# Natural logarithm of 2

(Redirected from Ln(2))

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

${\displaystyle \ln(2)\approx 0.693\,147\,180\,559\,945\,309\,417\,232\,121\,458.}$

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

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

The common logarithm in particular is ()

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

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

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

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 [1]

number digit frequency
0 59999962241
1 59999935189
2 59999701650
3 59999883740
4 59999893049
5 59999774875
6 60000238666
7 60000373799
8 60000148375
9 60000088416

## Series representations

### Rising alternate factorial

${\displaystyle \ln(2)=\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}=1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{5}}-{\frac {1}{6}}+\cdots .}$ This is the well-known "alternating harmonic series".
${\displaystyle \ln(2)={\tfrac {1}{2}}+{\tfrac {1}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)}}.}$
${\displaystyle \ln(2)={\tfrac {5}{8}}+{\tfrac {1}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)}}.}$
${\displaystyle \ln(2)={\tfrac {2}{3}}+{\tfrac {3}{4}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)}}.}$
${\displaystyle \ln(2)={\tfrac {131}{192}}+{\tfrac {3}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)(n+4)}}.}$
${\displaystyle \ln(2)={\tfrac {661}{960}}+{\tfrac {15}{4}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)(n+4)(n+5)}}.}$

### Binary rising constant factorial

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

### Other series representations

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

### Involving the Riemann Zeta function

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

(γ is the Euler–Mascheroni constant and ζ Riemann's zeta function.)

### BBP-type representations

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

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

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

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

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

${\displaystyle \ln(2)=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{2^{n}n}}+\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{3^{n}n}}.}$
${\displaystyle \ln(2)=\sum _{n=1}^{\infty }{\frac {1}{3^{n}n}}+\sum _{n=1}^{\infty }{\frac {1}{4^{n}n}}.}$
${\displaystyle \ln(2)={\frac {2}{5}}\sum _{k=0}^{\infty }{\frac {1}{25^{k}(2k+1)}}+{\frac {2}{7}}\sum _{k=0}^{\infty }{\frac {1}{49^{k}(2k+1)}}.}$

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

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

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

${\displaystyle \ln(2)=7\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{15^{n}n}}+3\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{80^{n}n}}+5\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{24^{n}n}}.}$
${\displaystyle \ln(2)=7\sum _{n=1}^{\infty }{\frac {1}{16^{n}n}}+3\sum _{n=1}^{\infty }{\frac {1}{81^{n}n}}+5\sum _{n=1}^{\infty }{\frac {1}{25^{n}n}}.}$
${\displaystyle \ln(2)={\frac {14}{31}}\sum _{k=0}^{\infty }{\frac {1}{961^{k}(2k+1)}}\,+\,{\frac {6}{161}}\sum _{k=0}^{\infty }{\frac {1}{25921^{k}(2k+1)}}\,+\,{\frac {10}{49}}\sum _{k=0}^{\infty }{\frac {1}{2401^{k}(2k+1)}}.}$

## Representation as integrals

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

(γ is the Euler–Mascheroni constant.)

## Other representations

The Pierce expansion is

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

The Engel expansion is

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

The cotangent expansion is

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

The simple continued fraction expansion is

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

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

${\displaystyle \ln(2)=\left[0;1,2,3,1,5,{\tfrac {2}{3}},7,{\tfrac {1}{2}},9,{\tfrac {2}{5}},...,2k-1,{\frac {2}{k}},...\right]}$,[2]
also expressible as
${\displaystyle \ln(2)={\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{3+{\cfrac {2}{2+{\cfrac {2}{5+{\cfrac {3}{2+{\cfrac {3}{7+{\cfrac {4}{2+\ddots }}}}}}}}}}}}}}}}={\cfrac {2}{3-{\cfrac {1^{2}}{9-{\cfrac {2^{2}}{15-{\cfrac {3^{2}}{21-\ddots }}}}}}}}}$

## 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

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

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 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

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

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

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

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 of q = 10 one 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 of ${\displaystyle \ln(2)}$. As of December 2018, it has been calculated to more digits than any other natural logarithm. [3] [4]

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