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, $$0^0$$ is taken to have the value $$1$$
 * $$\{x\}$$ denotes the fractional part of $$x$$
 * $$B_n(x)$$ is a Bernoulli polynomial.
 * $$B_n$$ is a Bernoulli number, and here, $$B_1=-\frac{1}{2}.$$
 * $$E_n$$ is an Euler number.
 * $$\zeta(s) $$ is the Riemann zeta function.
 * $$\Gamma(z)$$ is the gamma function.
 * $$\psi_n(z)$$ is a polygamma function.
 * $$\operatorname{Li}_s(z)$$ is a polylogarithm.
 * $$ n \choose k $$ is binomial coefficient
 * $$\exp(x)$$ denotes exponential of $$x$$

Sums of powers
See Faulhaber's formula. The first few values are:
 * $$\sum_{k=0}^m k^{n-1}=\frac{B_n(m+1)-B_n}{n}$$
 * $$\sum_{k=1}^m k=\frac{m(m+1)}{2}$$
 * $$\sum_{k=1}^m k^2=\frac{m(m+1)(2m+1)}{6}=\frac{m^3}{3}+\frac{m^2}{2}+\frac{m}{6}$$
 * $$\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. The first few values are:
 * $$\zeta(2n)=\sum^{\infty}_{k=1} \frac{1}{k^{2n}}=(-1)^{n+1} \frac{B_{2n} (2\pi)^{2n}}{2(2n)!} $$
 * $$\zeta(2)=\sum^{\infty}_{k=1} \frac{1}{k^2}=\frac{\pi^2}{6}$$ (the Basel problem)
 * $$\zeta(4)=\sum^{\infty}_{k=1} \frac{1}{k^4}=\frac{\pi^4}{90}$$
 * $$\zeta(6)=\sum^{\infty}_{k=1} \frac{1}{k^6}=\frac{\pi^6}{945}$$

Low-order polylogarithms
Finite sums:
 * $$\sum_{k=m}^{n} z^k = \frac{z^{m}-z^{n+1}}{1-z}$$, (geometric series)
 * $$\sum_{k=0}^{n} z^k = \frac{1-z^{n+1}}{1-z}$$
 * $$\sum_{k=1}^{n} z^k = \frac{1-z^{n+1}}{1-z}-1 = \frac{z-z^{n+1}}{1-z}$$
 * $$\sum_{k=1}^n k z^k = z\frac{1-(n+1)z^n+nz^{n+1}}{(1-z)^2}$$
 * $$\sum_{k=1}^n k^2 z^k = z\frac{1+z-(n+1)^2z^n+(2n^2+2n-1)z^{n+1}-n^2z^{n+2}}{(1-z)^3} $$
 * $$\sum_{k=1}^n k^m z^k = \left(z \frac{d}{dz}\right)^m \frac{1-z^{n+1}}{1-z}$$

Infinite sums, valid for $$|z|<1$$ (see polylogarithm): The following is a useful property to calculate low-integer-order polylogarithms recursively in closed form:
 * $$\operatorname{Li}_n(z)=\sum_{k=1}^{\infty} \frac{z^k}{k^n}$$
 * $$\frac{\mathrm{d}}{\mathrm{d}z}\operatorname{Li}_n(z)=\frac{\operatorname{Li}_{n-1}(z)}{z}$$
 * $$\operatorname{Li}_{1}(z)=\sum_{k=1}^\infty \frac{z^k}{k}=-\ln(1-z)$$
 * $$\operatorname{Li}_{0}(z)=\sum_{k=1}^\infty z^k=\frac{z}{1-z}$$
 * $$\operatorname{Li}_{-1}(z)=\sum_{k=1}^\infty k z^k=\frac{z}{(1-z)^2}$$
 * $$\operatorname{Li}_{-2}(z)=\sum_{k=1}^\infty k^2 z^k=\frac{z(1+z)}{(1-z)^3}$$
 * $$\operatorname{Li}_{-3}(z)=\sum_{k=1}^\infty k^3 z^k =\frac{z(1+4z+z^2)}{(1-z)^4}$$
 * $$\operatorname{Li}_{-4}(z)=\sum_{k=1}^\infty k^4 z^k =\frac{z(1+z)(1+10z+z^2)}{(1-z)^5}$$

Exponential function

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

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

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

 * $$\sum_{k=0}^\infty \frac{(-1)^k z^{2k+1}}{(2k+1)!}=\sin z$$
 * $$\sum_{k=0}^\infty \frac{z^{2k+1}}{(2k+1)!}=\sinh z$$
 * $$\sum_{k=0}^\infty \frac{(-1)^k z^{2k}}{(2k)!}=\cos z$$
 * $$\sum_{k=0}^\infty \frac{z^{2k}}{(2k)!}=\cosh z$$
 * $$\sum_{k=1}^\infty \frac{(-1)^{k-1}(2^{2k}-1)2^{2k}B_{2k}z^{2k-1}}{(2k)!}=\tan z, |z|<\frac{\pi}{2}$$
 * $$\sum_{k=1}^\infty \frac{(2^{2k}-1)2^{2k}B_{2k}z^{2k-1}}{(2k)!}=\tanh z, |z|<\frac{\pi}{2}$$
 * $$\sum_{k=0}^\infty \frac{(-1)^k2^{2k}B_{2k}z^{2k-1}}{(2k)!}=\cot z, |z|<\pi$$
 * $$\sum_{k=0}^\infty \frac{2^{2k}B_{2k}z^{2k-1}}{(2k)!}=\coth z, |z|<\pi$$
 * $$\sum_{k=0}^\infty \frac{(-1)^{k-1}(2^{2k}-2)B_{2k}z^{2k-1}}{(2k)!}=\csc z, |z|<\pi$$
 * $$\sum_{k=0}^\infty \frac{-(2^{2k}-2)B_{2k}z^{2k-1}}{(2k)!}=\operatorname{csch} z, |z|<\pi$$
 * $$\sum_{k=0}^\infty \frac{(-1)^kE_{2k}z^{2k}}{(2k)!}=\operatorname{sech} z, |z|<\frac{\pi}{2}$$
 * $$\sum_{k=0}^\infty \frac{E_{2k}z^{2k}}{(2k)!}=\sec z, |z| < \frac{\pi}{2}$$
 * $$\sum_{k=1}^\infty \frac{(-1)^{k-1} z^{2k}}{(2k)!}=\operatorname{ver}z$$ (versine)
 * $$\sum_{k=1}^\infty \frac{(-1)^{k-1} z^{2k}}{2(2k)!}=\operatorname{hav}z$$ (haversine)
 * $$\sum_{k=0}^\infty \frac{(2k)!z^{2k+1}}{2^{2k}(k!)^2(2k+1)}=\arcsin z, |z|\le1$$
 * $$\sum_{k=0}^\infty \frac{(-1)^k(2k)!z^{2k+1}}{2^{2k}(k!)^2(2k+1)}=\operatorname{arcsinh} {z}, |z| \le 1$$
 * $$\sum_{k=0}^\infty \frac{(-1)^kz^{2k+1}}{2k+1}=\arctan z, |z|<1$$
 * $$\sum_{k=0}^\infty \frac{z^{2k+1}}{2k+1}=\operatorname{arctanh} z, |z|<1$$
 * $$\ln2+\sum_{k=1}^\infty \frac{(-1)^{k-1}(2k)!z^{2k}}{2^{2k+1}k(k!)^2}=\ln\left(1+\sqrt{1+z^2}\right), |z| \le 1$$
 * $$\sum_{k=2}^\infty \left( k \cdot \operatorname{arctanh}\left(\frac{1}{k}\right) - 1 \right) = \frac{3-\ln(4 \pi)}{2} $$

Modified-factorial denominators

 * $$\sum^{\infty}_{k=0} \frac{(4k)!}{2^{4k} \sqrt{2} (2k)! (2k+1)!} z^k = \sqrt{\frac{1-\sqrt{1-z}}{z}}, |z|<1$$
 * $$\sum^{\infty}_{k=0} \frac{2^{2k} (k!)^2}{(k+1) (2k+1)!} z^{2k+2} = \left(\arcsin{z}\right)^2, |z|\le1 $$
 * $$\sum^{\infty}_{n=0} \frac{\prod_{k=0}^{n-1}(4k^2+\alpha^2)}{(2n)!} z^{2n} + \sum^{\infty}_{n=0} \frac{\alpha \prod_{k=0}^{n-1}[(2k+1)^2+\alpha^2]}{(2n+1)!} z^{2n+1} = e^{\alpha \arcsin{z}}, |z|\le1 $$

Binomial coefficients

 * $$(1+z)^\alpha = \sum_{k=0}^\infty {\alpha \choose k} z^k, |z|<1$$ (see )
 * $$\sum_{k=0}^\infty {{\alpha+k-1} \choose k} z^k = \frac{1}{(1-z)^\alpha}, |z|<1$$
 * $$\sum_{k=0}^\infty \frac{1}{k+1}{2k \choose k} z^k = \frac{1-\sqrt{1-4z}}{2z}, |z|\leq\frac{1}{4}$$, generating function of the Catalan numbers
 * $$\sum_{k=0}^\infty {2k \choose k} z^k = \frac{1}{\sqrt{1-4z}}, |z|<\frac{1}{4}$$, generating function of the Central binomial coefficients
 * $$\sum_{k=0}^\infty {2k + \alpha \choose k} z^k = \frac{1}{\sqrt{1-4z}}\left(\frac{1-\sqrt{1-4z}}{2z}\right)^\alpha, |z|<\frac{1}{4}$$

Harmonic numbers
(See harmonic numbers, themselves defined $H_n = \sum_{j=1}^{n} \frac{1}{j} $, and $$H(x)$$ generalized to the real numbers)


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

Binomial coefficients

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

Trigonometric functions
Sums of sines and cosines arise in Fourier series.


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

Rational functions

 * $$\sum_{n=a+1}^{\infty} \frac{a}{n^2 - a^2} = \frac{1}{2} H_{2a}$$
 * $$\sum_{n=0}^\infty\frac{1}{n^2+a^2}=\frac{1+a\pi\coth (a\pi)}{2a^2}$$
 * $$\sum_{n=0}^\infty\frac{(-1)^n}{n^2+a^2} = \frac{1 + a\pi \; \text{csch}(a\pi)}{2a^2}$$
 * $$\sum_{n=0}^\infty\frac{(2n+1)(-1)^n}{(2n+1)^2+a^2}= \frac{\pi}{4} \text{sech} \left( \frac{a \pi}{2} \right) $$
 * $$\displaystyle \sum_{n=0}^\infty \frac {1}{n^4+4a^4} = \dfrac{1}{8a^4}+\dfrac{\pi(\sinh(2\pi a)+\sin(2\pi a))}{8a^3(\cosh(2\pi a)-\cos(2\pi a))}$$
 * An infinite series of any rational function of $$n$$ can be reduced to a finite series of polygamma functions, by use of partial fraction decomposition, 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

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

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

Alternating harmonic series

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

Sum of reciprocal of factorials

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

Trigonometry and π

 * $$\sum^{\infty}_{k=0} \frac{(-1)^k}{(2k+1)!}=\frac{1}{1!}-\frac{1}{3!}+\frac{1}{5!}-\frac{1}{7!}+\frac{1}{9!}+\cdots=\sin 1$$
 * $$\sum^{\infty}_{k=0} \frac{(-1)^k}{(2k)!}=\frac{1}{0!}-\frac{1}{2!}+\frac{1}{4!}-\frac{1}{6!}+\frac{1}{8!}+\cdots=\cos 1$$
 * $$\sum^{\infty}_{k=1} \frac{1}{k^2+1}=\frac{1}{2}+\frac{1}{5}+\frac{1}{10}+\frac{1}{17}+\cdots=\frac{1}{2}(\pi \coth \pi - 1)$$
 * $$\sum^{\infty}_{k=1} \frac{(-1)^k}{k^2+1}=-\frac{1}{2}+\frac{1}{5}-\frac{1}{10}+\frac{1}{17}+\cdots=\frac{1}{2}(\pi \operatorname{csch} \pi - 1)$$
 * $$ 3 + \frac{4}{2\times3\times4} - \frac{4}{4\times5\times6} + \frac{4}{6\times7\times8} - \frac{4}{8\times9\times10} + \cdots = \pi$$

Reciprocal of tetrahedral numbers
Where $$Te_n=\sum^{n}_{k=1} T_k$$
 * $$\sum^{\infty}_{k=1} \frac{1}{Te_k}=\frac{1}{1}+\frac{1}{4}+\frac{1}{10}+\frac{1}{20}+\frac{1}{35}+\cdots=\frac{3}{2}$$

Exponential and logarithms

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