- Source: List of mathematical series
- Simbol takhingga
- Teorema Pythagoras
- 0,999...
- Branko Grünbaum
- Daftar singkatan matematis
- Kesultanan Utsmaniyah
- Daftar masalah matematika yang belum terpecahkan
- Hukum gerak Newton
- Rasio keuangan
- Srinivasa Ramanujan
- List of mathematical series
- Lists of mathematics topics
- Series (mathematics)
- List of mathematical proofs
- Glossary of mathematical symbols
- List of mathematical functions
- Mathematics
- List of mathematical constants
- List of mathematics reference tables
- Harmonic series (mathematics)
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}
(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}
∑
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}
∑
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)
∑
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}
∑
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
∑
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
∑
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}
∑
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}
∑
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 }
,
∑
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
∑
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 )}
∑
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}}
∑
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, 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}
∑
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)}
, that is
∀
n
>
1
{\displaystyle \forall n>1}
See also
Notes
References
Many books with a list of integrals also have a list of series.