List of mathematical series

See also: Series (mathematics) § Examples of numerical series, and Summation

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, ${\displaystyle 0^{0}}$ is taken to have the value ${\displaystyle 1}$
• ${\displaystyle \{x\}}$ denotes the fractional part of ${\displaystyle x}$
• ${\displaystyle B_{n}(x)}$ is a Bernoulli polynomial.
• ${\displaystyle B_{n}}$ is a Bernoulli number, and here, ${\displaystyle B_{1}=-{\frac {1}{2}}.}$
• ${\displaystyle E_{n}}$ is an Euler number.
• ${\displaystyle \zeta (s)}$ is the Riemann zeta function.
• ${\displaystyle \Gamma (z)}$ is the gamma function.
• ${\displaystyle \psi _{n}(z)}$ is a polygamma function.
• ${\displaystyle \operatorname {Li} _{s}(z)}$ is a polylogarithm.
• ${\displaystyle n \choose k}$ is binomial coefficient
• ${\displaystyle \exp(x)}$ denotes exponential of ${\displaystyle x}$

Sums of powers

See Faulhaber's formula.

• ${\displaystyle \sum _{k=0}^{m}k^{n-1}={\frac {B_{n}(m+1)-B_{n}}{n}}}$

The first few values are:

• ${\displaystyle \sum _{k=1}^{m}k={\frac {m(m+1)}{2}}}$
• ${\displaystyle \sum _{k=1}^{m}k^{2}={\frac {m(m+1)(2m+1)}{6}}={\frac {m^{3}}{3}}+{\frac {m^{2}}{2}}+{\frac {m}{6}}}$
• ${\displaystyle \sum _{k=1}^{m}k^{3}=\left[{\frac {m(m+1)}{2}}\right]^{2}={\frac {m^{4}}{4}}+{\frac {m^{3}}{2}}+{\frac {m^{2}}{4}}}$

See zeta constants.

• ${\displaystyle \zeta (2n)=\sum _{k=1}^{\infty }{\frac {1}{k^{2n}}}=(-1)^{n+1}{\frac {B_{2n}(2\pi )^{2n}}{2(2n)!}}}$

The first few values are:

• ${\displaystyle \zeta (2)=\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}={\frac {\pi ^{2}}{6}}}$ (the Basel problem)
• ${\displaystyle \zeta (4)=\sum _{k=1}^{\infty }{\frac {1}{k^{4}}}={\frac {\pi ^{4}}{90}}}$
• ${\displaystyle \zeta (6)=\sum _{k=1}^{\infty }{\frac {1}{k^{6}}}={\frac {\pi ^{6}}{945}}}$

Power series

Low-order polylogarithms

Finite sums:

• ${\displaystyle \sum _{k=m}^{n}z^{k}={\frac {z^{m}-z^{n+1}}{1-z}}}$, (geometric series)
• ${\displaystyle \sum _{k=0}^{n}z^{k}={\frac {1-z^{n+1}}{1-z}}}$
• ${\displaystyle \sum _{k=1}^{n}z^{k}={\frac {1-z^{n+1}}{1-z}}-1={\frac {z-z^{n+1}}{1-z}}}$
• ${\displaystyle \sum _{k=1}^{n}kz^{k}=z{\frac {1-(n+1)z^{n}+nz^{n+1}}{(1-z)^{2}}}}$
• ${\displaystyle \sum _{k=1}^{n}k^{2}z^{k}=z{\frac {1+z-(n+1)^{2}z^{n}+(2n^{2}+2n-1)z^{n+1}-n^{2}z^{n+2}}{(1-z)^{3}}}}$
• ${\displaystyle \sum _{k=0}^{n}k^{m}z^{k}=\left(z{\frac {d}{dz}}\right)^{m}{\frac {1-z^{n+1}}{1-z}}}$

Infinite sums, valid for ${\displaystyle |z|<1}$ (see polylogarithm):

• ${\displaystyle \operatorname {Li} _{n}(z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k^{n}}}}$

The following is a useful property to calculate low-integer-order polylogarithms recursively in closed form:

• ${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} z}}\operatorname {Li} _{n}(z)={\frac {\operatorname {Li} _{n-1}(z)}{z}}}$
• ${\displaystyle \operatorname {Li} _{1}(z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k}}=-\ln(1-z)}$
• ${\displaystyle \operatorname {Li} _{0}(z)=\sum _{k=1}^{\infty }z^{k}={\frac {z}{1-z}}}$
• ${\displaystyle \operatorname {Li} _{-1}(z)=\sum _{k=1}^{\infty }kz^{k}={\frac {z}{(1-z)^{2}}}}$
• ${\displaystyle \operatorname {Li} _{-2}(z)=\sum _{k=1}^{\infty }k^{2}z^{k}={\frac {z(1+z)}{(1-z)^{3}}}}$
• ${\displaystyle \operatorname {Li} _{-3}(z)=\sum _{k=1}^{\infty }k^{3}z^{k}={\frac {z(1+4z+z^{2})}{(1-z)^{4}}}}$
• ${\displaystyle \operatorname {Li} _{-4}(z)=\sum _{k=1}^{\infty }k^{4}z^{k}={\frac {z(1+z)(1+10z+z^{2})}{(1-z)^{5}}}}$

Exponential function

• ${\displaystyle \sum _{k=0}^{\infty }{\frac {z^{k}}{k!}}=e^{z}}$
• ${\displaystyle \sum _{k=0}^{\infty }k{\frac {z^{k}}{k!}}=ze^{z}}$ (cf. mean of Poisson distribution)
• ${\displaystyle \sum _{k=0}^{\infty }k^{2}{\frac {z^{k}}{k!}}=(z+z^{2})e^{z}}$ (cf. second moment of Poisson distribution)
• ${\displaystyle \sum _{k=0}^{\infty }k^{3}{\frac {z^{k}}{k!}}=(z+3z^{2}+z^{3})e^{z}}$
• ${\displaystyle \sum _{k=0}^{\infty }k^{4}{\frac {z^{k}}{k!}}=(z+7z^{2}+6z^{3}+z^{4})e^{z}}$
• ${\displaystyle \sum _{k=0}^{\infty }k^{n}{\frac {z^{k}}{k!}}=z{\frac {d}{dz}}\sum _{k=0}^{\infty }k^{n-1}{\frac {z^{k}}{k!}}\,\!=e^{z}T_{n}(z)}$

where ${\displaystyle T_{n}(z)}$ is the Touchard polynomials.

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

• ${\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k+1}}{(2k+1)!}}=\sin z}$
• ${\displaystyle \sum _{k=0}^{\infty }{\frac {z^{2k+1}}{(2k+1)!}}=\sinh z}$
• ${\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k}}{(2k)!}}=\cos z}$
• ${\displaystyle \sum _{k=0}^{\infty }{\frac {z^{2k}}{(2k)!}}=\cosh z}$
• ${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}(2^{2k}-1)2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\tan z,|z|<{\frac {\pi }{2}}}$
• ${\displaystyle \sum _{k=1}^{\infty }{\frac {(2^{2k}-1)2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\tanh z,|z|<{\frac {\pi }{2}}}$
• ${\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\cot z,|z|<\pi }$
• ${\displaystyle \sum _{k=0}^{\infty }{\frac {2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\coth z,|z|<\pi }$
• ${\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k-1}(2^{2k}-2)B_{2k}z^{2k-1}}{(2k)!}}=\csc z,|z|<\pi }$
• ${\displaystyle \sum _{k=0}^{\infty }{\frac {-(2^{2k}-2)B_{2k}z^{2k-1}}{(2k)!}}=\operatorname {csch} z,|z|<\pi }$
• ${\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}E_{2k}z^{2k}}{(2k)!}}=\operatorname {sech} z,|z|<{\frac {\pi }{2}}}$
• ${\displaystyle \sum _{k=0}^{\infty }{\frac {E_{2k}z^{2k}}{(2k)!}}=\sec z,|z|<{\frac {\pi }{2}}}$
• ${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}z^{2k}}{(2k)!}}=\operatorname {ver} z}$ (versine)
• ${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}z^{2k}}{2(2k)!}}=\operatorname {hav} z}$^[1] (haversine)
• ${\displaystyle \sum _{k=0}^{\infty }{\frac {(2k)!z^{2k+1}}{2^{2k}(k!)^{2}(2k+1)}}=\arcsin z,|z|\leq 1}$
• ${\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}(2k)!z^{2k+1}}{2^{2k}(k!)^{2}(2k+1)}}=\operatorname {arcsinh} {z},|z|\leq 1}$
• ${\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k+1}}{2k+1}}=\arctan z,|z|<1}$
• ${\displaystyle \sum _{k=0}^{\infty }{\frac {z^{2k+1}}{2k+1}}=\operatorname {arctanh} z,|z|<1}$
• ${\displaystyle \ln 2+\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}(2k)!z^{2k}}{2^{2k+1}k(k!)^{2}}}=\ln \left(1+{\sqrt {1+z^{2}}}\right),|z|\leq 1}$
• ${\displaystyle \sum _{k=2}^{\infty }\left(k\cdot \operatorname {arctanh} \left({\frac {1}{k}}\right)-1\right)={\frac {3-\ln(4\pi )}{2}}}$

Modified-factorial denominators

• ${\displaystyle \sum _{k=0}^{\infty }{\frac {(4k)!}{2^{4k}{\sqrt {2}}(2k)!(2k+1)!}}z^{k}={\sqrt {\frac {1-{\sqrt {1-z}}}{z}}},|z|<1}$^[2]
• ${\displaystyle \sum _{k=0}^{\infty }{\frac {2^{2k}(k!)^{2}}{(k+1)(2k+1)!}}z^{2k+2}=\left(\arcsin {z}\right)^{2},|z|\leq 1}$^[2]
• ${\displaystyle \sum _{n=0}^{\infty }{\frac {\prod _{k=0}^{n-1}(4k^{2}+\alpha ^{2})}{(2n)!}}z^{2n}+\sum _{n=0}^{\infty }{\frac {\alpha \prod _{k=0}^{n-1}[(2k+1)^{2}+\alpha ^{2}]}{(2n+1)!}}z^{2n+1}=e^{\alpha \arcsin {z}},|z|\leq 1}$

Binomial coefficients

• ${\displaystyle (1+z)^{\alpha }=\sum _{k=0}^{\infty }{\alpha \choose k}z^{k},|z|<1}$ (see Binomial theorem § Newton's generalized binomial theorem)
• ^[3] ${\displaystyle \sum _{k=0}^{\infty }{{\alpha +k-1} \choose k}z^{k}={\frac {1}{(1-z)^{\alpha }}},|z|<1}$
• ^[3] ${\displaystyle \sum _{k=0}^{\infty }{\frac {1}{k+1}}{2k \choose k}z^{k}={\frac {1-{\sqrt {1-4z}}}{2z}},|z|\leq {\frac {1}{4}}}$, generating function of the Catalan numbers
• ^[3] ${\displaystyle \sum _{k=0}^{\infty }{2k \choose k}z^{k}={\frac {1}{\sqrt {1-4z}}},|z|<{\frac {1}{4}}}$, generating function of the Central binomial coefficients
• ^[3] ${\displaystyle \sum _{k=0}^{\infty }{2k+\alpha \choose k}z^{k}={\frac {1}{\sqrt {1-4z}}}\left({\frac {1-{\sqrt {1-4z}}}{2z}}\right)^{\alpha },|z|<{\frac {1}{4}}}$

Harmonic numbers

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

• ${\displaystyle \sum _{k=1}^{\infty }H_{k}z^{k}={\frac {-\ln(1-z)}{1-z}},|z|<1}$
• ${\displaystyle \sum _{k=1}^{\infty }{\frac {H_{k}}{k+1}}z^{k+1}={\frac {1}{2}}\left[\ln(1-z)\right]^{2},\qquad |z|<1}$
• ${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}H_{2k}}{2k+1}}z^{2k+1}={\frac {1}{2}}\arctan {z}\log {(1+z^{2})},\qquad |z|<1}$^[2]
• ${\displaystyle \sum _{n=0}^{\infty }\sum _{k=0}^{2n}{\frac {(-1)^{k}}{2k+1}}{\frac {z^{4n+2}}{4n+2}}={\frac {1}{4}}\arctan {z}\log {\frac {1+z}{1-z}},\qquad |z|<1}$^[2]
• ${\displaystyle \sum _{n=0}^{\infty }{\frac {x^{2}}{n^{2}(n+x)}}=x{\frac {\pi ^{2}}{6}}-H(x)}$

Binomial coefficients

Main article: Binomial coefficient

• ${\displaystyle \sum _{k=0}^{n}{n \choose k}=2^{n}}$
• ${\displaystyle \sum _{k=0}^{n}{n \choose k}^{2}={2n \choose n}}$
• ${\displaystyle \sum _{k=0}^{n}(-1)^{k}{n \choose k}=0,{\text{ where }}n\geq 1}$
• ${\displaystyle \sum _{k=0}^{n}{k \choose m}={n+1 \choose m+1}}$
• ${\displaystyle \sum _{k=0}^{n}{m+k-1 \choose k}={n+m \choose n}}$ (see Multiset)
• ${\displaystyle \sum _{k=0}^{n}{\alpha \choose k}{\beta \choose n-k}={\alpha +\beta \choose n},{\text{where}}\ \alpha +\beta \geq n}$ (see Vandermonde identity)
• ${\displaystyle \sum _{A\ \in \ {\mathcal {P}}(E)}1=2^{n}{\text{, where }}E{\text{ is a finite set, and card(}}E{\text{) = n}}}$
• ${\displaystyle \sum _{\begin{cases}(A,\ B)\ \in \ ({\mathcal {P}}(E))^{2}\\A\ \subset \ B\end{cases}}1=3^{n}{\text{, where }}E{\text{ is a finite set, and card(}}E{\text{) = n}}}$
• ${\displaystyle \sum _{A\ \in \ {\mathcal {P}}(E)}card(A)=n2^{n-1}{\text{, where }}E{\text{ is a finite set, and card(}}E{\text{) = n}}}$

Trigonometric functions

Sums of sines and cosines arise in Fourier series.

• ${\displaystyle \sum _{k=1}^{\infty }{\frac {\cos(k\theta )}{k}}=-{\frac {1}{2}}\ln(2-2\cos \theta )=-\ln \left(2\sin {\frac {\theta }{2}}\right),0<\theta <2\pi }$
• ${\displaystyle \sum _{k=1}^{\infty }{\frac {\sin(k\theta )}{k}}={\frac {\pi -\theta }{2}},0<\theta <2\pi }$
• ${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{k}}\cos(k\theta )={\frac {1}{2}}\ln(2+2\cos \theta )=\ln \left(2\cos {\frac {\theta }{2}}\right),0\leq \theta <\pi }$
• ${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{k}}\sin(k\theta )={\frac {\theta }{2}},-{\frac {\pi }{2}}\leq \theta \leq {\frac {\pi }{2}}}$
• ${\displaystyle \sum _{k=1}^{\infty }{\frac {\cos(2k\theta )}{2k}}=-{\frac {1}{2}}\ln(2\sin \theta ),0<\theta <\pi }$
• ${\displaystyle \sum _{k=1}^{\infty }{\frac {\sin(2k\theta )}{2k}}={\frac {\pi -2\theta }{4}},0<\theta <\pi }$
• ${\displaystyle \sum _{k=0}^{\infty }{\frac {\cos[(2k+1)\theta ]}{2k+1}}={\frac {1}{2}}\ln \left(\cot {\frac {\theta }{2}}\right),0<\theta <\pi }$
• ${\displaystyle \sum _{k=0}^{\infty }{\frac {\sin[(2k+1)\theta ]}{2k+1}}={\frac {\pi }{4}},0<\theta <\pi }$,^[4]
• ${\displaystyle \sum _{k=1}^{\infty }{\frac {\sin(2\pi kx)}{k}}=\pi \left({\dfrac {1}{2}}-\{x\}\right),\ x\in \mathbb {R} }$
• ${\displaystyle \sum \limits _{k=1}^{\infty }{\frac {\sin \left(2\pi kx\right)}{k^{2n-1}}}=(-1)^{n}{\frac {(2\pi )^{2n-1}}{2(2n-1)!}}B_{2n-1}(\{x\}),\ x\in \mathbb {R} ,\ n\in \mathbb {N} }$
• ${\displaystyle \sum \limits _{k=1}^{\infty }{\frac {\cos \left(2\pi kx\right)}{k^{2n}}}=(-1)^{n-1}{\frac {(2\pi )^{2n}}{2(2n)!}}B_{2n}(\{x\}),\ x\in \mathbb {R} ,\ n\in \mathbb {N} }$
• ${\displaystyle B_{n}(x)=-{\frac {n!}{2^{n-1}\pi ^{n}}}\sum _{k=1}^{\infty }{\frac {1}{k^{n}}}\cos \left(2\pi kx-{\frac {\pi n}{2}}\right),0<x<1}$^[5]
• ${\displaystyle \sum _{k=0}^{n}\sin(\theta +k\alpha )={\frac {\sin {\frac {(n+1)\alpha }{2}}\sin(\theta +{\frac {n\alpha }{2}})}{\sin {\frac {\alpha }{2}}}}}$
• ${\displaystyle \sum _{k=0}^{n}\cos(\theta +k\alpha )={\frac {\sin {\frac {(n+1)\alpha }{2}}\cos(\theta +{\frac {n\alpha }{2}})}{\sin {\frac {\alpha }{2}}}}}$
• ${\displaystyle \sum _{k=1}^{n-1}\sin {\frac {\pi k}{n}}=\cot {\frac {\pi }{2n}}}$
• ${\displaystyle \sum _{k=1}^{n-1}\sin {\frac {2\pi k}{n}}=0}$
• ${\displaystyle \sum _{k=0}^{n-1}\csc ^{2}\left(\theta +{\frac {\pi k}{n}}\right)=n^{2}\csc ^{2}(n\theta )}$^[6]
• ${\displaystyle \sum _{k=1}^{n-1}\csc ^{2}{\frac {\pi k}{n}}={\frac {n^{2}-1}{3}}}$
• ${\displaystyle \sum _{k=1}^{n-1}\csc ^{4}{\frac {\pi k}{n}}={\frac {n^{4}+10n^{2}-11}{45}}}$

Roots of unity

A ${\displaystyle n}$'th root of unity is a solution to the equation ${\displaystyle z^{n}=0}$ and they can be written like:

${\displaystyle z_{m}=\exp \left(i\cdot {\frac {2\pi m}{n}}\right),\quad m\in \{0,\cdots ,n-1\}}$

The following summation identities hold:

${\displaystyle \sum _{m\neq 0}^{n-1}{\frac {1}{1-z_{m}}}={\frac {n-1}{2}}}$

${\displaystyle \sum _{m\neq 0}^{n-1}{\frac {1}{(1-z_{m})^{2}}}={\frac {(n-1)(5-n)}{12}}}$

Let ${\displaystyle k}$ be an integer ${\displaystyle 0<k<n}$ then we also got:

${\displaystyle \sum _{m=0}^{n-1}z_{m}^{k}=0}$

${\displaystyle \sum _{m=0}^{n-1}(x-z_{m})^{k}=n\cdot x^{k}}$

${\displaystyle \sum _{m\neq 0}^{n-1}{\frac {(z_{m})^{k}}{1-z_{m}}}={\frac {2k-n-1}{2}}}$

${\displaystyle \sum _{m\neq 0}^{n-1}{\frac {(z_{m})^{k}}{(1-z_{m})^{2}}}={\frac {6k(n+2-k)-(n+1)(n+5)}{12}}}$

Rational functions

• ${\displaystyle \sum _{n=a+1}^{\infty }{\frac {a}{n^{2}-a^{2}}}={\frac {1}{2}}H_{2a}}$^[7]
• ${\displaystyle \sum _{n=0}^{\infty }{\frac {1}{n^{2}+a^{2}}}={\frac {1+a\pi \coth(a\pi )}{2a^{2}}}}$
• ${\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n^{2}+a^{2}}}={\frac {1+a\pi \;{\text{csch}}(a\pi )}{2a^{2}}}}$
• ${\displaystyle \sum _{n=0}^{\infty }{\frac {(2n+1)(-1)^{n}}{(2n+1)^{2}+a^{2}}}={\frac {\pi }{4}}{\text{sech}}\left({\frac {a\pi }{2}}\right)}$
• ${\displaystyle \displaystyle \sum _{n=0}^{\infty }{\frac {1}{n^{4}+4a^{4}}}={\dfrac {1}{8a^{4}}}+{\dfrac {\pi (\sinh(2\pi a)+\sin(2\pi a))}{8a^{3}(\cosh(2\pi a)-\cos(2\pi a))}}}$
• An infinite series of any rational function of ${\displaystyle n}$ 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

• ${\displaystyle \displaystyle {\dfrac {1}{\sqrt {p}}}\sum _{n=0}^{p-1}\exp \left({\frac {2\pi in^{2}q}{p}}\right)={\dfrac {e^{\pi i/4}}{\sqrt {2q}}}\sum _{n=0}^{2q-1}\exp \left(-{\frac {\pi in^{2}p}{2q}}\right)}$(see the Landsberg–Schaar relation)
• ${\displaystyle \displaystyle \sum _{n=-\infty }^{\infty }e^{-\pi n^{2}}={\frac {\sqrt[{4}]{\pi }}{\Gamma \left({\frac {3}{4}}\right)}}}$

Numeric series

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

Alternating harmonic series

• ${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k}}={\frac {1}{1}}-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+\cdots =\ln 2}$
• ${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{2k-1}}={\frac {1}{1}}-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}-\cdots ={\frac {\pi }{4}}}$

Alternating arithmetic series

Let ${\displaystyle S(a,b)}$ be defined as:

${\displaystyle S(a,b)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{ak+b}}}$

where ${\displaystyle a,b>0}$ are positive whole numbers. Then if ${\displaystyle gcd(a,b)=c}$ we can write ${\displaystyle a=c\alpha }$ and ${\displaystyle b=c\beta }$, where ${\displaystyle gcd(\alpha ,\beta )=1}$, and get:

${\displaystyle S(a,b)=S(c\alpha ,c\beta )=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{cdk+ce}}={\frac {1}{c}}\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{\alpha k+\beta }}={\frac {S(\alpha ,\beta )}{c}}}$

Now if ${\displaystyle b>a}$ we can, per Euclid's division lemma, write ${\displaystyle b=ca+d}$ where ${\displaystyle a>d>0}$ and then

${\displaystyle S(a,b)=S(a,ca+d)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{ak+ca+d}}=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{a(k+c)+d}}=\sum _{k=c}^{\infty }{\frac {(-1)^{k-c}}{ak+d}}}$

where we now can add the remaining rows back and subtract them to give us:

${\displaystyle S(a,b)=(-1)^{c}\left(\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{ak+d}}-\sum _{k=0}^{c-1}{\frac {(-1)^{k}}{ak+d}}\right)=(-1)^{c}\left(S(a,d)-\sum _{k=0}^{c-1}{\frac {(-1)^{k}}{ak+d}}\right)}$

what that means is that all the infinite choices of ${\displaystyle a}$ and ${\displaystyle b}$ can essentially be boiled down to the cases where ${\displaystyle gcd(a,b)=1}$ and ${\displaystyle a>b>0}$. If we assume those two things we can then write:

${\displaystyle S(a,b)={\frac {1}{a}}\left({\frac {\pi }{2\sin \left({\frac {\pi b}{a}}\right)}}-2\sum _{m=0}^{\lfloor {\frac {a}{2}}\rfloor }\cos \left(\pi {\frac {(2m+1)b}{a}}\right)\ln \left(\sin \left(\pi {\frac {2m+1}{2a}}\right)\right)\right)}$

and in the case of using a negative sign instead:

${\displaystyle S_{-}(a,b)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{ak-b}}}$

the same two rules apply from above apply and then we can do the following for the case with ${\displaystyle a>b>0}$ (since ${\displaystyle a>a-b>0}$):

${\displaystyle S_{-}(a,b)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{ak-b}}=-{\frac {1}{b}}+\sum _{k=0}^{\infty }{\frac {(-1)^{k+1}}{a(k+1)-b}}=-{\frac {1}{b}}-\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{ak+(a-b)}}=-S(a,a-b)-{\frac {1}{b}}}$

Let us test out the formula: ${\displaystyle S(3,2)={\frac {1}{3}}\left({\frac {\pi }{2\sin \left({\frac {2\pi }{3}}\right)}}-2\left(\cos \left({\frac {2\pi }{3}}\right)\ln \left(\sin \left({\frac {\pi }{6}}\right)\right)+\cos \left(2\pi \right)\ln \left(\sin \left({\frac {\pi }{2}}\right)\right)\right)\right)={\frac {\pi }{3{\sqrt {3}}}}-{\frac {\ln(2)}{3}}}$

Sum of reciprocal of factorials

• ${\displaystyle \sum _{k=0}^{\infty }{\frac {1}{k!}}={\frac {1}{0!}}+{\frac {1}{1!}}+{\frac {1}{2!}}+{\frac {1}{3!}}+{\frac {1}{4!}}+\cdots =e}$
• ${\displaystyle \sum _{k=0}^{\infty }{\frac {1}{(2k)!}}={\frac {1}{0!}}+{\frac {1}{2!}}+{\frac {1}{4!}}+{\frac {1}{6!}}+{\frac {1}{8!}}+\cdots ={\frac {1}{2}}\left(e+{\frac {1}{e}}\right)=\cosh 1}$
• ${\displaystyle \sum _{k=0}^{\infty }{\frac {1}{(3k)!}}={\frac {1}{0!}}+{\frac {1}{3!}}+{\frac {1}{6!}}+{\frac {1}{9!}}+{\frac {1}{12!}}+\cdots ={\frac {1}{3}}\left(e+{\frac {2}{\sqrt {e}}}\cos {\frac {\sqrt {3}}{2}}\right)}$
• ${\displaystyle \sum _{k=0}^{\infty }{\frac {1}{(4k)!}}={\frac {1}{0!}}+{\frac {1}{4!}}+{\frac {1}{8!}}+{\frac {1}{12!}}+{\frac {1}{16!}}+\cdots ={\frac {1}{2}}\left(\cos 1+\cosh 1\right)}$

Trigonometry and π

• ${\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k+1)!}}={\frac {1}{1!}}-{\frac {1}{3!}}+{\frac {1}{5!}}-{\frac {1}{7!}}+{\frac {1}{9!}}+\cdots =\sin 1}$
• ${\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k)!}}={\frac {1}{0!}}-{\frac {1}{2!}}+{\frac {1}{4!}}-{\frac {1}{6!}}+{\frac {1}{8!}}+\cdots =\cos 1}$
• ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k^{2}+1}}={\frac {1}{2}}+{\frac {1}{5}}+{\frac {1}{10}}+{\frac {1}{17}}+\cdots ={\frac {1}{2}}(\pi \coth \pi -1)}$
• ${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k}}{k^{2}+1}}=-{\frac {1}{2}}+{\frac {1}{5}}-{\frac {1}{10}}+{\frac {1}{17}}+\cdots ={\frac {1}{2}}(\pi \operatorname {csch} \pi -1)}$
• ${\displaystyle 3+{\frac {4}{2\times 3\times 4}}-{\frac {4}{4\times 5\times 6}}+{\frac {4}{6\times 7\times 8}}-{\frac {4}{8\times 9\times 10}}+\cdots =\pi }$

Reciprocal of tetrahedral numbers

• ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{Te_{k}}}={\frac {1}{1}}+{\frac {1}{4}}+{\frac {1}{10}}+{\frac {1}{20}}+{\frac {1}{35}}+\cdots ={\frac {3}{2}}}$

Where ${\displaystyle Te_{n}=\sum _{k=1}^{n}T_{k}}$

Exponential and logarithms

• ${\displaystyle \sum _{k=0}^{\infty }{\frac {1}{(2k+1)(2k+2)}}={\frac {1}{1\times 2}}+{\frac {1}{3\times 4}}+{\frac {1}{5\times 6}}+{\frac {1}{7\times 8}}+{\frac {1}{9\times 10}}+\cdots =\ln 2}$
• ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{2^{k}k}}={\frac {1}{2}}+{\frac {1}{8}}+{\frac {1}{24}}+{\frac {1}{64}}+{\frac {1}{160}}+\cdots =\ln 2}$
• ${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{2^{k}k}}+\sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{3^{k}k}}={\Bigg (}{\frac {1}{2}}+{\frac {1}{3}}{\Bigg )}-{\Bigg (}{\frac {1}{8}}+{\frac {1}{18}}{\Bigg )}+{\Bigg (}{\frac {1}{24}}+{\frac {1}{81}}{\Bigg )}-{\Bigg (}{\frac {1}{64}}+{\frac {1}{324}}{\Bigg )}+\cdots =\ln 2}$
• ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{3^{k}k}}+\sum _{k=1}^{\infty }{\frac {1}{4^{k}k}}={\Bigg (}{\frac {1}{3}}+{\frac {1}{4}}{\Bigg )}+{\Bigg (}{\frac {1}{18}}+{\frac {1}{32}}{\Bigg )}+{\Bigg (}{\frac {1}{81}}+{\frac {1}{192}}{\Bigg )}+{\Bigg (}{\frac {1}{324}}+{\frac {1}{1024}}{\Bigg )}+\cdots =\ln 2}$
• ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{n^{k}k}}=\ln \left({\frac {n}{n-1}}\right)}$, that is ${\displaystyle \forall n>1}$

See also

• Series (mathematics)
• List of integrals
• Summation § Identities
• Taylor series
• Binomial theorem
• Gregory's series
• On-Line Encyclopedia of Integer Sequences