Esta lista de séries matemáticas contém fórmulas para somas finitas e infinitas. Ela pode ser usada em conjunto com outras ferramentas para avaliar somas.
Polilogaritmos de ordem baixa
editar
Somas com uma quantidade finita de termos:
∑
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}}}
série geométrica )
∑
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}}}
Somas com uma infinidade de termos, válidas para
|
z
|
<
1
{\displaystyle |z|<1}
polilogaritmo ):
Li
n
(
z
)
=
∑
k
=
1
∞
z
k
k
n
{\displaystyle \operatorname {Li} _{n}(z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k^{n}}}}
A propriedade a seguir é útil para calcular polilogaritmos de ordem inteira baixa recursivamente de forma fechada :
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}}}}
∑
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}}
distribuição de Poisson )
∑
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}}
segundo momento da distribuição de Poisson)
∑
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)}
em que
T
n
(
z
)
{\displaystyle T_{n}(z)}
polinômios de Touchard .
editar
∑
k
=
0
∞
(
−
1
)
k
z
2
k
+
1
(
2
k
+
1
)
!
=
s
e
n
z
{\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k+1}}{(2k+1)!}}=\mathrm {sen} \,z}
∑
k
=
0
∞
z
2
k
+
1
(
2
k
+
1
)
!
=
s
e
n
h
z
{\displaystyle \sum _{k=0}^{\infty }{\frac {z^{2k+1}}{(2k+1)!}}=\mathrm {senh} \,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}
seno verso )
∑
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
)
=
a
r
c
s
e
n
z
,
|
z
|
≤
1
{\displaystyle \sum _{k=0}^{\infty }{\frac {(2k)!z^{2k+1}}{2^{2k}(k!)^{2}(2k+1)}}=\mathrm {arcsen} \,z,|z|\leq 1}
∑
k
=
0
∞
(
−
1
)
k
(
2
k
)
!
z
2
k
+
1
2
2
k
(
k
!
)
2
(
2
k
+
1
)
=
arcsenh
z
,
|
z
|
≤
1
{\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}(2k)!z^{2k+1}}{2^{2k}(k!)^{2}(2k+1)}}=\operatorname {arcsenh} {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}
Denominadores fatoriais modificados
editar
∑
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
=
(
a
r
c
s
e
n
z
)
2
,
|
z
|
≤
1
{\displaystyle \sum _{k=0}^{\infty }{\frac {2^{2k}(k!)^{2}}{(k+1)(2k+1)!}}z^{2k+2}=\left(\mathrm {arcsen} \,{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
a
r
c
s
e
n
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 \mathrm {arcsen} \,{z}},|z|\leq 1}
(
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}
teorema binomial )[ 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}
∑
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}}}
função geradora dos dos números de Catalan
∑
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}}}
coeficientes binomiais centrais
∑
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}}}
(Ver números harmônicos , que são definidos por
H
n
=
∑
j
=
1
n
1
j
{\textstyle H_{n}=\sum _{j=1}^{n}{\frac {1}{j}}}
∑
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}
∑
k
=
0
n
(
n
k
)
=
2
n
{\displaystyle \sum _{k=0}^{n}{n \choose k}=2^{n}}
∑
k
=
0
n
(
−
1
)
k
(
n
k
)
=
0
,
where
n
>
0
{\displaystyle \sum _{k=0}^{n}(-1)^{k}{n \choose k}=0,{\text{ where }}n>0}
∑
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}}
multiconjunto )
∑
k
=
0
n
(
α
k
)
(
β
n
−
k
)
=
(
α
+
β
n
)
{\displaystyle \sum _{k=0}^{n}{\alpha \choose k}{\beta \choose n-k}={\alpha +\beta \choose n}}
identidade de Vandermonde )
editar
Soma de senos e cossenos surgem nas séries de Fourier .
∑
k
=
1
∞
s
e
n
(
k
θ
)
k
=
π
−
θ
2
,
0
<
θ
<
2
π
{\displaystyle \sum _{k=1}^{\infty }{\frac {\mathrm {sen} \,(k\theta )}{k}}={\frac {\pi -\theta }{2}},0<\theta <2\pi }
∑
k
=
1
∞
cos
(
k
θ
)
k
=
−
1
2
ln
(
2
−
2
cos
θ
)
,
θ
∈
R
{\displaystyle \sum _{k=1}^{\infty }{\frac {\cos(k\theta )}{k}}=-{\frac {1}{2}}\ln(2-2\cos \theta ),\theta \in \mathbb {R} }
∑
k
=
0
∞
s
e
n
[
(
2
k
+
1
)
θ
]
2
k
+
1
=
π
4
,
0
<
θ
<
π
{\displaystyle \sum _{k=0}^{\infty }{\frac {\mathrm {sen} \,[(2k+1)\theta ]}{2k+1}}={\frac {\pi }{4}},0<\theta <\pi }
[ 4]
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
s
e
n
(
θ
+
k
α
)
=
s
e
n
(
n
+
1
)
α
2
s
e
n
(
θ
+
n
α
2
)
s
e
n
α
2
{\displaystyle \sum _{k=0}^{n}\mathrm {sen} \,(\theta +k\alpha )={\frac {\mathrm {sen} \,{\frac {(n+1)\alpha }{2}}\mathrm {sen} \,(\theta +{\frac {n\alpha }{2}})}{\mathrm {sen} \,{\frac {\alpha }{2}}}}}
∑
k
=
0
n
cos
(
θ
+
k
α
)
=
s
e
n
(
n
+
1
)
α
2
cos
(
θ
+
n
α
2
)
s
e
n
α
2
{\displaystyle \sum _{k=0}^{n}\cos(\theta +k\alpha )={\frac {\mathrm {sen} \,{\frac {(n+1)\alpha }{2}}\cos(\theta +{\frac {n\alpha }{2}})}{\mathrm {sen} \,{\frac {\alpha }{2}}}}}
∑
k
=
1
n
−
1
s
e
n
π
k
n
=
cot
π
2
n
{\displaystyle \sum _{k=1}^{n-1}\mathrm {sen} \,{\frac {\pi k}{n}}=\cot {\frac {\pi }{2n}}}
∑
k
=
1
n
−
1
s
e
n
2
π
k
n
=
0
{\displaystyle \sum _{k=1}^{n-1}\mathrm {sen} \,{\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}}}
∑
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
4
+
4
a
4
=
1
+
a
π
coth
(
a
π
)
8
a
4
{\displaystyle \displaystyle \sum _{n=0}^{\infty }{\frac {1}{n^{4}+4a^{4}}}={\dfrac {1+a\pi \coth(a\pi )}{8a^{4}}}}
Uma série infinita de qualquer função racional de
n
{\displaystyle n}
decomposição em frações parciais .[ 8] tempo constante , mesmo quando a série contém um grande número de termos.
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)}
relação de Landsberg-Schaar )
∑
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)}}}
↑ Weisstein, Eric W. «Haversine» . Wolfram Research, Inc. Consultado em 6 de novembro de 2015 . Cópia arquivada em 10 de março de 2005 ↑ a b Wilf, Herbert R. (1994). generatingfunctionology (PDF) . Academic Press, Inc . [S.l.: s.n.] ↑ «Theoretical computer science cheat sheet» (PDF) ↑
Calculate the Fourier expansion of the function
f
(
x
)
=
π
4
{\displaystyle f(x)={\frac {\pi }{4}}}
0
<
x
<
π
{\displaystyle 0<x<\pi }
↑ «Bernoulli polynomials: Series representations (subsection 06/02)» . Wolfram Research . Consultado em 2 de junho de 2011 ↑ Hofbauer, Josef. «A simple proof of 1+1/2^2+1/3^2+...=PI^2/6 and related identities» (PDF) . Consultado em 2 de junho de 2011 ↑
Sondow, Jonathan; Weisstein, Eric W. «Riemann Zeta Function (eq. 52)» . MathWorld —A Wolfram Web Resource
↑ Abramowitz, Milton ; Stegun, Irene (1964). «6.4 Polygamma functions». Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables ISBN 0-486-61272-4 
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/83655857c974dd27c9b29de8cda04d7c65d334e3)
![{\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 \mathrm {arcsen} \,{z}},|z|\leq 1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/265f2fc1c18c2d6fce755d881d4aaac2f6fa0e9e)
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a1c2c3f140738f0c5c61f88f041f311fbda3a340)
![{\displaystyle \sum _{k=0}^{\infty }{\frac {\mathrm {sen} \,[(2k+1)\theta ]}{2k+1}}={\frac {\pi }{4}},0<\theta <\pi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/a8b6177936fe5df65bdf05b4e53ee14e169b21f5)
![{\displaystyle \displaystyle \sum _{n=-\infty }^{\infty }e^{-\pi n^{2}}={\frac {\sqrt[{4}]{\pi }}{\Gamma \left({\frac {3}{4}}\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4aee717a740629f569ad7c408608acb53f1ec4bd)
