# 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 (sequence A002162 in the OEIS) truncated at 30 decimal places is given by:
    $\ln 2\approx 0.693\,147\,180\,559\,945\,309\,417\,232\,121\,458.$
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 (OEIS: A007524)
    $\log _{10}2\approx 0.301\,029\,995\,663\,981\,195.$
The inverse of this number is the binary logarithm of 10:
    $\log _{2}10={\frac {1}{\log _{10}2}}\approx 3.321\,928\,095$ (OEIS: A020862).
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

    $\ln 2=\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}=1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{5}}-{\frac {1}{6}}+\cdots .$ This is the well-known "alternating harmonic series".
    $\ln 2={\frac {1}{2}}+{\frac {1}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)}}.$
    $\ln 2={\frac {5}{8}}+{\frac {1}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)}}.$
    $\ln 2={\frac {2}{3}}+{\frac {3}{4}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)}}.$
    $\ln 2={\frac {131}{192}}+{\frac {3}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)(n+4)}}.$
    $\ln 2={\frac {661}{960}}+{\frac {15}{4}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)(n+4)(n+5)}}.$
    $\ln 2={\frac {2}{3}}\left(1+{\frac {2}{4^{3}-4}}+{\frac {2}{8^{3}-8}}+{\frac {2}{12^{3}-12}}+\dots \right).$

### 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 of decagonal numbers)

### Involving the Riemann Zeta function

    $\sum _{n=1}^{\infty }{\frac {1}{n}}[\zeta (2n)-1]=\ln 2.$
    $\sum _{n=2}^{\infty }{\frac {1}{2^{n}}}[\zeta (n)-1]=\ln 2-{\frac {1}{2}}.$
    $\sum _{n=1}^{\infty }{\frac {1}{2n+1}}[\zeta (2n+1)-1]=1-\gamma -{\frac {\ln 2}{2}}.$
    $\sum _{n=1}^{\infty }{\frac {1}{2^{2n-1}(2n+1)}}\zeta (2n)=1-\ln 2.$
(γ is the Euler–Mascheroni constant and ζ Riemann'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 about Bailey–Borwein–Plouffe (BBP)-type representations.)
Applying the three general series for natural logarithm to 2 directly gives:
    $\ln 2=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n}}.$
    $\ln 2=\sum _{n=1}^{\infty }{\frac {1}{2^{n}n}}.$
    $\ln 2={\frac {2}{3}}\sum _{k=0}^{\infty }{\frac {1}{9^{k}(2k+1)}}.$
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 is OEIS: A091846
    $\ln 2=1-{\frac {1}{1\cdot 3}}+{\frac {1}{1\cdot 3\cdot 12}}-\cdots .$
The Engel expansion is OEIS: 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 is OEIS: A081785
    $\ln 2=\cot({\operatorname {arccot}(0)-\operatorname {arccot}(1)+\operatorname {arccot}(5)-\operatorname {arccot}(55)+\operatorname {arccot}(14187)-\cdots }).$
The simple continued fraction expansion is OEIS: 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.
This generalized 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 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
| 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 $^n$√c = ⁠1/n⁠ ln(c).
The logarithm of 2 is useful in the sense that the powers of 2 are rather densely distributed; finding powers 2$^i$ close to powers b$^j$ 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 p$^s$ = q$^t$ + d with some small d, then ⁠p$^s$/q$^t$⁠ = 1 + ⁠d/q$^t$⁠ and therefore
    $s\ln p-t\ln q=\ln \left(1+{\frac {d}{q^{t}}}\right)=\sum _{m=1}^{\infty }{\frac {(-1)^{m+1}}{m}}\left({\frac {d}{q^{t}}}\right)^{m}=\sum _{n=0}^{\infty }{\frac {2}{2n+1}}{\left({\frac {d}{2q^{t}+d}}\right)}^{2n+1}.$
Selecting q = 2 represents ln p by ln 2 and a series of a parameter ⁠d/q$^t$⁠ that one wishes to keep small for quick convergence. Taking 3$^2$ = 2$^3$ + 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/q$^t$⁠ |
| 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/q$^t$⁠ |
| 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 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.
| 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 |