List of mathematical series

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\}} denotes the fractional part of x {\displaystyle x}
  • B n ( x ) {\displaystyle B_{n}(x)} is a Bernoulli polynomial.
  • B n {\displaystyle B_{n}} is a Bernoulli number, and here, B 1 = 1 2 . {\displaystyle B_{1}=-{\frac {1}{2}}.}
  • E n {\displaystyle E_{n}} is an Euler number.
  • ζ ( s ) {\displaystyle \zeta (s)} is the Riemann zeta function.
  • Γ ( z ) {\displaystyle \Gamma (z)} is the gamma function.
  • ψ n ( z ) {\displaystyle \psi _{n}(z)} is a polygamma function.
  • Li s ( z ) {\displaystyle \operatorname {Li} _{s}(z)} is a polylogarithm.
  • ( n k ) {\displaystyle n \choose k} is binomial coefficient
  • exp ( x ) {\displaystyle \exp(x)} denotes exponential of x {\displaystyle x}

Sums of powers

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}}} (the 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

Low-order polylogarithms

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} (see 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

  • 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}} (cf. mean of 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}} (cf. 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)} is the Touchard polynomials.

Trigonometric, inverse trigonometric, hyperbolic, and inverse hyperbolic functions relationship

  • 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

  • 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

  • ( 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} (see 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}}} , generating function of the 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

(See harmonic numbers, themselves defined H n = j = 1 n 1 j {\textstyle H_{n}=\sum _{j=1}^{n}{\frac {1}{j}}} , and H ( x ) {\displaystyle H(x)} generalized to the real numbers)

  • 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

  • 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}} (see 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} (see 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

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

  • 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} can be reduced to a finite series of polygamma functions, by use of partial fraction decomposition,[8] as explained 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

  • 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)} (see the 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

These numeric series can be found by plugging in numbers from the series listed above.

Alternating harmonic series

  • 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

  • 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

  • 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

  • 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

  • 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}

See also

Notes

  1. ^ Weisstein, Eric W. "Haversine". MathWorld. Wolfram Research, Inc. Archived from the original on 2005-03-10. Retrieved 2015-11-06.
  2. ^ a b c d Wilf, Herbert R. (1994). generatingfunctionology (PDF). Academic Press, Inc.
  3. ^ a b c d "Theoretical computer science cheat sheet" (PDF).
  4. ^ Calculate the Fourier expansion of the function f ( x ) = π 4 {\displaystyle f(x)={\frac {\pi }{4}}} on the interval 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}}}
  5. ^ "Bernoulli polynomials: Series representations (subsection 06/02)". Wolfram Research. Retrieved 2 June 2011.
  6. ^ Hofbauer, Josef. "A simple proof of 1 + 1/22 + 1/32 + ··· = π2/6 and related identities" (PDF). Retrieved 2 June 2011.
  7. ^ Sondow, Jonathan; Weisstein, Eric W. "Riemann Zeta Function (eq. 52)". MathWorld—A Wolfram Web Resource.
  8. ^ Abramowitz, Milton; Stegun, Irene (1964). "6.4 Polygamma functions". Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. p. 260. ISBN 0-486-61272-4.

References

  • 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}}}