Natural logarithm of 2

In mathematics, the natural logarithm of 2 is the unique real number argument such that the exponential function equals two. It appears frequently in various formulas and is also given by the alternating harmonic series. The decimal value of the natural logarithm of 2 Template:OEIS truncated at 30 decimal places is given by:

$\ln 2\approx 0.693147180559945309417232121458.$

The logarithm of 2 in other bases is obtained with the formula

${\log }_{b}2=\frac{\ln 2}{\ln b}.$

The common logarithm in particular is (Template:OEIS2C)

${\log }_{10}2\approx 0.301029995663981195.$

The inverse of this number is the binary logarithm of 10:

${\log }_{2}10=\frac{1}{{\log }_{10}2}\approx 3.321928095$ (Template:OEIS2C).

By the Lindemann–Weierstrass theorem, the natural logarithm of any natural number other than 0 and 1 (more generally, of any positive algebraic number other than 1) is a transcendental number. It is also contained in the ring of algebraic periods.

Series representations

Rising alternate factorial

$\ln 2=\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}}{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots .$ This is the well-known "alternating harmonic series".

$\ln 2=\frac{1}{2}+\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}}{n(n+1)}.$

$\ln 2=\frac{5}{8}+\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}}{n(n+1)(n+2)}.$

$\ln 2=\frac{2}{3}+\frac{3}{4}\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}}{n(n+1)(n+2)(n+3)}.$

$\ln 2=\frac{131}{192}+\frac{3}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}}{n(n+1)(n+2)(n+3)(n+4)}.$

$\ln 2=\frac{661}{960}+\frac{15}{4}\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}}{n(n+1)(n+2)(n+3)(n+4)(n+5)}.$

$\ln 2=\frac{2}{3}(1+\frac{2}{4^3-4}+\frac{2}{8^3-8}+\frac{2}{12^3-12}+\dots ).$

Binary rising constant factorial

$\ln 2=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2^{n}n}.$

$\ln 2=1-\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2^{n}n(n+1)}.$

$\ln 2=\frac{1}{2}+2\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2^{n}n(n+1)(n+2)}.$

$\ln 2=\frac{5}{6}-6\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2^{n}n(n+1)(n+2)(n+3)}.$

$\ln 2=\frac{7}{12}+24\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2^{n}n(n+1)(n+2)(n+3)(n+4)}.$

$\ln 2=\frac{47}{60}-120\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2^{n}n(n+1)(n+2)(n+3)(n+4)(n+5)}.$

Other series representations

$\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1)(2n+2)}=\ln 2.$

$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(4n^2-1)}=2\ln 2-1.$

$\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n}{n(4n^2-1)}=\ln 2-1.$

$\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n}{n(9n^2-1)}=2\ln 2-\frac{3}{2}.$

$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{4n^2-2n}=\ln 2.$

$\sum _{n=1}^{\mathrm{\infty }}\frac{2(-1{)}^{n+1}(2n-1)+1}{8n^2-4n}=\ln 2.$

$\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{3n+1}=\frac{\ln 2}{3}+\frac{\pi }{3\sqrt{3}}.$

$\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{3n+2}=-\frac{\ln 2}{3}+\frac{\pi }{3\sqrt{3}}.$

$\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(3n+1)(3n+2)}=\frac{2\ln 2}{3}.$

$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\sum _{k=1}^n{k}^2}=18-24\ln 2$ using $\underset{N\to \mathrm{\infty }}{lim}\sum _{n=N}^{2N}\frac{1}{n}=\ln 2$

$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{4n^2-3n}=\ln 2+\frac{\pi }{6}$ (sums of the reciprocals of decagonal numbers)

Involving the Riemann Zeta function

$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}[\zeta (2n)-1]=\ln 2.$

$\sum _{n=2}^{\mathrm{\infty }}\frac{1}{2^n}[\zeta (n)-1]=\ln 2-\frac{1}{2}.$

$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2n+1}[\zeta (2n+1)-1]=1-\gamma -\frac{\ln 2}{2}.$

$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2^{2n-1}(2n+1)}\zeta (2n)=1-\ln 2.$

(Template:Math is the Euler–Mascheroni constant and Template:Math Riemann's zeta function.)

BBP-type representations

$\ln 2=\frac{2}{3}+\frac{1}{2}\sum _{k=1}^{\mathrm{\infty }}(\frac{1}{2k}+\frac{1}{4k+1}+\frac{1}{8k+4}+\frac{1}{16k+12})\frac{1}{16^k}.$

(See more about Bailey–Borwein–Plouffe (BBP)-type representations.)

Applying the three general series for natural logarithm to 2 directly gives:

$\ln 2=\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}}{n}.$

$\ln 2=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2^{n}n}.$

$\ln 2=\frac{2}{3}\sum _{k=0}^{\mathrm{\infty }}\frac{1}{9^k(2k+1)}.$

Applying them to $2=\frac{3}{2}\cdot \frac{4}{3}$ gives:

$\ln 2=\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}}{2^{n}n}+\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}}{3^{n}n}.$

$\ln 2=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{3^{n}n}+\sum _{n=1}^{\mathrm{\infty }}\frac{1}{4^{n}n}.$

$\ln 2=\frac{2}{5}\sum _{k=0}^{\mathrm{\infty }}\frac{1}{25^k(2k+1)}+\frac{2}{7}\sum _{k=0}^{\mathrm{\infty }}\frac{1}{49^k(2k+1)}.$

Applying them to $2=(\sqrt{2}{)}^2$ gives:

$\ln 2=2\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}}{(\sqrt{2}+1{)}^{n}n}.$

$\ln 2=2\sum _{n=1}^{\mathrm{\infty }}\frac{1}{(2+\sqrt{2}{)}^{n}n}.$

$\ln 2=\frac{4}{3+2\sqrt{2}}\sum _{k=0}^{\mathrm{\infty }}\frac{1}{(17+12\sqrt{2}{)}^k(2k+1)}.$

Applying them to $2={\left(\frac{16}{15}\right)}^7\cdot {\left(\frac{81}{80}\right)}^3\cdot {\left(\frac{25}{24}\right)}^5$ gives:

$\ln 2=7\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}}{15^{n}n}+3\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}}{80^{n}n}+5\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}}{24^{n}n}.$

$\ln 2=7\sum _{n=1}^{\mathrm{\infty }}\frac{1}{16^{n}n}+3\sum _{n=1}^{\mathrm{\infty }}\frac{1}{81^{n}n}+5\sum _{n=1}^{\mathrm{\infty }}\frac{1}{25^{n}n}.$

$\ln 2=\frac{14}{31}\sum _{k=0}^{\mathrm{\infty }}\frac{1}{961^k(2k+1)}+\frac{6}{161}\sum _{k=0}^{\mathrm{\infty }}\frac{1}{25921^k(2k+1)}+\frac{10}{49}\sum _{k=0}^{\mathrm{\infty }}\frac{1}{2401^k(2k+1)}.$

Representation as integrals

The natural logarithm of 2 occurs frequently as the result of integration. Some explicit formulas for it include:

${\int }_0^1\frac{dx}{1+x}={\int }_1^2\frac{dx}{x}=\ln 2$

${\int }_0^{\mathrm{\infty }}{e}^{-x}\frac{1-e^{-x}}{x}dx=\ln 2$

${\int }_0^{\mathrm{\infty }}{2}^{-x}dx=\frac{1}{\ln 2}$

${\int }_0^{\frac{\pi }{3}}\tan xdx=2{\int }_0^{\frac{\pi }{4}}\tan xdx=\ln 2$

$-\frac{1}{\pi i}{\int }_0^{\mathrm{\infty }}\frac{\ln x\ln \ln x}{(x+1{)}^2}dx=\ln 2$

Other representations

The Pierce expansion is Template:OEIS2C

$\ln 2=1-\frac{1}{1\cdot 3}+\frac{1}{1\cdot 3\cdot 12}-\cdots .$

The Engel expansion is Template:OEIS2C

$\ln 2=\frac{1}{2}+\frac{1}{2\cdot 3}+\frac{1}{2\cdot 3\cdot 7}+\frac{1}{2\cdot 3\cdot 7\cdot 9}+\cdots .$

The cotangent expansion is Template:OEIS2C

$\ln 2=\cot (\mathrm{arccot}(0)-\mathrm{arccot}(1)+\mathrm{arccot}(5)-\mathrm{arccot}(55)+\mathrm{arccot}(14187)-\cdots ).$

The simple continued fraction expansion is Template:OEIS2C

$\ln 2=[0;1,2,3,1,6,3,1,1,2,1,1,1,1,3,10,1,1,1,2,1,1,1,1,3,2,3,1,\mathrm{...}]$,

which yields rational approximations, the first few of which are 0, 1, 2/3, 7/10, 9/13 and 61/88.

This generalized continued fraction:

$\ln 2=[0;1,2,3,1,5,\frac{2}{3},7,\frac{1}{2},9,\frac{2}{5},\mathrm{...},2k-1,\frac{2}{k},\mathrm{...}]$,

also expressible as

$\ln 2=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{2}{2+\frac{2}{5+\frac{3}{2+\frac{3}{7+\frac{4}{2+\ddots }}}}}}}}=\frac{2}{3-\frac{1^2}{9-\frac{2^2}{15-\frac{3^2}{21-\ddots }}}}$

The following continued fraction representation (J.L.Lagrange) gives (asymptotically) 1.53 new correct decimal places per cycle:

$\ln 2=\frac{\frac{1}{2}}{L_0},L_k=2k+1+\frac{k+1}{2+\frac{k+1}{L_{k+1}}}$

or

$\ln 2=\frac{2}{G_0},G_k=6k+3-\frac{(k+1{)}^2}{G_{k+1}}$

Bootstrapping other logarithms

Given a value of Template:Math, a scheme of computing the logarithms of other integers is to tabulate the logarithms of the prime numbers and in the next layer the logarithms of the composite numbers Template:Math based on their factorizations

$c=2^i{3}^j{5}^k{7}^l\cdots \to \ln (c)=i\ln (2)+j\ln (3)+k\ln (5)+l\ln (7)+\cdots$

This employs

Prime	Approximate natural logarithm	OEIS
2	Template:Val	Template:OEIS link
3	Template:Val	Template:OEIS link
5	Template:Val	Template:OEIS link
7	Template:Val	Template:OEIS link
11	Template:Val	Template:OEIS link
13	Template:Val	Template:OEIS link
17	Template:Val	Template:OEIS link
19	Template:Val	Template:OEIS link
23	Template:Val	Template:OEIS link
29	Template:Val	Template:OEIS link
31	Template:Val	Template:OEIS link
37	Template:Val	Template:OEIS link
41	Template:Val	Template:OEIS link
43	Template:Val	Template:OEIS link
47	Template:Val	Template:OEIS link
53	Template:Val	Template:OEIS link
59	Template:Val	Template:OEIS link
61	Template:Val	Template:OEIS link
67	Template:Val	Template:OEIS link
71	Template:Val	Template:OEIS link
73	Template:Val	Template:OEIS link
79	Template:Val	Template:OEIS link
83	Template:Val	Template:OEIS link
89	Template:Val	Template:OEIS link
97	Template:Val	Template:OEIS link

In a third layer, the logarithms of rational numbers Template:Math are computed with Template:Math, and logarithms of roots via Template:Math.

The logarithm of 2 is useful in the sense that the powers of 2 are rather densely distributed; finding powers Template:Math close to powers Template:Math of other numbers Template:Math is comparatively easy, and series representations of Template:Math are found by coupling 2 to Template:Math with logarithmic conversions.

Example

If Template:Math with some small Template:Math, then Template:Math and therefore

$s\ln p-t\ln q=\ln (1+\frac{d}{q^t})=\sum _{m=1}^{\mathrm{\infty }}\frac{(-1{)}^{m+1}}{m}{\left(\frac{d}{q^t}\right)}^m=\sum _{n=0}^{\mathrm{\infty }}\frac{2}{2n+1}{\left(\frac{d}{2q^t+d}\right)}^{2n+1}.$

Selecting Template:Math represents Template:Math by Template:Math and a series of a parameter Template:Math that one wishes to keep small for quick convergence. Taking Template:Math, for example, generates

$2\ln 3=3\ln 2-\sum _{k\ge 1}\frac{(-1{)}^k}{8^{k}k}=3\ln 2+\sum _{n=0}^{\mathrm{\infty }}\frac{2}{2n+1}{\left(\frac{1}{2\cdot 8+1}\right)}^{2n+1}.$

This is actually the third line in the following table of expansions of this type:

Template:Math	Template:Math	Template:Math	Template:Math	Template:Math
1	3	1	2	Template:Sfrac = Template:0Template:Val…
1	3	2	2	−Template:Sfrac = −Template:Val…
2	3	3	2	Template:Sfrac = Template:0Template:Val…
5	3	8	2	−Template:Sfrac = −Template:Val…
12	3	19	2	Template:Sfrac = Template:0Template:Val…
1	5	2	2	Template:Sfrac = Template:0Template:Val…
3	5	7	2	−Template:Sfrac = −Template:Val…
1	7	2	2	Template:Sfrac = Template:0Template:Val…
1	7	3	2	−Template:Sfrac = −Template:Val…
5	7	14	2	Template:Sfrac = Template:0Template:Val…
1	11	3	2	Template:Sfrac = Template:0Template:Val…
2	11	7	2	−Template:Sfrac = −Template:Val…
11	11	38	2	Template:Sfrac = Template:0Template:Val…
1	13	3	2	Template:Sfrac = Template:0Template:Val…
1	13	4	2	−Template:Sfrac = −Template:Val…
3	13	11	2	Template:Sfrac = Template:0Template:Val…
7	13	26	2	−Template:Sfrac = −Template:Val…
10	13	37	2	Template:Sfrac = Template:0Template:Val…
1	17	4	2	Template:Sfrac = Template:0Template:Val…
1	19	4	2	Template:Sfrac = Template:0Template:Val…
4	19	17	2	−Template:Sfrac = −Template:Val…
1	23	4	2	Template:Sfrac = Template:0Template:Val…
1	23	5	2	−Template:Sfrac = −Template:Val…
2	23	9	2	Template:Sfrac = Template:0Template:Val…
1	29	4	2	Template:Sfrac = Template:0Template:Val…
1	29	5	2	−Template:Sfrac = −Template:Val…
7	29	34	2	Template:Sfrac = Template:0Template:Val…
1	31	5	2	−Template:Sfrac = −Template:Val…
1	37	5	2	Template:Sfrac = Template:0Template:Val…
4	37	21	2	−Template:Sfrac = −Template:Val…
5	37	26	2	Template:Sfrac = Template:0Template:Val…
1	41	5	2	Template:Sfrac = Template:0Template:Val…
2	41	11	2	−Template:Sfrac = −Template:Val…
3	41	16	2	Template:Sfrac = Template:0Template:Val…
1	43	5	2	Template:Sfrac = Template:0Template:Val…
2	43	11	2	−Template:Sfrac = −Template:Val…
5	43	27	2	Template:Sfrac = Template:0Template:Val…
7	43	38	2	−Template:Sfrac = −Template:Val…

Starting from the natural logarithm of Template:Math one might use these parameters:

Template:Math	Template:Math	Template:Math	Template:Math	Template:Math
10	2	3	10	Template:Sfrac = Template:0Template:Val…
21	3	10	10	Template:Sfrac = Template:0Template:Val…
3	5	2	10	Template:Sfrac = Template:0Template:Val…
10	5	7	10	−Template:Sfrac = −Template:Val…
6	7	5	10	Template:Sfrac = Template:0Template:Val…
13	7	11	10	−Template:Sfrac = −Template:Val…
1	11	1	10	Template:Sfrac = Template:0Template:Val…
1	13	1	10	Template:Sfrac = Template:0Template:Val…
8	13	9	10	−Template:Sfrac = −Template:Val…
9	13	10	10	Template:Sfrac = Template:0Template:Val…
1	17	1	10	Template:Sfrac = Template:0Template:Val…
4	17	5	10	−Template:Sfrac = −Template:Val…
9	17	11	10	Template:Sfrac = Template:0Template:Val…
3	19	4	10	−Template:Sfrac = −Template:Val…
4	19	5	10	Template:Sfrac = Template:0Template:Val…
7	19	9	10	−Template:Sfrac = −Template:Val…
2	23	3	10	−Template:Sfrac = −Template:Val…
3	23	4	10	Template:Sfrac = Template:0Template:Val…
2	29	3	10	−Template:Sfrac = −Template:Val…
2	31	3	10	−Template:Sfrac = −Template:Val…

Known digits

This is a table of recent records in calculating digits of Template:Math. As of December 2018, it has been calculated to more digits than any other natural logarithm of a natural number, except that of 1.

Date	Name	Number of digits
January 7, 2009	A.Yee & R.Chan	15,500,000,000
February 4, 2009	A.Yee & R.Chan	31,026,000,000
February 21, 2011	Alexander Yee	50,000,000,050
May 14, 2011	Shigeru Kondo	100,000,000,000
February 28, 2014	Shigeru Kondo	200,000,000,050
July 12, 2015	Ron Watkins	250,000,000,000
January 30, 2016	Ron Watkins	350,000,000,000
April 18, 2016	Ron Watkins	500,000,000,000
December 10, 2018	Michael Kwok	600,000,000,000
April 26, 2019	Jacob Riffee	1,000,000,000,000
August 19, 2020	Seungmin Kim	1,200,000,000,100
September 9, 2021	William Echols	1,500,000,000,000
February 12, 2024	Jordan Ranous	3,000,000,000,000
November 15, 2025	Mamdouh Barakat	3,100,000,000,000

See also

• Rule of 72#Continuous compounding, in which Template:Math figures prominently
• Half-life#Formulas for half-life in exponential decay, in which Template:Math figures prominently
• Erdős–Moser equation: all solutions must come from a convergent of Template:Math.

References

• Template:Cite journal
• Template:Cite journal
• Template:Cite journal
• Template:Cite journal
• Template:Cite journal
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External links

• Template:MathWorld
• Template:Cite web