Матеріал з Вікіпедії — вільної енциклопедії.
Немає
перевірених версій цієї сторінки; ймовірно, її ще
не перевіряли на відповідність правилам проекту.
Цей список містить формули для рядів та сум .
∑
k
=
0
n
k
s
=
(
n
+
1
)
s
+
1
s
+
1
+
∑
k
=
1
s
B
k
s
−
k
+
1
(
s
k
)
(
n
+
1
)
s
−
k
+
1
{\displaystyle \sum _{k=0}^{n}k^{s}={\frac {(n+1)^{s+1}}{s+1}}+\sum _{k=1}^{s}{\frac {B_{k}}{s-k+1}}{s \choose k}(n+1)^{s-k+1}}
— формула Фольхабера [en]
Значення при
n
=
1
,
2
,
…
,
12
:
{\displaystyle n=1,2,\dots ,12:}
∑
k
=
1
n
k
=
n
(
n
+
1
)
2
=
(
n
+
1
)
3
−
n
3
−
1
6
{\displaystyle \sum _{k=1}^{n}k={\frac {n(n+1)}{2}}={\frac {(n+1)^{3}-n^{3}-1}{6}}\,\!}
∑
k
=
p
q
k
=
p
+
(
p
+
1
)
+
(
p
+
2
)
+
(
p
+
3
)
+
…
+
(
q
−
1
)
+
q
=
(
p
+
q
)
(
q
−
p
+
1
)
2
{\displaystyle \sum _{k=p}^{q}k=p+(p+1)+(p+2)+(p+3)+\ldots +(q-1)+q={\frac {(p+q)(q-p+1)}{2}}}
∑
k
=
1
n
2
k
=
2
+
4
+
6
+
8
+
10
+
12
+
14
+
16
+
…
+
(
2
n
−
2
)
+
2
n
=
n
(
n
+
1
)
{\displaystyle \sum _{k=1}^{n}2k=2+4+6+8+10+12+14+16+\ldots +(2n-2)+2n=n(n+1)}
∑
k
=
1
n
k
2
=
n
(
n
+
1
)
(
2
n
+
1
)
6
=
n
3
3
+
n
2
2
+
n
6
{\displaystyle \sum _{k=1}^{n}k^{2}={\frac {n(n+1)(2n+1)}{6}}={\frac {n^{3}}{3}}+{\frac {n^{2}}{2}}+{\frac {n}{6}}}
∑
k
=
1
n
k
3
=
(
n
(
n
+
1
)
2
)
2
=
n
4
4
+
n
3
2
+
n
2
4
=
(
∑
k
=
1
n
k
)
2
{\displaystyle \sum _{k=1}^{n}k^{3}=\left({\frac {n(n+1)}{2}}\right)^{2}={\frac {n^{4}}{4}}+{\frac {n^{3}}{2}}+{\frac {n^{2}}{4}}=\left(\sum _{k=1}^{n}k\right)^{2}}
— сума перших
n
{\displaystyle n}
кубів [en]
∑
k
=
1
n
k
4
=
n
(
n
+
1
)
(
2
n
+
1
)
(
3
n
2
+
3
n
−
1
)
30
{\displaystyle \sum _{k=1}^{n}k^{4}={\frac {n(n+1)(2n+1)(3n^{2}+3n-1)}{30}}}
∑
k
=
1
n
k
5
=
n
2
(
n
+
1
)
2
(
2
n
2
+
2
n
−
1
)
12
{\displaystyle \sum _{k=1}^{n}k^{5}={\frac {n^{2}(n+1)^{2}(2n^{2}+2n-1)}{12}}}
∑
k
=
1
n
k
6
=
n
(
n
+
1
)
(
2
n
+
1
)
(
3
n
4
+
6
n
3
−
3
n
+
1
)
42
{\displaystyle \sum _{k=1}^{n}k^{6}={\frac {n(n+1)(2n+1)(3n^{4}+6n^{3}-3n+1)}{42}}}
∑
k
=
1
n
k
7
=
n
2
(
n
+
1
)
2
(
3
n
4
+
6
n
3
−
n
2
−
4
n
+
2
)
24
{\displaystyle \sum _{k=1}^{n}k^{7}={\frac {n^{2}(n+1)^{2}(3n^{4}+6n^{3}-n^{2}-4n+2)}{24}}}
∑
k
=
1
n
k
8
=
n
(
n
+
1
)
(
2
n
+
1
)
(
5
n
6
+
15
n
5
+
5
n
4
−
15
n
3
−
n
2
+
9
n
−
3
)
24
{\displaystyle \sum _{k=1}^{n}k^{8}={\frac {n(n+1)(2n+1)(5n^{6}+15n^{5}+5n^{4}-15n^{3}-n^{2}+9n-3)}{24}}}
∑
k
=
1
n
k
9
=
n
2
(
n
+
1
)
2
(
n
2
+
n
−
1
)
(
2
n
4
+
4
n
3
−
n
2
−
3
n
+
3
)
20
{\displaystyle \sum _{k=1}^{n}k^{9}={\frac {n^{2}(n+1)^{2}(n^{2}+n-1)(2n^{4}+4n^{3}-n^{2}-3n+3)}{20}}}
∑
k
=
1
n
k
10
=
n
(
n
+
1
)
(
2
n
+
1
)
(
n
2
+
n
−
1
)
(
3
n
6
+
9
n
5
+
2
n
4
−
11
n
3
+
3
n
2
+
10
n
−
5
)
66
{\displaystyle \sum _{k=1}^{n}k^{10}={\frac {n(n+1)(2n+1)(n^{2}+n-1)(3n^{6}+9n^{5}+2n^{4}-11n^{3}+3n^{2}+10n-5)}{66}}}
∑
k
=
1
n
k
11
=
n
2
(
n
+
1
)
2
(
2
n
8
+
8
n
7
+
4
n
6
−
16
n
5
−
5
n
4
+
26
n
3
−
3
n
2
−
20
n
+
10
)
24
{\displaystyle \sum _{k=1}^{n}k^{11}={\frac {n^{2}(n+1)^{2}(2n^{8}+8n^{7}+4n^{6}-16n^{5}-5n^{4}+26n^{3}-3n^{2}-20n+10)}{24}}}
∑
k
=
1
n
k
12
=
n
(
n
+
1
)
(
2
n
+
1
)
(
105
n
10
+
525
n
9
+
525
n
8
−
1050
n
7
−
1190
n
6
+
2310
n
5
+
1420
n
4
−
3285
n
3
−
287
n
2
+
2073
n
−
691
)
2730
{\displaystyle \sum _{k=1}^{n}k^{12}={\frac {n(n+1)(2n+1)(105n^{10}+525n^{9}+525n^{8}-1050n^{7}-1190n^{6}+2310n^{5}+1420n^{4}-3285n^{3}-287n^{2}+2073n-691)}{2730}}}
Суми степенів непарних чисел:
∑
k
=
1
n
(
2
k
−
1
)
=
1
+
3
+
5
+
7
+
9
+
…
+
(
2
n
−
3
)
+
(
2
n
−
1
)
=
n
2
{\displaystyle \sum _{k=1}^{n}(2k-1)=1+3+5+7+9+\ldots +(2n-3)+(2n-1)=n^{2}}
∑
k
=
1
n
(
2
k
−
1
)
2
=
1
2
+
3
2
+
5
2
+
7
2
+
9
2
+
…
+
(
2
n
−
3
)
2
+
(
2
n
−
1
)
2
=
n
(
4
n
2
−
1
)
3
{\displaystyle \sum _{k=1}^{n}(2k-1)^{2}=1^{2}+3^{2}+5^{2}+7^{2}+9^{2}+\ldots +(2n-3)^{2}+(2n-1)^{2}={\frac {n(4n^{2}-1)}{3}}}
∑
k
=
1
n
(
2
k
−
1
)
3
=
1
3
+
3
3
+
5
3
+
7
3
+
9
3
+
…
+
(
2
n
−
3
)
3
+
(
2
n
−
1
)
3
=
n
2
(
2
n
2
−
1
)
{\displaystyle \sum _{k=1}^{n}(2k-1)^{3}=1^{3}+3^{3}+5^{3}+7^{3}+9^{3}+\ldots +(2n-3)^{3}+(2n-1)^{3}=n^{2}(2n^{2}-1)}
Окремі значення дзета-функції Рімана [en] :
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)!}}}
Значення при
n
=
1
,
2
,
3
:
{\displaystyle n=1,2,3:}
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}}}
— ряд обернених квадратів
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}}}
∑
k
=
1
∞
(
−
1
)
k
+
1
k
2
s
=
(
1
−
1
2
2
s
−
1
)
∑
k
=
1
∞
1
k
2
s
{\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{k^{2s}}}=\left(1-{\frac {1}{2^{2s-1}}}\right)\sum _{k=1}^{\infty }{\frac {1}{k^{2s}}}}
∑
k
=
0
∞
1
(
2
k
+
1
)
2
=
π
2
8
{\displaystyle \sum _{k=0}^{\infty }{\frac {1}{(2k+1)^{2}}}={\frac {\pi ^{2}}{8}}}
∑
k
=
0
∞
1
(
2
k
+
1
)
4
=
π
4
96
{\displaystyle \sum _{k=0}^{\infty }{\frac {1}{(2k+1)^{4}}}={\frac {\pi ^{4}}{96}}}
∑
k
=
0
∞
1
(
2
k
+
1
)
6
=
π
6
960
{\displaystyle \sum _{k=0}^{\infty }{\frac {1}{(2k+1)^{6}}}={\frac {\pi ^{6}}{960}}}
Скінченні суми:
∑
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}}}
— геометрична прогресія
∑
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}}}
Нескінченні суми, виконується при
|
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}}}}
Корисна властивість для рекурсивного обчислення полілогарифмів:
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}}}
Полілогарифми малих по модулю цілих порядків:
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}}
∑
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}}
∑
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)}
де
T
n
(
z
)
{\displaystyle T_{n}(z)}
— поліном Тушара [en] .
Тригонометричні, обернені тригонометричні, гіперболічні та обернені гіперболічні функції[ ред. | ред. код ]
∑
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
)
!
=
sh
z
{\displaystyle \sum _{k=0}^{\infty }{\frac {z^{2k+1}}{(2k+1)!}}=\operatorname {sh} 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
)
!
=
ch
z
{\displaystyle \sum _{k=0}^{\infty }{\frac {z^{2k}}{(2k)!}}=\operatorname {ch} z}
∑
k
=
1
∞
(
−
1
)
k
−
1
(
2
2
k
−
1
)
2
2
k
B
2
k
z
2
k
−
1
(
2
k
)
!
=
tg
z
,
|
z
|
<
π
2
{\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}(2^{2k}-1)2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\operatorname {tg} z,|z|<{\frac {\pi }{2}}}
∑
k
=
1
∞
(
2
2
k
−
1
)
2
2
k
B
2
k
z
2
k
−
1
(
2
k
)
!
=
th
z
,
|
z
|
<
π
2
{\displaystyle \sum _{k=1}^{\infty }{\frac {(2^{2k}-1)2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\operatorname {th} z,|z|<{\frac {\pi }{2}}}
∑
k
=
0
∞
(
−
1
)
k
2
2
k
B
2
k
z
2
k
−
1
(
2
k
)
!
=
ctg
z
,
|
z
|
<
π
{\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\operatorname {ctg} z,|z|<\pi }
∑
k
=
0
∞
2
2
k
B
2
k
z
2
k
−
1
(
2
k
)
!
=
cth
z
,
|
z
|
<
π
{\displaystyle \sum _{k=0}^{\infty }{\frac {2^{2k}B_{2k}z^{2k-1}}{(2k)!}}=\operatorname {cth} 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}
— синус-верзус
∑
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] — гаверсинус
∑
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|\leqslant 1}
∑
k
=
0
∞
(
−
1
)
k
(
2
k
)
!
z
2
k
+
1
2
2
k
(
k
!
)
2
(
2
k
+
1
)
=
arsh
z
,
|
z
|
⩽
1
{\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}(2k)!z^{2k+1}}{2^{2k}(k!)^{2}(2k+1)}}=\operatorname {arsh} {z},|z|\leqslant 1}
∑
k
=
0
∞
(
−
1
)
k
z
2
k
+
1
2
k
+
1
=
arctg
z
,
|
z
|
<
1
{\displaystyle \sum _{k=0}^{\infty }{\frac {(-1)^{k}z^{2k+1}}{2k+1}}=\operatorname {arctg} z,|z|<1}
∑
k
=
0
∞
z
2
k
+
1
2
k
+
1
=
arth
z
,
|
z
|
<
1
{\displaystyle \sum _{k=0}^{\infty }{\frac {z^{2k+1}}{2k+1}}=\operatorname {arth} z,|z|<1}
∑
k
=
1
∞
(
−
1
)
k
−
1
(
2
k
)
!
z
2
k
2
2
k
+
1
k
(
k
!
)
2
=
ln
(
1
+
1
+
z
2
2
)
,
|
z
|
⩽
1
{\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k-1}(2k)!z^{2k}}{2^{2k+1}k(k!)^{2}}}=\ln \left({\frac {1+{\sqrt {1+z^{2}}}}{2}}\right),|z|\leqslant 1}
∑
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|\leqslant 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|\leqslant 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}
∑
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}}}
— генератриса чисел Каталана
∑
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}}}
— генератриса центральних біноміальних коефіцієнтів
∑
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
arctg
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}}\operatorname {arctg} {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
arctg
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}}\operatorname {arctg} {z}\log {\frac {1+z}{1-z}},\qquad |z|<1}
[ 2]
∑
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
,
{\displaystyle \sum _{k=0}^{n}(-1)^{k}{n \choose k}=0,}
де
n
≠
0
{\displaystyle n\neq 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}}
∑
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}}
— тотожність Вандермонда
∑
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
(
ctg
θ
2
)
,
0
<
θ
<
π
{\displaystyle \sum _{k=0}^{\infty }{\frac {\cos[(2k+1)\theta ]}{2k+1}}={\frac {1}{2}}\ln \left(\operatorname {ctg} {\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 }
,[ 3]
∑
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}
[ 4]
∑
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
=
ctg
π
2
n
{\displaystyle \sum _{k=1}^{n-1}\sin {\frac {\pi k}{n}}=\operatorname {ctg} {\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 )}
[ 5]
∑
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}}}
Ряд раціональних функцій від
n
{\displaystyle n}
можна звести до скінченної суми полігамма-функцій за допомогою розкладання на прості дроби .[ 6] Це також можна застосувати до обчислення скінченних сум раціональних функцій за сталий час , навіть якщо сума містить велику кількість членів.
∑
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
cth
(
a
π
)
2
a
2
{\displaystyle \sum _{n=0}^{\infty }{\frac {1}{n^{2}+a^{2}}}={\frac {1+a\pi \operatorname {cth} (a\pi )}{2a^{2}}}}
∑
n
=
0
∞
1
n
4
+
4
a
4
=
1
8
a
4
+
π
(
sh
(
2
π
a
)
+
sin
(
2
π
a
)
)
8
a
3
(
ch
(
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 (\operatorname {sh} (2\pi a)+\sin(2\pi a))}{8a^{3}(\operatorname {ch} (2\pi a)-\cos(2\pi a))}}}
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)}
— співвідношення Ландсберга–Шара [en]
∑
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)}}}
Ці числові ряди можна знайти за допомогою рядів, наведених вище.
∑
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}}}
∑
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
)
=
ch
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}}(e+{\frac {1}{e}})=\operatorname {ch} 1}
∑
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
cth
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 \operatorname {cth} \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 }
∑
k
=
1
∞
1
T
k
=
1
1
+
1
3
+
1
6
+
1
10
+
1
15
+
⋯
=
2
{\displaystyle \sum _{k=1}^{\infty }{\frac {1}{T_{k}}}={\frac {1}{1}}+{\frac {1}{3}}+{\frac {1}{6}}+{\frac {1}{10}}+{\frac {1}{15}}+\cdots =2}
де
T
n
=
∑
k
=
1
n
k
{\displaystyle T_{n}=\sum _{k=1}^{n}k}
—
n
{\displaystyle n}
-те трикутне число
∑
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}}}
де
T
e
n
=
∑
k
=
1
n
T
k
{\displaystyle Te_{n}=\sum _{k=1}^{n}T_{k}}
—
n
{\displaystyle n}
-те тетраедричне число
∑
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}
↑ Weisstein, Eric W. Haversine . MathWorld . Wolfram Research, Inc. Архів оригіналу за 10 березня 2005.
↑ а б в г Wilf, Herbert R. (1994). generatingfunctionology (PDF) . Academic Press, Inc.
↑
Знайдіть розклад в ряд Фур’є функції
f
(
x
)
=
π
4
{\displaystyle f(x)={\frac {\pi }{4}}}
на інтервалі
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
- непарне
0
,
n
- парне
d
n
=
0
,
n
∈
N
{\displaystyle \Rightarrow {\begin{cases}c_{n}={\begin{cases}{\frac {1}{n}},\quad n{\text{- непарне}}\\0,\quad n{\text{- парне}}\end{cases}}\\d_{n}=0,\quad n\in \mathbb {N} \end{cases}}}
↑ Bernoulli polynomials: Series representations (subsection 06/02) . Wolfram Research .
↑ Hofbauer, Josef. A simple proof of 1 + 1/22 + 1/32 + ··· = π 2 /6 and related identities (PDF) .
↑ Abramowitz, Milton; Stegun, Irene (1964). 6.4 Polygamma functions . Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables . с. 260 . ISBN 0-486-61272-4 .