This list of mathematical series contains formulae for finite and infinite sums. It can be used in conjunction with other tools for evaluating sums.
Here, 0 0 {\displaystyle 0^{0}} is taken to have the value 1 {\displaystyle 1} { x } {\displaystyle \{x\}} x {\displaystyle x} B n ( x ) {\displaystyle B_{n}(x)} Bernoulli polynomial . B n {\displaystyle B_{n}} Bernoulli number , and here, B 1 = − 1 2 . {\displaystyle B_{1}=-{\frac {1}{2}}.} E n {\displaystyle E_{n}} Euler number . ζ ( s ) {\displaystyle \zeta (s)} Riemann zeta function . Γ ( z ) {\displaystyle \Gamma (z)} gamma function . ψ n ( z ) {\displaystyle \psi _{n}(z)} polygamma function . Li s ( z ) {\displaystyle \operatorname {Li} _{s}(z)} polylogarithm . ( n k ) {\displaystyle n \choose k} binomial coefficient exp ( x ) {\displaystyle \exp(x)} exponential of x {\displaystyle x} Sums of powers [ edit ] See Faulhaber's formula .
∑ 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}}} The first few values are:
∑ 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}}} See zeta constants .
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)!}}} The first few values are:
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}}} Basel problem ) 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}}} Power series [ edit ] Low-order polylogarithms [ edit ] Finite sums:
∑ k = m n z k = z m − z n + 1 1 − z {\displaystyle \sum _{k=m}^{n}z^{k}={\frac {z^{m}-z^{n+1}}{1-z}}} geometric series ) ∑ 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}}} ∑ k = 1 n z k = 1 − z n + 1 1 − z − 1 = z − z n + 1 1 − z {\displaystyle \sum _{k=1}^{n}z^{k}={\frac {1-z^{n+1}}{1-z}}-1={\frac {z-z^{n+1}}{1-z}}} ∑ 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}}} Infinite sums, valid for | z | < 1 {\displaystyle |z|<1} polylogarithm ):
Li n ( z ) = ∑ k = 1 ∞ z k k n {\displaystyle \operatorname {Li} _{n}(z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k^{n}}}} The following is a useful property to calculate low-integer-order polylogarithms recursively in closed form :
d d z Li n ( z ) = Li n − 1 ( z ) z {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} z}}\operatorname {Li} _{n}(z)={\frac {\operatorname {Li} _{n-1}(z)}{z}}} 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}}}} Exponential function [ edit ] ∑ 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}} Poisson distribution ) ∑ 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}} second moment of Poisson distribution) ∑ 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)} where T n ( z ) {\displaystyle T_{n}(z)} Touchard polynomials .
Trigonometric, inverse trigonometric, hyperbolic, and inverse hyperbolic functions relationship [ edit ] ∑ 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 ) ! = sech z , | z | < π 2 {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}E_{2k}z^{2k}}{(2k)!}}=\operatorname {sech} z,|z|<{\frac {\pi }{2}}} ∑ k = 0 ∞ E 2 k z 2 k ( 2 k ) ! = sec z , | z | < π 2 {\displaystyle \sum _{k=0}^{\infty }{\frac {E_{2k}z^{2k}}{(2k)!}}=\sec 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} versine ) ∑ 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] haversine ) ∑ 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} ∑ k = 2 ∞ ( k ⋅ arctanh ( 1 k ) − 1 ) = 3 − ln ( 4 π ) 2 {\displaystyle \sum _{k=2}^{\infty }\left(k\cdot \operatorname {arctanh} \left({\frac {1}{k}}\right)-1\right)={\frac {3-\ln(4\pi )}{2}}} Modified-factorial denominators [ edit ] ∑ 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} Binomial coefficients [ edit ] ( 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} Binomial theorem § Newton's generalized binomial theorem )[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}}} generating function of the Catalan numbers [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}}} Central binomial coefficients [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}}} Harmonic numbers [ edit ] (See harmonic numbers , themselves defined H n = ∑ j = 1 n 1 j {\textstyle H_{n}=\sum _{j=1}^{n}{\frac {1}{j}}} H ( x ) {\displaystyle H(x)}
∑ 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] ∑ n = 0 ∞ x 2 n 2 ( n + x ) = x π 2 6 − H ( x ) {\displaystyle \sum _{n=0}^{\infty }{\frac {x^{2}}{n^{2}(n+x)}}=x{\frac {\pi ^{2}}{6}}-H(x)} Binomial coefficients [ edit ] ∑ k = 0 n ( n k ) = 2 n {\displaystyle \sum _{k=0}^{n}{n \choose k}=2^{n}} ∑ k = 0 n ( n k ) 2 = ( 2 n n ) {\displaystyle \sum _{k=0}^{n}{n \choose k}^{2}={2n \choose n}} ∑ k = 0 n ( − 1 ) k ( n k ) = 0 , where n ≥ 1 {\displaystyle \sum _{k=0}^{n}(-1)^{k}{n \choose k}=0,{\text{ where }}n\geq 1} ∑ 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}} Multiset ) ∑ k = 0 n ( α k ) ( β n − k ) = ( α + β n ) , where α + β ≥ n {\displaystyle \sum _{k=0}^{n}{\alpha \choose k}{\beta \choose n-k}={\alpha +\beta \choose n},{\text{where}}\ \alpha +\beta \geq n} Vandermonde identity ) ∑ A ∈ P ( E ) 1 = 2 n , where E is a finite set, and card( E ) = n {\displaystyle \sum _{A\ \in \ {\mathcal {P}}(E)}1=2^{n}{\text{, where }}E{\text{ is a finite set, and card(}}E{\text{) = n}}} ∑ { ( A , B ) ∈ ( P ( E ) ) 2 A ⊂ B 1 = 3 n , where E is a finite set, and card( E ) = n {\displaystyle \sum _{\begin{cases}(A,\ B)\ \in \ ({\mathcal {P}}(E))^{2}\\A\ \subset \ B\end{cases}}1=3^{n}{\text{, where }}E{\text{ is a finite set, and card(}}E{\text{) = n}}} ∑ A ∈ P ( E ) c a r d ( A ) = n 2 n − 1 , where E is a finite set, and card( E ) = n {\displaystyle \sum _{A\ \in \ {\mathcal {P}}(E)}card(A)=n2^{n-1}{\text{, where }}E{\text{ is a finite set, and card(}}E{\text{) = n}}} Trigonometric functions [ edit ] Sums of sines and cosines arise in Fourier series .
∑ k = 1 ∞ cos ( k θ ) k = − 1 2 ln ( 2 − 2 cos θ ) = − ln ( 2 sin θ 2 ) , 0 < θ < 2 π {\displaystyle \sum _{k=1}^{\infty }{\frac {\cos(k\theta )}{k}}=-{\frac {1}{2}}\ln(2-2\cos \theta )=-\ln \left(2\sin {\frac {\theta }{2}}\right),0<\theta <2\pi } ∑ 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 ∞ ( − 1 ) k − 1 k cos ( k θ ) = 1 2 ln ( 2 + 2 cos θ ) = ln ( 2 cos θ 2 ) , 0 ≤ θ < π {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{k}}\cos(k\theta )={\frac {1}{2}}\ln(2+2\cos \theta )=\ln \left(2\cos {\frac {\theta }{2}}\right),0\leq \theta <\pi } ∑ k = 1 ∞ ( − 1 ) k − 1 k sin ( k θ ) = θ 2 , − π 2 ≤ θ ≤ π 2 {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{k}}\sin(k\theta )={\frac {\theta }{2}},-{\frac {\pi }{2}}\leq \theta \leq {\frac {\pi }{2}}} ∑ k = 1 ∞ cos ( 2 k θ ) 2 k = − 1 2 ln ( 2 sin θ ) , 0 < θ < π {\displaystyle \sum _{k=1}^{\infty }{\frac {\cos(2k\theta )}{2k}}=-{\frac {1}{2}}\ln(2\sin \theta ),0<\theta <\pi } ∑ k = 1 ∞ sin ( 2 k θ ) 2 k = π − 2 θ 4 , 0 < θ < π {\displaystyle \sum _{k=1}^{\infty }{\frac {\sin(2k\theta )}{2k}}={\frac {\pi -2\theta }{4}},0<\theta <\pi } ∑ k = 0 ∞ cos [ ( 2 k + 1 ) θ ] 2 k + 1 = 1 2 ln ( cot θ 2 ) , 0 < θ < π {\displaystyle \sum _{k=0}^{\infty }{\frac {\cos[(2k+1)\theta ]}{2k+1}}={\frac {1}{2}}\ln \left(\cot {\frac {\theta }{2}}\right),0<\theta <\pi } ∑ 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 } [4] ∑ k = 1 ∞ sin ( 2 π k x ) k = π ( 1 2 − { x } ) , x ∈ R {\displaystyle \sum _{k=1}^{\infty }{\frac {\sin(2\pi kx)}{k}}=\pi \left({\dfrac {1}{2}}-\{x\}\right),\ x\in \mathbb {R} } ∑ k = 1 ∞ sin ( 2 π k x ) k 2 n − 1 = ( − 1 ) n ( 2 π ) 2 n − 1 2 ( 2 n − 1 ) ! B 2 n − 1 ( { x } ) , x ∈ R , n ∈ N {\displaystyle \sum \limits _{k=1}^{\infty }{\frac {\sin \left(2\pi kx\right)}{k^{2n-1}}}=(-1)^{n}{\frac {(2\pi )^{2n-1}}{2(2n-1)!}}B_{2n-1}(\{x\}),\ x\in \mathbb {R} ,\ n\in \mathbb {N} } ∑ k = 1 ∞ cos ( 2 π k x ) k 2 n = ( − 1 ) n − 1 ( 2 π ) 2 n 2 ( 2 n ) ! B 2 n ( { x } ) , x ∈ R , n ∈ N {\displaystyle \sum \limits _{k=1}^{\infty }{\frac {\cos \left(2\pi kx\right)}{k^{2n}}}=(-1)^{n-1}{\frac {(2\pi )^{2n}}{2(2n)!}}B_{2n}(\{x\}),\ x\in \mathbb {R} ,\ n\in \mathbb {N} } 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} [5] ∑ 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 = 0 n cos ( θ + k α ) = sin ( n + 1 ) α 2 cos ( θ + n α 2 ) sin α 2 {\displaystyle \sum _{k=0}^{n}\cos(\theta +k\alpha )={\frac {\sin {\frac {(n+1)\alpha }{2}}\cos(\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 )} [6] ∑ 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}}} Rational functions [ edit ] ∑ 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}} [7] ∑ 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 n 2 + a 2 = 1 + a π csch ( a π ) 2 a 2 {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n^{2}+a^{2}}}={\frac {1+a\pi \;{\text{csch}}(a\pi )}{2a^{2}}}} ∑ n = 0 ∞ ( 2 n + 1 ) ( − 1 ) n ( 2 n + 1 ) 2 + a 2 = π 4 sech ( a π 2 ) {\displaystyle \sum _{n=0}^{\infty }{\frac {(2n+1)(-1)^{n}}{(2n+1)^{2}+a^{2}}}={\frac {\pi }{4}}{\text{sech}}\left({\frac {a\pi }{2}}\right)} ∑ n = 0 ∞ 1 n 4 + 4 a 4 = 1 8 a 4 + π ( sinh ( 2 π a ) + sin ( 2 π a ) ) 8 a 3 ( cosh ( 2 π a ) − cos ( 2 π a ) ) {\displaystyle \displaystyle \sum _{n=0}^{\infty }{\frac {1}{n^{4}+4a^{4}}}={\dfrac {1}{8a^{4}}}+{\dfrac {\pi (\sinh(2\pi a)+\sin(2\pi a))}{8a^{3}(\cosh(2\pi a)-\cos(2\pi a))}}} An infinite series of any rational function of n {\displaystyle n} polygamma functions , by use of partial fraction decomposition ,[8] here . This fact can also be applied to finite series of rational functions, allowing the result to be computed in constant time even when the series contains a large number of terms. Exponential function [ edit ] 1 p ∑ n = 0 p − 1 exp ( 2 π i n 2 q p ) = e π i / 4 2 q ∑ n = 0 2 q − 1 exp ( − π i n 2 p 2 q ) {\displaystyle \displaystyle {\dfrac {1}{\sqrt {p}}}\sum _{n=0}^{p-1}\exp \left({\frac {2\pi in^{2}q}{p}}\right)={\dfrac {e^{\pi i/4}}{\sqrt {2q}}}\sum _{n=0}^{2q-1}\exp \left(-{\frac {\pi in^{2}p}{2q}}\right)} Landsberg–Schaar relation ) ∑ n = − ∞ ∞ e − π n 2 = π 4 Γ ( 3 4 ) {\displaystyle \displaystyle \sum _{n=-\infty }^{\infty }e^{-\pi n^{2}}={\frac {\sqrt[{4}]{\pi }}{\Gamma \left({\frac {3}{4}}\right)}}} Numeric series [ edit ] These numeric series can be found by plugging in numbers from the series listed above.
Alternating harmonic series [ edit ] ∑ k = 1 ∞ ( − 1 ) k + 1 k = 1 1 − 1 2 + 1 3 − 1 4 + ⋯ = ln 2 {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k}}={\frac {1}{1}}-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+\cdots =\ln 2} ∑ k = 1 ∞ ( − 1 ) k + 1 2 k − 1 = 1 1 − 1 3 + 1 5 − 1 7 + 1 9 − ⋯ = π 4 {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{2k-1}}={\frac {1}{1}}-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots ={\frac {\pi }{4}}} Sum of reciprocal of factorials [ edit ] ∑ k = 0 ∞ 1 k ! = 1 0 ! + 1 1 ! + 1 2 ! + 1 3 ! + 1 4 ! + ⋯ = e {\displaystyle \sum _{k=0}^{\infty }{\frac {1}{k!}}={\frac {1}{0!}}+{\frac {1}{1!}}+{\frac {1}{2!}}+{\frac {1}{3!}}+{\frac {1}{4!}}+\cdots =e} ∑ k = 0 ∞ 1 ( 2 k ) ! = 1 0 ! + 1 2 ! + 1 4 ! + 1 6 ! + 1 8 ! + ⋯ = 1 2 ( e + 1 e ) = cosh 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {1}{(2k)!}}={\frac {1}{0!}}+{\frac {1}{2!}}+{\frac {1}{4!}}+{\frac {1}{6!}}+{\frac {1}{8!}}+\cdots ={\frac {1}{2}}\left(e+{\frac {1}{e}}\right)=\cosh 1} ∑ k = 0 ∞ 1 ( 3 k ) ! = 1 0 ! + 1 3 ! + 1 6 ! + 1 9 ! + 1 12 ! + ⋯ = 1 3 ( e + 2 e cos 3 2 ) {\displaystyle \sum _{k=0}^{\infty }{\frac {1}{(3k)!}}={\frac {1}{0!}}+{\frac {1}{3!}}+{\frac {1}{6!}}+{\frac {1}{9!}}+{\frac {1}{12!}}+\cdots ={\frac {1}{3}}\left(e+{\frac {2}{\sqrt {e}}}\cos {\frac {\sqrt {3}}{2}}\right)} ∑ k = 0 ∞ 1 ( 4 k ) ! = 1 0 ! + 1 4 ! + 1 8 ! + 1 12 ! + 1 16 ! + ⋯ = 1 2 ( cos 1 + cosh 1 ) {\displaystyle \sum _{k=0}^{\infty }{\frac {1}{(4k)!}}={\frac {1}{0!}}+{\frac {1}{4!}}+{\frac {1}{8!}}+{\frac {1}{12!}}+{\frac {1}{16!}}+\cdots ={\frac {1}{2}}\left(\cos 1+\cosh 1\right)} Trigonometry and π [ edit ] ∑ k = 0 ∞ ( − 1 ) k ( 2 k + 1 ) ! = 1 1 ! − 1 3 ! + 1 5 ! − 1 7 ! + 1 9 ! + ⋯ = sin 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k+1)!}}={\frac {1}{1!}}-{\frac {1}{3!}}+{\frac {1}{5!}}-{\frac {1}{7!}}+{\frac {1}{9!}}+\cdots =\sin 1} ∑ k = 0 ∞ ( − 1 ) k ( 2 k ) ! = 1 0 ! − 1 2 ! + 1 4 ! − 1 6 ! + 1 8 ! + ⋯ = cos 1 {\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k)!}}={\frac {1}{0!}}-{\frac {1}{2!}}+{\frac {1}{4!}}-{\frac {1}{6!}}+{\frac {1}{8!}}+\cdots =\cos 1} ∑ k = 1 ∞ 1 k 2 + 1 = 1 2 + 1 5 + 1 10 + 1 17 + ⋯ = 1 2 ( π coth π − 1 ) {\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k^{2}+1}}={\frac {1}{2}}+{\frac {1}{5}}+{\frac {1}{10}}+{\frac {1}{17}}+\cdots ={\frac {1}{2}}(\pi \coth \pi -1)} ∑ k = 1 ∞ ( − 1 ) k k 2 + 1 = − 1 2 + 1 5 − 1 10 + 1 17 + ⋯ = 1 2 ( π csch π − 1 ) {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k}}{k^{2}+1}}=-{\frac {1}{2}}+{\frac {1}{5}}-{\frac {1}{10}}+{\frac {1}{17}}+\cdots ={\frac {1}{2}}(\pi \operatorname {csch} \pi -1)} 3 + 4 2 × 3 × 4 − 4 4 × 5 × 6 + 4 6 × 7 × 8 − 4 8 × 9 × 10 + ⋯ = π {\displaystyle 3+{\frac {4}{2\times 3\times 4}}-{\frac {4}{4\times 5\times 6}}+{\frac {4}{6\times 7\times 8}}-{\frac {4}{8\times 9\times 10}}+\cdots =\pi } Reciprocal of tetrahedral numbers [ edit ] ∑ k = 1 ∞ 1 T e k = 1 1 + 1 4 + 1 10 + 1 20 + 1 35 + ⋯ = 3 2 {\displaystyle \sum _{k=1}^{\infty }{\frac {1}{Te_{k}}}={\frac {1}{1}}+{\frac {1}{4}}+{\frac {1}{10}}+{\frac {1}{20}}+{\frac {1}{35}}+\cdots ={\frac {3}{2}}} Where T e n = ∑ k = 1 n T k {\displaystyle Te_{n}=\sum _{k=1}^{n}T_{k}}
Exponential and logarithms [ edit ] ∑ k = 0 ∞ 1 ( 2 k + 1 ) ( 2 k + 2 ) = 1 1 × 2 + 1 3 × 4 + 1 5 × 6 + 1 7 × 8 + 1 9 × 10 + ⋯ = ln 2 {\displaystyle \sum _{k=0}^{\infty }{\frac {1}{(2k+1)(2k+2)}}={\frac {1}{1\times 2}}+{\frac {1}{3\times 4}}+{\frac {1}{5\times 6}}+{\frac {1}{7\times 8}}+{\frac {1}{9\times 10}}+\cdots =\ln 2} ∑ k = 1 ∞ 1 2 k k = 1 2 + 1 8 + 1 24 + 1 64 + 1 160 + ⋯ = ln 2 {\displaystyle \sum _{k=1}^{\infty }{\frac {1}{2^{k}k}}={\frac {1}{2}}+{\frac {1}{8}}+{\frac {1}{24}}+{\frac {1}{64}}+{\frac {1}{160}}+\cdots =\ln 2} ∑ k = 1 ∞ ( − 1 ) k + 1 2 k k + ∑ k = 1 ∞ ( − 1 ) k + 1 3 k k = ( 1 2 + 1 3 ) − ( 1 8 + 1 18 ) + ( 1 24 + 1 81 ) − ( 1 64 + 1 324 ) + ⋯ = ln 2 {\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{2^{k}k}}+\sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{3^{k}k}}={\Bigg (}{\frac {1}{2}}+{\frac {1}{3}}{\Bigg )}-{\Bigg (}{\frac {1}{8}}+{\frac {1}{18}}{\Bigg )}+{\Bigg (}{\frac {1}{24}}+{\frac {1}{81}}{\Bigg )}-{\Bigg (}{\frac {1}{64}}+{\frac {1}{324}}{\Bigg )}+\cdots =\ln 2} ∑ k = 1 ∞ 1 3 k k + ∑ k = 1 ∞ 1 4 k k = ( 1 3 + 1 4 ) + ( 1 18 + 1 32 ) + ( 1 81 + 1 192 ) + ( 1 324 + 1 1024 ) + ⋯ = ln 2 {\displaystyle \sum _{k=1}^{\infty }{\frac {1}{3^{k}k}}+\sum _{k=1}^{\infty }{\frac {1}{4^{k}k}}={\Bigg (}{\frac {1}{3}}+{\frac {1}{4}}{\Bigg )}+{\Bigg (}{\frac {1}{18}}+{\frac {1}{32}}{\Bigg )}+{\Bigg (}{\frac {1}{81}}+{\frac {1}{192}}{\Bigg )}+{\Bigg (}{\frac {1}{324}}+{\frac {1}{1024}}{\Bigg )}+\cdots =\ln 2} ∑ k = 1 ∞ 1 n k k = ln ( n n − 1 ) {\displaystyle \sum _{k=1}^{\infty }{\frac {1}{n^{k}k}}=\ln \left({\frac {n}{n-1}}\right)} ∀ n > 1 {\displaystyle \forall n>1} See also [ edit ] ^ Weisstein, Eric W. "Haversine" . MathWorld Wolfram Research, Inc. Archived from the original on 2005-03-10. Retrieved 2015-11-06 .^ a b c d Wilf, Herbert R. (1994). generatingfunctionology (PDF) . Academic Press, Inc. ^ a b c d "Theoretical computer science cheat sheet" (PDF) .^ Calculate the Fourier expansion of the function f ( x ) = π 4 {\displaystyle f(x)={\frac {\pi }{4}}} 0 < x < π {\displaystyle 0<x<\pi } π 4 = ∑ n = 0 ∞ c n sin [ n x ] + d n cos [ n x ] {\displaystyle {\frac {\pi }{4}}=\sum _{n=0}^{\infty }c_{n}\sin[nx]+d_{n}\cos[nx]} ⇒ { c n = { 1 n ( n odd ) 0 ( n even ) d n = 0 ( ∀ n ) {\displaystyle \Rightarrow {\begin{cases}c_{n}={\begin{cases}{\frac {1}{n}}\quad (n{\text{ odd}})\\0\quad (n{\text{ even}})\end{cases}}\\d_{n}=0\quad (\forall n)\end{cases}}} ^ "Bernoulli polynomials: Series representations (subsection 06/02)" . Wolfram Research . Retrieved 2 June 2011 .^ Hofbauer, Josef. "A simple proof of 1 + 1/22 + 1/32 + ··· = π 2 /6 and related identities" (PDF) . Retrieved 2 June 2011 . ^ Sondow, Jonathan; Weisstein, Eric W. "Riemann Zeta Function (eq. 52)" . MathWorld —A Wolfram Web Resource ^ Abramowitz, Milton ; Stegun, Irene (1964). "6.4 Polygamma functions" . Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables 260 . ISBN 0-486-61272-4 References [ edit ] Many books with a list of integrals also have a list of series. π 4 = ∑ k = 0 ∞ ( − 1 ) k 2 k + 1 {\displaystyle {\frac {\pi }{4}}=\sum _{k=0}^{\infty }{\frac {\left(-1\right)^{k}}{2k+1}}} 
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