uredi
Konačne sume:
∑
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}}}
geometrijski red )
∑
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}}}
Beskonačne sume, važe za
|
z
|
<
1
{\displaystyle |z|<1}
Li
n
(
z
)
=
∑
k
=
1
∞
z
k
k
n
{\displaystyle \operatorname {Li} _{n}(z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k^{n}}}}
Naredne formule su korisne za rekurzivno izračunavanje polilogaritama nižeg celobrojnog reda u zatvorenoj formi:
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}}}}
uredi
∑
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}}
Poasonove raspodele )
∑
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}}
Poasonove raspodele )
∑
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)}
uredi
∑
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}
∑
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]
∑
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}}}
uredi
∑
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
=
(
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}
[ 2]
∑
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}
uredi
(
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}
[ 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}
[ 3]
∑
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}}}
Katalanovih brojeva [ 3]
∑
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}}}
[ 3]
∑
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}}}
∑
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}
[ 2]
∑
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)}
uredi
Sume sinusa i kosinusa rastu po Furijeovom redu .
∑
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 }
[ 4]
∑
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<x<1}
[ 5]
∑
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 )}
[ 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}}}
uredi
∑
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
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))}}}
Beskonačni red bilo koje racionalne funkcije u zavisnosti od
n
{\displaystyle n}
[ 8]
uredi
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)}
∑
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“ . MathWorld Wolfram Research, Inc. . ↑ 2,0 2,1 2,2 2,3 Wilf, Herbert R. (1994). generatingfunctionology ↑ 3,0 3,1 3,2 3,3 „Theoretical computer science cheat sheet“ . ↑
Calculate the Fourier expansion of the function
f
(
x
)
=
π
4
{\displaystyle f(x)={\frac {\pi }{4}}}
0
<
x
<
π
{\displaystyle 0<x<\pi }
π
4
=
∑
n
=
0
∞
c
n
sin
[
n
x
]
+
d
n
cos
[
n
x
]
{\displaystyle {\frac {\pi }{4}}=\sum _{n=0}^{\infty }c_{n}\sin[nx]+d_{n}\cos[nx]}
⇒
{
c
n
=
{
1
n
(
n
odd
)
0
(
n
even
)
d
n
=
0
(
∀
n
)
{\displaystyle \Rightarrow {\begin{cases}c_{n}={\begin{cases}{\frac {1}{n}}\quad (n{\text{ odd}})\\0\quad (n{\text{ even}})\end{cases}}\\d_{n}=0\quad (\forall n)\end{cases}}}
↑ „Bernoulli polynomials: Series representations (subsection 06/02)“ Приступљено 2 June 2011 . ↑ Hofbauer, Josef. „A simple proof of 1 + 1/22 + 1/32 + ··· = Шаблон:Pi 2 /6 and related identities“ Приступљено 2 June 2011 . ↑
Sondow, Jonathan; Weisstein, Eric W.. „Riemann Zeta Function (eq. 52)“ .
↑ Abramowitz, Milton; Stegun, Irene (1964). „6.4 Polygamma functions“ . Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables 260 . 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 \arcsin {z}},|z|\leq 1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7690094e2c29c30c517059014511d42f93f0912a)
![{\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 {\cos[(2k+1)\theta ]}{2k+1}}={\frac {1}{2}}\ln \left(\cot {\frac {\theta }{2}}\right),0<\theta <\pi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/d1991f46f491715b581a7037b4125c14fe65025c)
![{\displaystyle \sum _{k=0}^{\infty }{\frac {\sin[(2k+1)\theta ]}{2k+1}}={\frac {\pi }{4}},0<\theta <\pi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/faf41e942b227c04e70af874e45bddb621602e58)
![{\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)
![{\displaystyle {\frac {\pi }{4}}=\sum _{n=0}^{\infty }c_{n}\sin[nx]+d_{n}\cos[nx]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a5b6fd91cf5e77255955c2b09cdc203bcb5bf73)
