Natural logarithm of 2

The decimal value of the natural logarithm of 2 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.

Rising alternate factorial

 * $$\ln 2 = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}=1-\frac12+\frac13-\frac14+\frac15-\frac16+\cdots.$$ 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 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
 * $$ \ln 2 = 1 -\frac{1}{1\cdot 3}+\frac{1}{1\cdot 3\cdot 12} -\cdots. $$

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

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.

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
 * $$c=2^i3^j5^k7^l\cdots\rightarrow \ln(c)=i\ln(2)+j\ln(3)+k\ln(5)+l\ln(7)+\cdots$$

This employs

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 $n} = 1⁄n ln(c)$.

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

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

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

Starting from the natural logarithm of $s$ one might use these parameters:

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