Natural logarithm of 2

The decimal value of thenatural logarithmof2(sequenceA002162in theOEIS) is approximately

$\ln 2\approx 0.693\,147\,180\,559\,945\,309\,417\,232\,121\,458.$

The logarithm of 2 in other bases is obtained with theformula

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

Thecommon logarithmin particular is (OEIS:A007524)

$\log _{10}2\approx 0.301\,029\,995\,663\,981\,195.$

The inverse of this number is thebinary logarithmof 10:

$\log _{2}10={\frac {1}{\log _{10}2}}\approx 3.321\,928\,095$(OEIS:A020862).

By theLindemann–Weierstrass theorem,the natural logarithm of anynatural numberother than 0 and 1 (more generally, of any positivealgebraic numberother than 1) is atranscendental number.

Series representations

Rising alternate factorial

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

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

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

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

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

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

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

Binary rising constant factorial

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

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

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

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

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

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

Other series representations

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

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

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

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

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

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

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

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

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

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

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

Involving the Riemann Zeta function

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

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

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

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

(γis theEuler–Mascheroni constantandζRiemann's zeta function.)

BBP-type representations

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Representation as integrals

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

$\int _{0}^{1}{\frac {dx}{1+x}}=\int _{1}^{2}{\frac {dx}{x}}=\ln 2$

$\int _{0}^{\infty }e^{-x}{\frac {1-e^{-x}}{x}}\,dx=\ln 2$

$\int _{0}^{\infty }2^{-x}dx={\frac {1}{\ln 2}}$

$\int _{0}^{\frac {\pi }{3}}\tan x\,dx=2\int _{0}^{\frac {\pi }{4}}\tan x\,dx=\ln 2$

$-{\frac {1}{\pi i}}\int _{0}^{\infty }{\frac {\ln x\ln \ln x}{(x+1)^{2}}}\,dx=\ln 2$

Other representations

The Pierce expansion isOEIS:A091846

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

TheEngel expansionisOEIS:A059180

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

The cotangent expansion isOEIS:A081785

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

The simplecontinued fractionexpansion isOEIS:A016730

$\ln 2=\left[0;1,2,3,1,6,3,1,1,2,1,1,1,1,3,10,1,1,1,2,1,1,1,1,3,2,3,1,...\right]$,

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

Thisgeneralized continued fraction:

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

also expressible as

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

Bootstrapping other logarithms

Given a value ofln 2,a scheme of computing the logarithms of otherintegersis to tabulate the logarithms of theprime numbersand in the next layer the logarithms of thecompositenumberscbased on theirfactorizations

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

This employs

prime	approximate natural logarithm	OEIS
2	0.693147180559945309417232121458	A002162
3	1.09861228866810969139524523692	A002391
5	1.60943791243410037460075933323	A016628
7	1.94591014905531330510535274344	A016630
11	2.39789527279837054406194357797	A016634
13	2.56494935746153673605348744157	A016636
17	2.83321334405621608024953461787	A016640
19	2.94443897916644046000902743189	A016642
23	3.13549421592914969080675283181	A016646
29	3.36729582998647402718327203236	A016652
31	3.43398720448514624592916432454	A016654
37	3.61091791264422444436809567103	A016660
41	3.71357206670430780386676337304	A016664
43	3.76120011569356242347284251335	A016666
47	3.85014760171005858682095066977	A016670
53	3.97029191355212183414446913903	A016676
59	4.07753744390571945061605037372	A016682
61	4.11087386417331124875138910343	A016684
67	4.20469261939096605967007199636	A016690
71	4.26267987704131542132945453251	A016694
73	4.29045944114839112909210885744	A016696
79	4.36944785246702149417294554148	A016702
83	4.41884060779659792347547222329	A016706
89	4.48863636973213983831781554067	A016712
97	4.57471097850338282211672162170	A016720

In a third layer, the logarithms of rational numbersr=⁠a/b⁠are computed withln(r) = ln(a) − ln(b),and logarithms of roots vialnⁿ√c=⁠1/n⁠ln(c).

The logarithm of 2 is useful in the sense that the powers of 2 are rather densely distributed; finding powers2ⁱclose to powersb^jof other numbersbis comparatively easy, and series representations ofln(b)are found by coupling 2 tobwithlogarithmic conversions.

Example

Ifp^s=q^t+dwith some smalld,then⁠p^s/q^t⁠= 1 +⁠d/q^t⁠and therefore

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

Selectingq= 2representslnpbyln 2and a series of a parameter⁠d/q^t⁠that one wishes to keep small for quick convergence. Taking3²= 2³+ 1,for example, generates

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

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

s	p	t	q	⁠d/qt⁠
1	3	1	2	⁠1/2⁠=−0.50000000…
1	3	2	2	−⁠1/4⁠= −0.25000000…
2	3	3	2	⁠1/8⁠=−0.12500000…
5	3	8	2	−⁠13/256⁠= −0.05078125…
12	3	19	2	⁠7153/524288⁠=−0.01364326…
1	5	2	2	⁠1/4⁠=−0.25000000…
3	5	7	2	−⁠3/128⁠= −0.02343750…
1	7	2	2	⁠3/4⁠=−0.75000000…
1	7	3	2	−⁠1/8⁠= −0.12500000…
5	7	14	2	⁠423/16384⁠=−0.02581787…
1	11	3	2	⁠3/8⁠=−0.37500000…
2	11	7	2	−⁠7/128⁠= −0.05468750…
11	11	38	2	⁠10433763667/274877906944⁠=−0.03795781…
1	13	3	2	⁠5/8⁠=−0.62500000…
1	13	4	2	−⁠3/16⁠= −0.18750000…
3	13	11	2	⁠149/2048⁠=−0.07275391…
7	13	26	2	−⁠4360347/67108864⁠= −0.06497423…
10	13	37	2	⁠419538377/137438953472⁠=−0.00305254…
1	17	4	2	⁠1/16⁠=−0.06250000…
1	19	4	2	⁠3/16⁠=−0.18750000…
4	19	17	2	−⁠751/131072⁠= −0.00572968…
1	23	4	2	⁠7/16⁠=−0.43750000…
1	23	5	2	−⁠9/32⁠= −0.28125000…
2	23	9	2	⁠17/512⁠=−0.03320312…
1	29	4	2	⁠13/16⁠=−0.81250000…
1	29	5	2	−⁠3/32⁠= −0.09375000…
7	29	34	2	⁠70007125/17179869184⁠=−0.00407495…
1	31	5	2	−⁠1/32⁠= −0.03125000…
1	37	5	2	⁠5/32⁠=−0.15625000…
4	37	21	2	−⁠222991/2097152⁠= −0.10633039…
5	37	26	2	⁠2235093/67108864⁠=−0.03330548…
1	41	5	2	⁠9/32⁠=−0.28125000…
2	41	11	2	−⁠367/2048⁠= −0.17919922…
3	41	16	2	⁠3385/65536⁠=−0.05165100…
1	43	5	2	⁠11/32⁠=−0.34375000…
2	43	11	2	−⁠199/2048⁠= −0.09716797…
5	43	27	2	⁠12790715/134217728⁠=−0.09529825…
7	43	38	2	−⁠3059295837/274877906944⁠= −0.01112965…

Starting from the natural logarithm ofq= 10one might use these parameters:

s	p	t	q	⁠d/qt⁠
10	2	3	10	⁠3/125⁠=−0.02400000…
21	3	10	10	⁠460353203/10000000000⁠=−0.04603532…
3	5	2	10	⁠1/4⁠=−0.25000000…
10	5	7	10	−⁠3/128⁠= −0.02343750…
6	7	5	10	⁠17649/100000⁠=−0.17649000…
13	7	11	10	−⁠3110989593/100000000000⁠= −0.03110990…
1	11	1	10	⁠1/10⁠=−0.10000000…
1	13	1	10	⁠3/10⁠=−0.30000000…
8	13	9	10	−⁠184269279/1000000000⁠= −0.18426928…
9	13	10	10	⁠604499373/10000000000⁠=−0.06044994…
1	17	1	10	⁠7/10⁠=−0.70000000…
4	17	5	10	−⁠16479/100000⁠= −0.16479000…
9	17	11	10	⁠18587876497/100000000000⁠=−0.18587876…
3	19	4	10	−⁠3141/10000⁠= −0.31410000…
4	19	5	10	⁠30321/100000⁠=−0.30321000…
7	19	9	10	−⁠106128261/1000000000⁠= −0.10612826…
2	23	3	10	−⁠471/1000⁠= −0.47100000…
3	23	4	10	⁠2167/10000⁠=−0.21670000…
2	29	3	10	−⁠159/1000⁠= −0.15900000…
2	31	3	10	−⁠39/1000⁠= −0.03900000…

Known digits

This is a table of recent records in calculating digits ofln 2.As of December 2018, it has been calculated to more digits than any other natural logarithmof 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

See also

• Rule of 72#Continuous compounding,in whichln 2figures prominently
• Half-life#Formulas for half-life in exponential decay,in whichln 2figures prominently
• Erdős–Moser equation:all solutions must come from aconvergentofln 2.

References

• Brent, Richard P. (1976)."Fast multiple-precision evaluation of elementary functions".J. ACM.23(2): 242–251.doi:10.1145/321941.321944.MR0395314.S2CID6761843.
• Uhler, Horace S. (1940)."Recalculation and extension of the modulus and of the logarithms of 2, 3, 5, 7 and 17".Proc. Natl. Acad. Sci. U.S.A.26(3): 205–212.Bibcode:1940PNAS...26..205U.doi:10.1073/pnas.26.3.205.MR0001523.PMC1078033.PMID16588339.
• Sweeney, Dura W. (1963)."On the computation of Euler's constant".Mathematics of Computation.17(82): 170–178.doi:10.1090/S0025-5718-1963-0160308-X.MR0160308.
• Chamberland, Marc (2003)."Binary BBP-formulae for logarithms and generalized Gaussian–Mersenne primes"(PDF).Journal of Integer Sequences.6:03.3.7.Bibcode:2003JIntS...6...37C.MR2046407.Archived fromthe original(PDF)on 2011-06-06.Retrieved2010-04-29.
• Gourévitch, Boris; Guillera Goyanes, Jesús (2007)."Construction of binomial sums forπand polylogarithmic constants inspired by BBP formulas "(PDF).Applied Math. E-Notes.7:237–246.MR2346048.
• Wu, Qiang (2003)."On the linear independence measure of logarithms of rational numbers".Mathematics of Computation.72(242): 901–911.doi:10.1090/S0025-5718-02-01442-4.

External links

• Weisstein, Eric W."Natural logarithm of 2".MathWorld.
• Gourdon, Xavier; Sebah, Pascal."The logarithm constant:log 2".