Berikut adalah daftar deret matematika yang berisi tentang rumus untuk penjumlahan terhingga dan tak terhingga. Ini dapat digunakan bersama-sama dengan alat-alat lain untuk menghitung penjumlahan.
Penjumlahan terhingga
∑
k
=
i
n
z
k
=
z
i
−
z
n
+
1
1
−
z
{\displaystyle \sum _{k=i}^{n}z^{k}={\frac {z^{i}-z^{n+1}}{1-z}}}
, (deret geometrik )
∑
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}}}
Penjumlahan tak terhingga, sah untuk
|
z
|
<
1
{\displaystyle \left|z\right|<1}
(lihat polilogaritma )
Li
n
(
z
)
=
∑
k
=
1
∞
z
k
k
n
{\displaystyle \operatorname {Li} _{n}(z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k^{n}}}}
Berikut ini adalah sebuah sifat yang berguna untuk menghitung polilogaritma urutan bilangan bulat rendah secara rekursif dalam bentuk tertutup :
d
d
z
Li
n
(
z
)
=
Li
n
−
1
(
z
)
z
{\displaystyle {\frac {d}{dz}}\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}}
(bandingkan rata-rata distribusi 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}}
(bandingkan momen kedua distribusi 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)}
dengan
T
n
(
z
)
{\displaystyle T_{n}(z)}
adalah polinomial Touchard .
Fungsi trigonometrik, trigonometrik invers, hiperbolik, dan hiperbolik invers
sunting
∑
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}
(versinus )
∑
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] (haversinus )
∑
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}
Penyebut faktorial yang dimodifikasi
sunting
∑
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}
(
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}
(lihat teorema binomial )
∑
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}}}
, menghasilkan fungsi bilangan Catalan [ 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}}}
, menghasilkan fungsi koefisien binomial pusat [ 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}}}
[ 3]
(Lihat bilangan harmonik yang didefinisikan
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}
[ 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,}
dengan
n
≥
1
{\displaystyle 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}}
(lihat multihimpunan )
∑
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}}
(lihat identitas Vandermonde )
∑
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}}
[ 6]
∑
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
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))}}}
Suatu deret tak terhingga dari setiap fungsi rasional
n
{\displaystyle n}
dapat direduksi menjadi suatu deret terhingga dari fungsi poligamma , dengan menggunakan dekomposisi pecahan parsial .[ 7] Fakta ini juga berlaku pada deret terhingga dari fungsi rasional, yang memungkinkan hasilnya dihitung dalam waktu konstanta bahkan jika deret tersebut memiliki banyak suku.
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)}
(lihat relasi 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" . MathWorld . Wolfram Research, Inc. Diarsipkan dari versi asli tanggal 2005-03-10. Diakses tanggal 2015-11-06 .
^ a b c d Wilf, Herbert R. (1994). generatingfunctionology (PDF) . Academic Press, Inc.
^ a b c d "Theoretical computer science cheat sheet" (PDF) .
^ "Bernoulli polynomials: Series representations (subsection 06/02)" . Wolfram Research . Diakses tanggal 2 June 2011 .
^ Hofbauer, Josef. "A simple proof of
1
+
1
2
2
+
1
3
2
+
⋯
=
π
2
6
{\textstyle 1+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+\dots ={\frac {\pi ^{2}}{6}}}
and related identities" (PDF) . Diakses tanggal 2 June 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 . hlm. 260 . ISBN 0-486-61272-4 .
Banyak buku-buku dengan sebuah daftar integral juga memiliki sebuah daftar deret.