} ∑ k = 1 n k 2 z k = z 1 + z − ( n + 1 ) 2 z n + ( 2 n 2 + 2 n − 1 ) z n + 1 − n 2 z n + 2 ( 1 − z ) 3 {\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}}}} ∑ k = 1 n k m z k = ( z d d z ) m 1 − z n + 1 1 − z {\displaystyle \sum _{k=1}^{n}k^{m}z^{k}=\left(z{\frac {d}{dz}}\right)^{m}{\frac {1-z^{n+1}}{1-z}}} 無限項求和，其中 | z | < 1 {\displaystyle |z|<1} （參見多重對數函數）︰ Li n ( z ) = ∑ k = 1 ∞ z k k n {\displaystyle \operatorname {Li} _{n}(z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k^{n}}}} 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︰ d d z Li n ( z ) = Li n − 1 ( z ) z {\displaystyle {\frac {d}{dz}}\operatorname {Li} _{n}(z)={\frac {\operatorname {Li} _{n-1}(z)}{z}}} 前幾項分別為︰ Li 1 ( z ) = ∑ k = 1 ∞ z k k = − ln ( 1 − z ) {\displaystyle \operatorname {Li} _{1}(z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k}}=-\ln(1-z)} Li 0 ( z ) = ∑ k = 1 ∞ z k = z 1 − z {\displaystyle \operatorname {Li} _{0}(z)=\sum _{k=1}^{\infty }z^{k}={\frac {z}{1-z}}} Li − 1 ( z ) = ∑ k = 1 ∞ k z k = z ( 1 − z ) 2 {\displaystyle \operatorname {Li} _{-1}(z)=\sum _{k=1}^{\infty }kz^{k}={\frac {z}{(1-z)^{2}}}} Li − 2 ( z ) = ∑ k = 1 ∞ k 2 z k = z ( 1 + z ) ( 1 − z ) 3 {\displaystyle \operatorname {Li} _{-2}(z)=\sum _{k=1}^{\infty }k^{2}z^{k}={\frac {z(1+z)}{(1-z)^{3}}}} Li − 3 ( z ) = ∑ k = 1 ∞ k 3 z k = z ( 1 + 4 z + z 2 ) ( 1 − z ) 4 {\displaystyle \operatorname {Li} _{-3}(z)=\sum _{k=1}^{\infty }k^{3}z^{k}={\frac {z(1+4z+z^{2})}{(1-z)^{4}}}} Li − 4 ( z ) = ∑ k = 1 ∞ k 4 z k = z ( 1 + z ) ( 1 + 10 z + z 2 ) ( 1 − z ) 5 {\displaystyle \operatorname {Li} _{-4}(z)=\sum _{k=1}^{\infty }k^{4}z^{k}={\frac {z(1+z)(1+10z+z^{2})}{(1-z)^{5}}}} Remove ads指數函數 ∑ k = 0 ∞ z k k ! = e z {\displaystyle \sum _{k=0}^{\infty }{\frac {z^{k}}{k!}}=e^{z}} ∑ k = 0 ∞ k z k k ! = z e z {\displaystyle \sum _{k=0}^{\infty }k{\frac {z^{k}}{k!}}=ze^{z}} （參見Poisson分佈的數學期望） ∑ k = 0 ∞ k 2 z k k ! = ( z + z 2 ) e z {\displaystyle \sum _{k=0}^{\infty }k^{2}{\frac {z^{k}}{k!}}=(z+z^{2})e^{z}} （參見Poisson分佈的二階矩） ∑ k = 0 ∞ k 3 z k k ! = ( z + 3 z 2 + z 3 ) e z {\displaystyle \sum _{k=0}^{\infty }k^{3}{\frac {z^{k}}{k!}}=(z+3z^{2}+z^{3})e^{z}} ∑ k = 0 ∞ k 4 z k k ! = ( z + 7 z 2 + 6 z 3 + z 4 ) e z {\displaystyle \sum _{k=0}^{\infty }k^{4}{\frac {z^{k}}{k!}}=(z+7z^{2}+6z^{3}+z^{4})e^{z}} ∑ k = 0 ∞ k n z k k ! = z d d z ∑ k = 0 ∞ k n − 1 z k k ! = e z T n ( z ) {\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)} 其中， T n ( z ) {\displaystyle T_{n}(z)} 表示圖沙德多項式。 Remove ads三角函數、反三角函數、雙曲函數及反雙曲函數 ∑ k = 0 ∞ ( − 1 ) k z 2 k + 1 ( 2 k + 1 ) ! = sin z {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k+1}}{(2k+1)!}}=\sin z} ∑ k = 0 ∞ z 2 k + 1 ( 2 k + 1 ) ! = sinh z {\displaystyle \sum _{k=0}^{\infty }{\frac {z^{2k+1}}{(2k+1)!}}=\sinh z} ∑ k = 0 ∞ ( − 1 ) k z 2 k ( 2 k ) ! = cos z {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k}}{(2k)!}}=\cos z} ∑ k = 0 ∞ z 2 k ( 2 k ) ! = cosh z {\displaystyle \sum _{k=0}^{\infty }{\frac {z^{2k}}{(2k)!}}=\cosh z} ∑ k = 1 ∞ ( − 1 ) k − 1 ( 2 2 k − 1 ) 2 2 k B 2 k z 2 k − 1 ( 2 k ) ! = tan z , | z | < π 2 {\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}}} ∑ k = 1 ∞ ( 2 2 k − 1 ) 2 2 k B 2 k z 2 k − 1 ( 2 k ) ! = tanh z , | z | < π 2 {\displaystyle \sum _{k=1}^{\infty }{\frac {(2^{2k}-1)2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\tanh z,|z|<{\frac {\pi }{2}}} ∑ k = 0 ∞ ( − 1 ) k 2 2 k B 2 k z 2 k − 1 ( 2 k ) ! = cot z , | z | < π {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\cot z,|z|<\pi } ∑ k = 0 ∞ 2 2 k B 2 k z 2 k − 1 ( 2 k ) ! = coth z , | z | < π {\displaystyle \sum _{k=0}^{\infty }{\frac {2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\coth z,|z|<\pi } ∑ k = 0 ∞ ( − 1 ) k − 1 ( 2 2 k − 2 ) B 2 k z 2 k − 1 ( 2 k ) ! = csc z , | z | < π {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k-1}(2^{2k}-2)B_{2k}z^{2k-1}}{(2k)!}}=\csc z,|z|<\pi } ∑ k = 0 ∞ − ( 2 2 k − 2 ) B 2 k z 2 k − 1 ( 2 k ) ! = csch z , | z | < π {\displaystyle \sum _{k=0}^{\infty }{\frac {-(2^{2k}-2)B_{2k}z^{2k-1}}{(2k)!}}=\operatorname {csch} z,|z|<\pi } ∑ k = 0 ∞ ( − 1 ) k E 2 k z 2 k ( 2 k ) ! = sec z , | z | < π 2 {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}E_{2k}z^{2k}}{(2k)!}}=\sec z,|z|<{\frac {\pi }{2}}} ∑ k = 0 ∞ E 2 k z 2 k ( 2 k ) ! = sech z , | z | < π 2 {\displaystyle \sum _{k=0}^{\infty }{\frac {E_{2k}z^{2k}}{(2k)!}}=\operatorname {sech} z,|z|<{\frac {\pi }{2}}} ∑ k = 1 ∞ ( − 1 ) k − 1 z 2 k ( 2 k ) ! = ver z {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}z^{2k}}{(2k)!}}=\operatorname {ver} z} （正矢） ∑ k = 1 ∞ ( − 1 ) k − 1 z 2 k 2 ( 2 k ) ! = hav z {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}z^{2k}}{2(2k)!}}=\operatorname {hav} z} [1]（半正矢） ∑ k = 0 ∞ ( 2 k ) ! z 2 k + 1 2 2 k ( k ! ) 2 ( 2 k + 1 ) = arcsin z , | z | ≤ 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {(2k)!z^{2k+1}}{2^{2k}(k!)^{2}(2k+1)}}=\arcsin z,|z|\leq 1} ∑ k = 0 ∞ ( − 1 ) k ( 2 k ) ! z 2 k + 1 2 2 k ( k ! ) 2 ( 2 k + 1 ) = arcsinh z , | z | ≤ 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}(2k)!z^{2k+1}}{2^{2k}(k!)^{2}(2k+1)}}=\operatorname {arcsinh} {z},|z|\leq 1} ∑ k = 0 ∞ ( − 1 ) k z 2 k + 1 2 k + 1 = arctan z , | z | < 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k+1}}{2k+1}}=\arctan z,|z|<1} ∑ k = 0 ∞ z 2 k + 1 2 k + 1 = arctanh z , | z | < 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {z^{2k+1}}{2k+1}}=\operatorname {arctanh} z,|z|<1} ln 2 + ∑ k = 1 ∞ ( − 1 ) k − 1 ( 2 k ) ! z 2 k 2 2 k + 1 k ( k ! ) 2 = ln ( 1 + 1 + z 2 ) , | z | ≤ 1 {\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} Remove ads修正的分母階乘 ∑ k = 0 ∞ ( 4 k ) ! 2 4 k 2 ( 2 k ) ! ( 2 k + 1 ) ! z k = 1 − 1 − z z , | z | < 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {(4k)!}{2^{4k}{\sqrt {2}}(2k)!(2k+1)!}}z^{k}={\sqrt {\frac {1-{\sqrt {1-z}}}{z}}},|z|<1} [2] ∑ k = 0 ∞ 2 2 k ( k ! ) 2 ( k + 1 ) ( 2 k + 1 ) ! z 2 k + 2 = ( arcsin z ) 2 , | z | ≤ 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {2^{2k}(k!)^{2}}{(k+1)(2k+1)!}}z^{2k+2}=\left(\arcsin {z}\right)^{2},|z|\leq 1} [2] ∑ n = 0 ∞ ∏ k = 0 n − 1 ( 4 k 2 + α 2 ) ( 2 n ) ! z 2 n + ∑ n = 0 ∞ α ∏ k = 0 n − 1 [ ( 2 k + 1 ) 2 + α 2 ] ( 2 n + 1 ) ! z 2 n + 1 = e α arcsin z , | z | ≤ 1 {\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} Remove ads二項式系數 ( 1 + z ) α = ∑ k = 0 ∞ ( α k ) z k , | z | < 1 {\displaystyle (1+z)^{\alpha }=\sum _{k=0}^{\infty }{\alpha \choose k}z^{k},|z|<1} （參見二項式定理） [3] ∑ k = 0 ∞ ( α + k − 1 k ) z k = 1 ( 1 − z ) α , | z | < 1 {\displaystyle \sum _{k=0}^{\infty }{{\alpha +k-1} \choose k}z^{k}={\frac {1}{(1-z)^{\alpha }}},|z|<1} [3] ∑ k = 0 ∞ 1 k + 1 ( 2 k k ) z k = 1 − 1 − 4 z 2 z , | z | ≤ 1 4 {\displaystyle \sum _{k=0}^{\infty }{\frac {1}{k+1}}{2k \choose k}z^{k}={\frac {1-{\sqrt {1-4z}}}{2z}},|z|\leq {\frac {1}{4}}} （卡塔蘭數的母函數） [3] ∑ k = 0 ∞ ( 2 k k ) z k = 1 1 − 4 z , | z | < 1 4 {\displaystyle \sum _{k=0}^{\infty }{2k \choose k}z^{k}={\frac {1}{\sqrt {1-4z}}},|z|<{\frac {1}{4}}} （中心二項式系數的母函數） [3] ∑ k = 0 ∞ ( 2 k + α k ) z k = 1 1 − 4 z ( 1 − 1 − 4 z 2 z ) α , | z | < 1 4 {\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}}} Remove ads調和數 ∑ k = 1 ∞ H k z k = − ln ( 1 − z ) 1 − z , | z | < 1 {\displaystyle \sum _{k=1}^{\infty }H_{k}z^{k}={\frac {-\ln(1-z)}{1-z}},|z|<1} ∑ k = 1 ∞ H k k + 1 z k + 1 = 1 2 [ ln ( 1 − z ) ] 2 , | z | < 1 {\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} ∑ k = 1 ∞ ( − 1 ) k − 1 H 2 k 2 k + 1 z 2 k + 1 = 1 2 arctan z log ( 1 + z 2 ) , | z | < 1 {\displaystyle \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} [2] ∑ n = 0 ∞ ∑ k = 0 2 n ( − 1 ) k 2 k + 1 z 4 n + 2 4 n + 2 = 1 4 arctan z log 1 + z 1 − z , | z | < 1 {\displaystyle \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} [2] Remove ads二項式系數 ∑ k = 0 n ( n k ) = 2 n {\displaystyle \sum _{k=0}^{n}{n \choose k}=2^{n}} ∑ k = 0 n ( − 1 ) k ( n k ) = 0 , 其中 n > 0 {\displaystyle \sum _{k=0}^{n}(-1)^{k}{n \choose k}=0,{\text{ 其中 }}n>0} ∑ k = 0 n ( k m ) = ( n + 1 m + 1 ) {\displaystyle \sum _{k=0}^{n}{k \choose m}={n+1 \choose m+1}} ∑ k = 0 n ( m + k − 1 k ) = ( n + m n ) {\displaystyle \sum _{k=0}^{n}{m+k-1 \choose k}={n+m \choose n}} （參見多重集） ∑ k = 0 n ( α k ) ( β n − k ) = ( α + β n ) {\displaystyle \sum _{k=0}^{n}{\alpha \choose k}{\beta \choose n-k}={\alpha +\beta \choose n}} （參見Vandermonde恆等式） 三角函數 正弦及餘弦的求和詳見Fourier級數。 ∑ k = 1 ∞ sin ( k θ ) k = π − θ 2 , 0 < θ < 2 π {\displaystyle \sum _{k=1}^{\infty }{\frac {\sin(k\theta )}{k}}={\frac {\pi -\theta }{2}},0<\theta <2\pi } ∑ k = 1 ∞ cos ( k θ ) k = − 1 2 ln ( 2 − 2 cos θ ) , θ ∈ R {\displaystyle \sum _{k=1}^{\infty }{\frac {\cos(k\theta )}{k}}=-{\frac {1}{2}}\ln(2-2\cos \theta ),\theta \in \mathbb {R} } ∑ k = 0 ∞ sin [ ( 2 k + 1 ) θ ] 2 k + 1 = π 4 , 0 < θ < π {\displaystyle \sum _{k=0}^{\infty }{\frac {\sin[(2k+1)\theta ]}{2k+1}}={\frac {\pi }{4}},0<\theta <\pi } B n ( x ) = − n ! 2 n − 1 π n ∑ k = 1 ∞ 1 k n cos ( 2 π k x − π n 2 ) , 0 < x < 1 {\displaystyle 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} [4] ∑ k = 0 n sin ( θ + k α ) = sin ( n + 1 ) α 2 sin ( θ + n α 2 ) sin α 2 {\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}}}}} ∑ k = 1 n − 1 sin π k n = cot π 2 n {\displaystyle \sum _{k=1}^{n-1}\sin {\frac {\pi k}{n}}=\cot {\frac {\pi }{2n}}} ∑ k = 1 n − 1 sin 2 π k n = 0 {\displaystyle \sum _{k=1}^{n-1}\sin {\frac {2\pi k}{n}}=0} ∑ k = 0 n − 1 csc 2 ( θ + π k n ) = n 2 csc 2 ( n θ ) {\displaystyle \sum _{k=0}^{n-1}\csc ^{2}\left(\theta +{\frac {\pi k}{n}}\right)=n^{2}\csc ^{2}(n\theta )} [5] ∑ k = 1 n − 1 csc 2 π k n = n 2 − 1 3 {\displaystyle \sum _{k=1}^{n-1}\csc ^{2}{\frac {\pi k}{n}}={\frac {n^{2}-1}{3}}} ∑ k = 1 n − 1 csc 4 π k n = n 4 + 10 n 2 − 11 45 {\displaystyle \sum _{k=1}^{n-1}\csc ^{4}{\frac {\pi k}{n}}={\frac {n^{4}+10n^{2}-11}{45}}} 有理函數 ∑ n = a + 1 ∞ a n 2 − a 2 = 1 2 H 2 a {\displaystyle \sum _{n=a+1}^{\infty }{\frac {a}{n^{2}-a^{2}}}={\frac {1}{2}}H_{2a}} [6] ∑ n = 0 ∞ 1 n 2 + a 2 = 1 + a π coth ( a π ) 2 a 2 {\displaystyle \sum _{n=0}^{\infty }{\frac {1}{n^{2}+a^{2}}}={\frac {1+a\pi \coth(a\pi )}{2a^{2}}}} ∑ n = 0 ∞ 1 n 4 + 4 a 4 = 1 + a π coth ( a π ) 8 a 4 {\displaystyle \displaystyle \sum _{n=0}^{\infty }{\frac {1}{n^{4}+4a^{4}}}={\dfrac {1+a\pi \coth(a\pi )}{8a^{4}}}} 使用部分分式分解方法，能夠將任何關於 n {\displaystyle n} 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。 參閱 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 註釋Loading content...Loading related searches...Wikiwand - on Seamless Wikipedia browsing. 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, (等比數列)
（參見多重對數函數）︰
![{\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}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/83655857c974dd27c9b29de8cda04d7c65d334e3)
![{\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}](http://wikimedia.org/api/rest_v1/media/math/render/svg/7690094e2c29c30c517059014511d42f93f0912a)
![{\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}](http://wikimedia.org/api/rest_v1/media/math/render/svg/a1c2c3f140738f0c5c61f88f041f311fbda3a340)
![{\displaystyle \sum _{k=0}^{\infty }{\frac {\sin[(2k+1)\theta ]}{2k+1}}={\frac {\pi }{4}},0<\theta <\pi }](http://wikimedia.org/api/rest_v1/media/math/render/svg/faf41e942b227c04e70af874e45bddb621602e58)
