Na Galipedia, a Wikipedia en galego.
O valor decimal do logaritmo natural de 2 (secuencia A002162 na OEIS ) é aproximadamente
ln
2
≈
0.693
147
180
559
945
309
417
232
121
458.
{\displaystyle \ln 2\approx 0.693\,147\,180\,559\,945\,309\,417\,232\,121\,458.}
O logaritmo decimal é (secuencia A007524 na OEIS )
log
10
2
≈
0.301
029
995
663
981
195.
{\displaystyle \log _{10}2\approx 0.301\,029\,995\,663\,981\,195.}
Segundo o teorema de Lindemann-Weierstrass , o logaritmo natural de calquera número natural que non sexa 0 e 1 (máis xeralmente, de calquera número alxébrico positivo que non sexa 1) é un número transcendental .
ln
2
=
∑
n
=
1
∞
(
−
1
)
n
+
1
n
=
1
−
1
2
+
1
3
−
1
4
+
1
5
−
1
6
+
⋯
.
{\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}=1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{5}}-{\frac {1}{6}}+\cdots .}
serie harmónica alterna ".
ln
2
=
1
2
+
1
2
∑
n
=
1
∞
(
−
1
)
n
+
1
n
(
n
+
1
)
.
{\displaystyle \ln 2={\frac {1}{2}}+{\frac {1}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)}}.}
ln
2
=
5
8
+
1
2
∑
n
=
1
∞
(
−
1
)
n
+
1
n
(
n
+
1
)
(
n
+
2
)
.
{\displaystyle \ln 2={\frac {5}{8}}+{\frac {1}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)}}.}
ln
2
=
2
3
+
3
4
∑
n
=
1
∞
(
−
1
)
n
+
1
n
(
n
+
1
)
(
n
+
2
)
(
n
+
3
)
.
{\displaystyle \ln 2={\frac {2}{3}}+{\frac {3}{4}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)}}.}
ln
2
=
131
192
+
3
2
∑
n
=
1
∞
(
−
1
)
n
+
1
n
(
n
+
1
)
(
n
+
2
)
(
n
+
3
)
(
n
+
4
)
.
{\displaystyle \ln 2={\frac {131}{192}}+{\frac {3}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)(n+4)}}.}
ln
2
=
661
960
+
15
4
∑
n
=
1
∞
(
−
1
)
n
+
1
n
(
n
+
1
)
(
n
+
2
)
(
n
+
3
)
(
n
+
4
)
(
n
+
5
)
.
{\displaystyle \ln 2={\frac {661}{960}}+{\frac {15}{4}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)(n+4)(n+5)}}.}
ln
2
=
2
3
.
(
1
+
2
4
3
−
4
+
2
8
3
−
8
+
2
12
3
−
12
+
.
.
.
.
.
.
.
.
.
)
.
{\displaystyle \ln 2={\frac {2}{3}}.(1+{\frac {2}{4^{3}-4}}+{\frac {2}{8^{3}-8}}+{\frac {2}{12^{3}-12}}+.........).}
ln
2
=
∑
n
=
1
∞
1
2
n
n
.
{\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {1}{2^{n}n}}.}
ln
2
=
1
−
∑
n
=
1
∞
1
2
n
n
(
n
+
1
)
.
{\displaystyle \ln 2=1-\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)}}.}
ln
2
=
1
2
+
2
∑
n
=
1
∞
1
2
n
n
(
n
+
1
)
(
n
+
2
)
.
{\displaystyle \ln 2={\frac {1}{2}}+2\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)}}.}
ln
2
=
5
6
−
6
∑
n
=
1
∞
1
2
n
n
(
n
+
1
)
(
n
+
2
)
(
n
+
3
)
.
{\displaystyle \ln 2={\frac {5}{6}}-6\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)(n+3)}}.}
ln
2
=
7
12
+
24
∑
n
=
1
∞
1
2
n
n
(
n
+
1
)
(
n
+
2
)
(
n
+
3
)
(
n
+
4
)
.
{\displaystyle \ln 2={\frac {7}{12}}+24\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)(n+3)(n+4)}}.}
ln
2
=
47
60
−
120
∑
n
=
1
∞
1
2
n
n
(
n
+
1
)
(
n
+
2
)
(
n
+
3
)
(
n
+
4
)
(
n
+
5
)
.
{\displaystyle \ln 2={\frac {47}{60}}-120\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)(n+3)(n+4)(n+5)}}.}
∑
n
=
0
∞
1
(
2
n
+
1
)
(
2
n
+
2
)
=
ln
2.
{\displaystyle \sum _{n=0}^{\infty }{\frac {1}{(2n+1)(2n+2)}}=\ln 2.}
∑
n
=
1
∞
1
n
(
4
n
2
−
1
)
=
2
ln
2
−
1.
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n(4n^{2}-1)}}=2\ln 2-1.}
∑
n
=
1
∞
(
−
1
)
n
n
(
4
n
2
−
1
)
=
ln
2
−
1.
{\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n(4n^{2}-1)}}=\ln 2-1.}
∑
n
=
1
∞
(
−
1
)
n
n
(
9
n
2
−
1
)
=
2
ln
2
−
3
2
.
{\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n(9n^{2}-1)}}=2\ln 2-{\frac {3}{2}}.}
∑
n
=
1
∞
1
4
n
2
−
2
n
=
ln
2.
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{4n^{2}-2n}}=\ln 2.}
∑
n
=
1
∞
2
(
−
1
)
n
+
1
(
2
n
−
1
)
+
1
8
n
2
−
4
n
=
ln
2.
{\displaystyle \sum _{n=1}^{\infty }{\frac {2(-1)^{n+1}(2n-1)+1}{8n^{2}-4n}}=\ln 2.}
∑
n
=
0
∞
(
−
1
)
n
3
n
+
1
=
ln
2
3
+
π
3
3
.
{\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3n+1}}={\frac {\ln 2}{3}}+{\frac {\pi }{3{\sqrt {3}}}}.}
∑
n
=
0
∞
(
−
1
)
n
3
n
+
2
=
−
ln
2
3
+
π
3
3
.
{\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3n+2}}=-{\frac {\ln 2}{3}}+{\frac {\pi }{3{\sqrt {3}}}}.}
∑
n
=
0
∞
(
−
1
)
n
(
3
n
+
1
)
(
3
n
+
2
)
=
2
ln
2
3
.
{\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(3n+1)(3n+2)}}={\frac {2\ln 2}{3}}.}
∑
n
=
1
∞
1
∑
k
=
1
n
k
2
=
18
−
24
ln
2
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{\sum _{k=1}^{n}k^{2}}}=18-24\ln 2}
lim
N
→
∞
∑
n
=
N
2
N
1
n
=
ln
2
{\displaystyle \lim _{N\rightarrow \infty }\sum _{n=N}^{2N}{\frac {1}{n}}=\ln 2}
∑
n
=
1
∞
1
4
n
2
−
3
n
=
ln
2
+
π
6
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{4n^{2}-3n}}=\ln 2+{\frac {\pi }{6}}}
∑
n
=
1
∞
1
n
[
ζ
(
2
n
)
−
1
]
=
ln
2.
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n}}[\zeta (2n)-1]=\ln 2.}
∑
n
=
2
∞
1
2
n
[
ζ
(
n
)
−
1
]
=
ln
2
−
1
2
.
{\displaystyle \sum _{n=2}^{\infty }{\frac {1}{2^{n}}}[\zeta (n)-1]=\ln 2-{\frac {1}{2}}.}
∑
n
=
1
∞
1
2
n
+
1
[
ζ
(
2
n
+
1
)
−
1
]
=
1
−
γ
−
ln
2
2
.
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{2n+1}}[\zeta (2n+1)-1]=1-\gamma -{\frac {\ln 2}{2}}.}
∑
n
=
1
∞
1
2
2
n
−
1
(
2
n
+
1
)
ζ
(
2
n
)
=
1
−
ln
2.
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{2^{2n-1}(2n+1)}}\zeta (2n)=1-\ln 2.}
(γ ζ
ln
2
=
2
3
+
1
2
∑
k
=
1
∞
(
1
2
k
+
1
4
k
+
1
+
1
8
k
+
4
+
1
16
k
+
12
)
1
16
k
.
{\displaystyle \ln 2={\frac {2}{3}}+{\frac {1}{2}}\sum _{k=1}^{\infty }\left({\frac {1}{2k}}+{\frac {1}{4k+1}}+{\frac {1}{8k+4}}+{\frac {1}{16k+12}}\right){\frac {1}{16^{k}}}.}
(Consulte máis información sobre as representacións de tipo Bailey–Borwein–Plouffe (BBP) .)
Aplicando as tres series xerais para o logaritmo natural a 2 directamente dá:
ln
2
=
∑
n
=
1
∞
(
−
1
)
n
−
1
n
.
{\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n}}.}
ln
2
=
∑
n
=
1
∞
1
2
n
n
.
{\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {1}{2^{n}n}}.}
ln
2
=
2
3
∑
k
=
0
∞
1
9
k
(
2
k
+
1
)
.
{\displaystyle \ln 2={\frac {2}{3}}\sum _{k=0}^{\infty }{\frac {1}{9^{k}(2k+1)}}.}
O logaritmo natural de 2 ocorre con frecuencia como resultado da integración. Algunhas fórmulas explícitas para iso inclúen:
∫
0
1
d
x
1
+
x
=
∫
1
2
d
x
x
=
ln
2
{\displaystyle \int _{0}^{1}{\frac {dx}{1+x}}=\int _{1}^{2}{\frac {dx}{x}}=\ln 2}
∫
0
∞
e
−
x
1
−
e
−
x
x
d
x
=
ln
2
{\displaystyle \int _{0}^{\infty }e^{-x}{\frac {1-e^{-x}}{x}}\,dx=\ln 2}
∫
0
∞
2
−
x
d
x
=
1
ln
2
{\displaystyle \int _{0}^{\infty }2^{-x}dx={\frac {1}{\ln 2}}}
∫
0
π
3
tan
x
d
x
=
2
∫
0
π
4
tan
x
d
x
=
ln
2
{\displaystyle \int _{0}^{\frac {\pi }{3}}\tan x\,dx=2\int _{0}^{\frac {\pi }{4}}\tan x\,dx=\ln 2}
−
1
π
i
∫
0
∞
ln
x
ln
ln
x
(
x
+
1
)
2
d
x
=
ln
2
{\displaystyle -{\frac {1}{\pi i}}\int _{0}^{\infty }{\frac {\ln x\ln \ln x}{(x+1)^{2}}}\,dx=\ln 2}
A expansión de Pierce é (secuencia A091846 na OEIS )
ln
2
=
1
−
1
1
⋅
3
+
1
1
⋅
3
⋅
12
−
⋯
.
{\displaystyle \ln 2=1-{\frac {1}{1\cdot 3}}+{\frac {1}{1\cdot 3\cdot 12}}-\cdots .}
A expansión de Engel é (secuencia A059180 na OEIS )
ln
2
=
1
2
+
1
2
⋅
3
+
1
2
⋅
3
⋅
7
+
1
2
⋅
3
⋅
7
⋅
9
+
⋯
.
{\displaystyle \ln 2={\frac {1}{2}}+{\frac {1}{2\cdot 3}}+{\frac {1}{2\cdot 3\cdot 7}}+{\frac {1}{2\cdot 3\cdot 7\cdot 9}}+\cdots .}
A expansión cotanxente é (secuencia A081785 na OEIS )
ln
2
=
cot
(
arccot
(
0
)
−
arccot
(
1
)
+
arccot
(
5
)
−
arccot
(
55
)
+
arccot
(
14187
)
−
⋯
)
.
{\displaystyle \ln 2=\cot({\operatorname {arccot}(0)-\operatorname {arccot}(1)+\operatorname {arccot}(5)-\operatorname {arccot}(55)+\operatorname {arccot}(14187)-\cdots }).}
A expansión como fracción continua simple é (secuencia A016730 na OEIS )
ln
2
=
[
0
;
1
,
2
,
3
,
1
,
6
,
3
,
1
,
1
,
2
,
1
,
1
,
1
,
1
,
3
,
10
,
1
,
1
,
1
,
2
,
1
,
1
,
1
,
1
,
3
,
2
,
3
,
1
,
.
.
.
]
{\displaystyle \ln 2=\left[0;1,2,3,1,6,3,1,1,2,1,1,1,1,3,10,1,1,1,2,1,1,1,1,3,2,3,1,...\right]}
que produce aproximacións racionais, as primeiras das cales son 0, 1, 2/3, 7/10, 9/13 e 61/88.
Esta fracción continua xeneralizada :
ln
2
=
[
0
;
1
,
2
,
3
,
1
,
5
,
2
3
,
7
,
1
2
,
9
,
2
5
,
.
.
.
,
2
k
−
1
,
2
k
,
.
.
.
]
{\displaystyle \ln 2=\left[0;1,2,3,1,5,{\tfrac {2}{3}},7,{\tfrac {1}{2}},9,{\tfrac {2}{5}},...,2k-1,{\frac {2}{k}},...\right]}
[ 1] tamén expresábel como
ln
2
=
1
1
+
1
2
+
1
3
+
2
2
+
2
5
+
3
2
+
3
7
+
4
2
+
⋱
=
2
3
−
1
2
9
−
2
2
15
−
3
2
21
−
⋱
{\displaystyle \ln 2={\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{3+{\cfrac {2}{2+{\cfrac {2}{5+{\cfrac {3}{2+{\cfrac {3}{7+{\cfrac {4}{2+\ddots }}}}}}}}}}}}}}}}={\cfrac {2}{3-{\cfrac {1^{2}}{9-{\cfrac {2^{2}}{15-{\cfrac {3^{2}}{21-\ddots }}}}}}}}}
Esta é unha táboa de rexistros recentes no cálculo de díxitos de ln 2 . Desde decembro de 2018, calculáronse máis díxitos que os de calquera outro logaritmo neperiano [ 2] [ 3]
Data
Nome
Número de díxitos
7 de xaneiro de 2009
A.Yee e R.Chan
15.500.000.000
4 de febreiro de 2009
A.Yee e R.Chan
31.026.000.000
21 de febreiro de 2011
Alexander Yee
50.000.000.050
14 de maio de 2011
Shigeru Kondo
100.000.000.000
28 de febreiro de 2014
Shigeru Kondo
200.000.000.050
12 de xullo de 2015
Ron Watkins
250.000.000.000
30 de xaneiro de 2016
Ron Watkins
350.000.000.000
18 de abril de 2016
Ron Watkins
500.000.000.000
10 de decembro de 2018
Michael Kwok
600.000.000.000
26 de abril de 2019
Jacob Riffee
1.000.000.000.000
19 de agosto de 2020
Seungmin Kim [ 4] [ 5]
1.200.000.000.100
9 de setembro de 2021
William Echols [ 6] [ 7]
1.500.000.000.000
![{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n}}[\zeta (2n)-1]=\ln 2.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9ef59c49175d34e1aee22a9a50f45fabf050db5f)
![{\displaystyle \sum _{n=2}^{\infty }{\frac {1}{2^{n}}}[\zeta (n)-1]=\ln 2-{\frac {1}{2}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b61c76e20b8727253043baa534c03f6a85808e72)
![{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{2n+1}}[\zeta (2n+1)-1]=1-\gamma -{\frac {\ln 2}{2}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69aa41d922c642991e8a73fb2c8567ec80d89632)
![{\displaystyle \ln 2=\left[0;1,2,3,1,6,3,1,1,2,1,1,1,1,3,10,1,1,1,2,1,1,1,1,3,2,3,1,...\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5f8173c7c5f057cbbf1d0292b9cf10e9eee6e208)
![{\displaystyle \ln 2=\left[0;1,2,3,1,5,{\tfrac {2}{3}},7,{\tfrac {1}{2}},9,{\tfrac {2}{5}},...,2k-1,{\frac {2}{k}},...\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/81cd30af8050ce0bde521ec3ff78833704ba286a)
