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.

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

The first few values are:

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

See zeta constants.

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

The first few values are:

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

Power series

Low-order polylogarithms

Finite sums:

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

Infinite sums, valid for $|z|<1$ (see polylogarithm):

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

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

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

Exponential function

$\sum _{k=0}^{\infty }{\frac {z^{k}}{k!}}=e^{z}$
$\sum _{k=0}^{\infty }k{\frac {z^{k}}{k!}}=ze^{z}$ (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

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

Modified-factorial denominators

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

Binomial coefficients

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

Harmonic numbers

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

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

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.

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

Roots of unity

A $n$ 'th root of unity is a solution to the equation $z^{n}=1$ and they can be written like:

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

The following summation identities hold:

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

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

Let $k$ be an integer $0<k<n$ then we also got:

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

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

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

\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

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

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

Numeric series

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

Alternating harmonic series

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

Alternating arithmetic series

Let $S(a,b)$ be defined as:

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

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

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

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

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

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

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

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

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

and in the case of using a negative sign instead:

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

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

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

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

Sum of reciprocal of factorials

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

Trigonometry and π

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

Reciprocal of tetrahedral numbers

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

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

Exponential and logarithms

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