# Natural logarithm of 2

The decimal value of the natural logarithm of 2 (sequence A002162 in the OEIS) 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 the formula

$\log _{b}(2)={\frac {\ln(2)}{\ln(b)}}.$ The common logarithm in particular is ()

$\log _{10}(2)\approx 0.301\,029\,995\,663\,981\,195.$ The inverse of this number is the binary logarithm of 10:

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

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.

Although not proven, empirical evidence suggests that ln(2) is a normal number. The decimal counts in the first 600 billion digits are as follows 

numberdigit frequency
059999962241
159999935189
259999701650
359999883740
459999893049
559999774875
660000238666
760000373799
860000148375
960000088416

## 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".
$\ln(2)={\tfrac {1}{2}}+{\tfrac {1}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)}}.$ $\ln(2)={\tfrac {5}{8}}+{\tfrac {1}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)}}.$ $\ln(2)={\tfrac {2}{3}}+{\tfrac {3}{4}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)}}.$ $\ln(2)={\tfrac {131}{192}}+{\tfrac {3}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)(n+4)}}.$ $\ln(2)={\tfrac {661}{960}}+{\tfrac {15}{4}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)(n+4)(n+5)}}.$ ### Binary rising constant factorial

$\ln(2)=\sum _{n=1}^{\infty }{\frac {1}{2^{n}n}}.$ $\ln(2)=1-\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)}}.$ $\ln(2)={\tfrac {1}{2}}+2\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)}}.$ $\ln(2)={\tfrac {5}{6}}-6\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)(n+3)}}.$ $\ln(2)={\tfrac {7}{12}}+24\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)(n+3)(n+4)}}.$ $\ln(2)={\tfrac {47}{60}}-120\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)(n+3)(n+4)(n+5)}}.$ ### Other series representations

$\sum _{n=0}^{\infty }{\frac {1}{(2n+1)(2n+2)}}=\ln(2).$ $\sum _{n=1}^{\infty }{\frac {1}{n(4n^{2}-1)}}=2\ln(2)-1.$ $\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n(4n^{2}-1)}}=\ln(2)-1.$ $\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n(9n^{2}-1)}}=2\ln(2)-{\tfrac {3}{2}}.$ $\sum _{n=1}^{\infty }{\frac {1}{4n^{2}-2n}}=\ln(2).$ $\sum _{n=1}^{\infty }{\frac {2(-1)^{n+1}(2n-1)+1}{8n^{2}-4n}}=\ln(2).$ $\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3n+1}}={\frac {\ln(2)}{3}}+{\frac {\pi }{3{\sqrt {3}}}}.$ $\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3n+2}}=-{\frac {\ln(2)}{3}}+{\frac {\pi }{3{\sqrt {3}}}}.$ $\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(3n+1)(3n+2)}}={\frac {2\ln(2)}{3}}.$ $\sum _{n=1}^{\infty }{\frac {1}{\sum _{k=1}^{n}k^{2}}}=18-24\ln(2)$ using $\lim _{N\rightarrow \infty }\sum _{n=N}^{2N}{\frac {1}{n}}=\ln(2)$ $\sum _{n=1}^{\infty }{\frac {1}{4k^{2}-3k}}=\ln(2)+{\frac {\pi }{6}}$ (sums of the reciprocals of decagonal numbers)

### Involving the Riemann Zeta function

$\sum _{n=2}^{\infty }{\frac {1}{2^{n}}}[\zeta (n)-1]=\ln(2)-{\tfrac {1}{2}}.$ $\sum _{n=2}^{\infty }{\frac {1}{2n+1}}[\zeta (n)-1]=1-\gamma -{\frac {\ln(2)}{2}}.$ $\sum _{n=1}^{\infty }{\frac {1}{2^{2n-1}(2n+1)}}\zeta (2n)=1-\ln(2).$ (γ is the Euler–Mascheroni constant and ζ Riemann's zeta function.)

### BBP-type representations

$\ln(2)={\tfrac {2}{3}}+{\tfrac {1}{2}}\sum _{k=1}^{\infty }\left({\frac {1}{2k}}+{\frac {1}{4k+1}}+{\frac {1}{8k+4}}+{\frac {1}{16k+12}}\right){\frac {1}{16^{k}}}.$ (See more about Bailey–Borwein–Plouffe (BBP)-type representations.)

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

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

$\int _{0}^{1}{\frac {dx}{1+x}}=\ln(2),{\text{ or, equivalently, }}\int _{1}^{2}{\frac {dx}{x}}=\ln(2).$ $\int _{1}^{\infty }{\frac {dx}{(1+x^{2})(1+x)^{2}}}={\frac {1-\ln(2)}{4}}.$ $\int _{0}^{\infty }{\frac {dx}{1+e^{nx}}}={\frac {\ln(2)}{n}};\int _{0}^{\infty }{\frac {dx}{3+e^{nx}}}={\frac {2\ln(2)}{3n}}.$ $\int _{0}^{\infty }\left({\frac {1}{e^{x}-1}}-{\frac {2}{e^{2x}-1}}\right)\,dx=\ln(2).$ $\int _{0}^{\infty }e^{-x}{\frac {1-e^{-x}}{x}}\,dx=\ln(2).$ $\int _{0}^{1}\ln \left({\frac {x^{2}-1}{x\ln(x)}}\right)\,dx=-1+\ln(2)+\gamma .$ $\int _{0}^{\frac {\pi }{3}}\tan(x)\,dx=2\int _{0}^{\frac {\pi }{4}}\tan(x)\,dx=\ln(2).$ $\int _{-{\frac {\pi }{4}}}^{\frac {\pi }{4}}\ln \left(\sin(x)+\cos(x)\right)\,dx=-{\frac {\pi \ln(2)}{4}}.$ $\int _{0}^{1}x^{2}\ln(1+x)\,dx={\frac {2\ln(2)}{3}}-{\frac {5}{18}}.$ $\int _{0}^{1}x\ln(1+x)\ln(1-x)\,dx={\tfrac {1}{4}}-\ln(2).$ $\int _{0}^{1}x^{3}\ln(1+x)\ln(1-x)\,dx={\tfrac {13}{96}}-{\frac {2\ln(2)}{3}}.$ $\int _{0}^{1}{\frac {\ln x}{(1+x)^{2}}}\,dx=-\ln(2).$ $\int _{0}^{1}{\frac {\ln(1+x)-x}{x^{2}}}\,dx=1-2\ln(2).$ $\int _{0}^{1}{\frac {dx}{x(1-\ln(x))(1-2\ln(x))}}=\ln(2).$ $\int _{1}^{\infty }{\frac {\ln \left(\ln(x)\right)}{x^{3}}}\,dx=-{\frac {\gamma +\ln(2)}{2}}.$ (γ is the Euler–Mascheroni constant.)

## Other representations

The Pierce expansion is

$\ln(2)=1-{\frac {1}{1\cdot 3}}+{\frac {1}{1\cdot 3\cdot 12}}-\cdots .$ The Engel expansion is

$\ln(2)={\frac {1}{2}}+{\frac {1}{2\cdot 3}}+{\frac {1}{2\cdot 3\cdot 7}}+{\frac {1}{2\cdot 3\cdot 7\cdot 9}}+\cdots .$ The cotangent expansion is

$\ln(2)=\cot({\operatorname {arccot} (0)-\operatorname {arccot} (1)+\operatorname {arccot} (5)-\operatorname {arccot} (55)+\operatorname {arccot} (14187)-\cdots }).$ The simple continued fraction expansion is

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

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

$\ln(2)=\left[0;1,2,3,1,5,{\tfrac {2}{3}},7,{\tfrac {1}{2}},9,{\tfrac {2}{5}},...,2k-1,{\frac {2}{k}},...\right]$ ,
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 of ln(2), 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 c based on their factorizations

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

primeapproximate natural logarithmOEIS
20.693147180559945309417232121458A002162
31.09861228866810969139524523692A002391
51.60943791243410037460075933323A016628
71.94591014905531330510535274344A016630
112.39789527279837054406194357797A016634
132.56494935746153673605348744157A016636
172.83321334405621608024953461787A016640
192.94443897916644046000902743189A016642
233.13549421592914969080675283181A016646
293.36729582998647402718327203236A016652
313.43398720448514624592916432454A016654
373.61091791264422444436809567103A016660
413.71357206670430780386676337304A016664
433.76120011569356242347284251335A016666
473.85014760171005858682095066977A016670
533.97029191355212183414446913903A016676
594.07753744390571945061605037372A016682
614.11087386417331124875138910343A016684
674.20469261939096605967007199636A016690
714.26267987704131542132945453251A016694
734.29045944114839112909210885744A016696
794.36944785246702149417294554148A016702
834.41884060779659792347547222329A016706
894.48863636973213983831781554067A016712
974.57471097850338282211672162170A016720

In a third layer, the logarithms of rational numbers r = a/b are computed with ln(r) = ln(a) − ln(b), and logarithms of roots via ln nc = 1/n ln(c).

The logarithm of 2 is useful in the sense that the powers of 2 are rather densely distributed; finding powers 2i close to powers bj of other numbers b is comparatively easy, and series representations of ln(b) are found by coupling 2 to b with logarithmic conversions.

### Example

If ps = qt + d with some small d, then ps/qt = 1 + d/qt and therefore

$s\ln(p)-t\ln(q)=\ln \left(1+{\frac {d}{q^{t}}}\right)=\sum _{m=1}^{\infty }(-1)^{m+1}{\frac {({\frac {d}{q^{t}}})^{m}}{m}}=\sum _{n=0}^{\infty }{\frac {2}{2n+1}}{\left({\frac {d}{2q^{t}+d}}\right)}^{2n+1}.$ Selecting q = 2 represents ln(p) by ln(2) and a series of a parameter d/qt that one wishes to keep small for quick convergence. Taking 32 = 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:

sptqd/qt
13121/2 = 0.50000000
13221/4 = −0.25000000
23321/8 = 0.12500000
538213/256 = −0.05078125
1231927153/524288 = 0.01364326
15221/4 = 0.25000000
35723/128 = −0.02343750
17223/4 = 0.75000000
17321/8 = −0.12500000
57142423/16384 = 0.02581787
111323/8 = 0.37500000
211727/128 = −0.05468750
111138210433763667/274877906944 = 0.03795781
113325/8 = 0.62500000
113423/16 = −0.18750000
313112149/2048 = 0.07275391
7132624360347/67108864 = −0.06497423
1013372419538377/137438953472 = 0.00305254
117421/16 = 0.06250000
119423/16 = 0.18750000
419172751/131072 = −0.00572968
123427/16 = 0.43750000
123529/32 = −0.28125000
2239217/512 = 0.03320312
1294213/16 = 0.81250000
129523/32 = −0.09375000
72934270007125/17179869184 = 0.00407495
131521/32 = −0.03125000
137525/32 = 0.15625000
437212222991/2097152 = −0.10633039
5372622235093/67108864 = 0.03330548
141529/32 = 0.28125000
241112367/2048 = −0.17919922
3411623385/65536 = 0.05165100
1435211/32 = 0.34375000
243112199/2048 = −0.09716797
54327212790715/134217728 = 0.09529825
7433823059295837/274877906944 = −0.01112965

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

sptqd/qt
1023103/125 = 0.02400000
2131010460353203/10000000000 = 0.04603532
352101/4 = 0.25000000
1057103/128 = −0.02343750
6751017649/100000 = 0.17649000
13711103110989593/100000000000 = −0.03110990
1111101/10 = 0.10000000
1131103/10 = 0.30000000
813910184269279/1000000000 = −0.18426928
9131010604499373/10000000000 = 0.06044994
1171107/10 = 0.70000000
41751016479/100000 = −0.16479000
917111018587876497/100000000000 = 0.18587876
3194103141/10000 = −0.31410000
41951030321/100000 = 0.30321000
719910106128261/1000000000 = −0.10612826
223310471/1000 = −0.47100000
3234102167/10000 = 0.21670000
229310159/1000 = −0.15900000
23131039/1000 = −0.03900000

## Known digits

This is a table of recent records in calculating digits of $\ln(2)$ . As of December 2018, it has been calculated to more digits than any other natural logarithm   of a natural number, except that of 1.

January 7, 2009A.Yee & R.Chan15,500,000,000
February 4, 2009A.Yee & R.Chan31,026,000,000
February 21, 2011Alexander Yee50,000,000,050
May 14, 2011Shigeru Kondo100,000,000,000
February 28, 2014Shigeru Kondo200,000,000,050
July 12, 2015Ron Watkins250,000,000,000
January 30, 2016Ron Watkins350,000,000,000
April 18, 2016Ron Watkins500,000,000,000
December 10, 2018Michael Kwok600,000,000,000
April 26, 2019 Jacob Riffee 1,000,000,000,000