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 ≈ 0.693 147 180 559 945 309 417 232 121 458. {\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
log b 2 = ln 2 ln b . {\displaystyle \log _{b}2={\frac {\ln 2}{\ln b}}.}
The common logarithm in particular is (OEIS: A007524)
log 10 2 ≈ 0.301 029 995 663 981 195. {\displaystyle \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 = 1 log 10 2 ≈ 3.321 928 095 {\displaystyle \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 = ∑ n = 1 ∞ ( − 1 ) n + 1 n = 1 − 1 2 + 1 3 − 1 4 + 1 5 − 1 6 + ⋯ . {\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".
ln 2 = 1 2 + 1 2 ∑ n = 1 ∞ ( − 1 ) n + 1 n ( n + 1 ) . {\displaystyle \ln 2={\frac {1}{2}}+{\frac {1}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)}}.}
ln 2 = 5 8 + 1 2 ∑ n = 1 ∞ ( − 1 ) n + 1 n ( n + 1 ) ( n + 2 ) . {\displaystyle \ln 2={\frac {5}{8}}+{\frac {1}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)}}.}
ln 2 = 2 3 + 3 4 ∑ n = 1 ∞ ( − 1 ) n + 1 n ( n + 1 ) ( n + 2 ) ( n + 3 ) . {\displaystyle \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 = 131 192 + 3 2 ∑ n = 1 ∞ ( − 1 ) n + 1 n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) . {\displaystyle \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 = 661 960 + 15 4 ∑ n = 1 ∞ ( − 1 ) n + 1 n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) ( n + 5 ) . {\displaystyle \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 = 2 3 ( 1 + 2 4 3 − 4 + 2 8 3 − 8 + 2 12 3 − 12 + … ) . {\displaystyle \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 = ∑ n = 1 ∞ 1 2 n n . {\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {1}{2^{n}n}}.}
ln 2 = 1 − ∑ n = 1 ∞ 1 2 n n ( n + 1 ) . {\displaystyle \ln 2=1-\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)}}.}
ln 2 = 1 2 + 2 ∑ n = 1 ∞ 1 2 n n ( n + 1 ) ( n + 2 ) . {\displaystyle \ln 2={\frac {1}{2}}+2\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)}}.}
ln 2 = 5 6 − 6 ∑ n = 1 ∞ 1 2 n n ( n + 1 ) ( n + 2 ) ( n + 3 ) . {\displaystyle \ln 2={\frac {5}{6}}-6\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)(n+3)}}.}
ln 2 = 7 12 + 24 ∑ n = 1 ∞ 1 2 n n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) . {\displaystyle \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 = 47 60 − 120 ∑ n = 1 ∞ 1 2 n n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) ( n + 5 ) . {\displaystyle \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
∑ n = 0 ∞ 1 ( 2 n + 1 ) ( 2 n + 2 ) = ln 2. {\displaystyle \sum _{n=0}^{\infty }{\frac {1}{(2n+1)(2n+2)}}=\ln 2.}
∑ n = 1 ∞ 1 n ( 4 n 2 − 1 ) = 2 ln 2 − 1. {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n(4n^{2}-1)}}=2\ln 2-1.}
∑ n = 1 ∞ ( − 1 ) n n ( 4 n 2 − 1 ) = ln 2 − 1. {\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n(4n^{2}-1)}}=\ln 2-1.}
∑ n = 1 ∞ ( − 1 ) n n ( 9 n 2 − 1 ) = 2 ln 2 − 3 2 . {\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n(9n^{2}-1)}}=2\ln 2-{\frac {3}{2}}.}
∑ n = 1 ∞ 1 4 n 2 − 2 n = ln 2. {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{4n^{2}-2n}}=\ln 2.}
∑ n = 1 ∞ 2 ( − 1 ) n + 1 ( 2 n − 1 ) + 1 8 n 2 − 4 n = ln 2. {\displaystyle \sum _{n=1}^{\infty }{\frac {2(-1)^{n+1}(2n-1)+1}{8n^{2}-4n}}=\ln 2.}
∑ n = 0 ∞ ( − 1 ) n 3 n + 1 = ln 2 3 + π 3 3 . {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3n+1}}={\frac {\ln 2}{3}}+{\frac {\pi }{3{\sqrt {3}}}}.}
∑ n = 0 ∞ ( − 1 ) n 3 n + 2 = − ln 2 3 + π 3 3 . {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3n+2}}=-{\frac {\ln 2}{3}}+{\frac {\pi }{3{\sqrt {3}}}}.}
∑ n = 0 ∞ ( − 1 ) n ( 3 n + 1 ) ( 3 n + 2 ) = 2 ln 2 3 . {\displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(3n+1)(3n+2)}}={\frac {2\ln 2}{3}}.}
∑ n = 1 ∞ 1 ∑ k = 1 n k 2 = 18 − 24 ln 2 {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{\sum _{k=1}^{n}k^{2}}}=18-24\ln 2} using lim N → ∞ ∑ n = N 2 N 1 n = ln 2 {\displaystyle \lim _{N\rightarrow \infty }\sum _{n=N}^{2N}{\frac {1}{n}}=\ln 2}
∑ n = 1 ∞ 1 4 n 2 − 3 n = ln 2 + π 6 {\displaystyle \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
∑ n = 1 ∞ 1 n [ ζ ( 2 n ) − 1 ] = ln 2. {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n}}[\zeta (2n)-1]=\ln 2.}
∑ n = 2 ∞ 1 2 n [ ζ ( n ) − 1 ] = ln 2 − 1 2 . {\displaystyle \sum _{n=2}^{\infty }{\frac {1}{2^{n}}}[\zeta (n)-1]=\ln 2-{\frac {1}{2}}.}
∑ n = 1 ∞ 1 2 n + 1 [ ζ ( 2 n + 1 ) − 1 ] = 1 − γ − ln 2 2 . {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{2n+1}}[\zeta (2n+1)-1]=1-\gamma -{\frac {\ln 2}{2}}.}
∑ n = 1 ∞ 1 2 2 n − 1 ( 2 n + 1 ) ζ ( 2 n ) = 1 − ln 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
ln 2 = 2 3 + 1 2 ∑ k = 1 ∞ ( 1 2 k + 1 4 k + 1 + 1 8 k + 4 + 1 16 k + 12 ) 1 16 k . {\displaystyle \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 = ∑ n = 1 ∞ ( − 1 ) n − 1 n . {\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n}}.}
ln 2 = ∑ n = 1 ∞ 1 2 n n . {\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {1}{2^{n}n}}.}
ln 2 = 2 3 ∑ k = 0 ∞ 1 9 k ( 2 k + 1 ) . {\displaystyle \ln 2={\frac {2}{3}}\sum _{k=0}^{\infty }{\frac {1}{9^{k}(2k+1)}}.}
Applying them to 2 = 3 2 ⋅ 4 3 {\displaystyle \textstyle 2={\frac {3}{2}}\cdot {\frac {4}{3}}} gives:
ln 2 = ∑ n = 1 ∞ ( − 1 ) n − 1 2 n n + ∑ n = 1 ∞ ( − 1 ) n − 1 3 n n . {\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}}.}
ln 2 = ∑ n = 1 ∞ 1 3 n n + ∑ n = 1 ∞ 1 4 n n . {\displaystyle \ln 2=\sum _{n=1}^{\infty }{\frac {1}{3^{n}n}}+\sum _{n=1}^{\infty }{\frac {1}{4^{n}n}}.}
ln 2 = 2 5 ∑ k = 0 ∞ 1 25 k ( 2 k + 1 ) + 2 7 ∑ k = 0 ∞ 1 49 k ( 2 k + 1 ) . {\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 2 = ( 2 ) 2 {\displaystyle \textstyle 2=({\sqrt {2}})^{2}} gives:
ln 2 = 2 ∑ n = 1 ∞ ( − 1 ) n − 1 ( 2 + 1 ) n n . {\displaystyle \ln 2=2\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{({\sqrt {2}}+1)^{n}n}}.}
ln 2 = 2 ∑ n = 1 ∞ 1 ( 2 + 2 ) n n . {\displaystyle \ln 2=2\sum _{n=1}^{\infty }{\frac {1}{(2+{\sqrt {2}})^{n}n}}.}
ln 2 = 4 3 + 2 2 ∑ k = 0 ∞ 1 ( 17 + 12 2 ) k ( 2 k + 1 ) . {\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 2 = ( 16 15 ) 7 ⋅ ( 81 80 ) 3 ⋅ ( 25 24 ) 5 {\displaystyle \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 ∑ n = 1 ∞ ( − 1 ) n − 1 15 n n + 3 ∑ n = 1 ∞ ( − 1 ) n − 1 80 n n + 5 ∑ n = 1 ∞ ( − 1 ) n − 1 24 n n . {\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}}.}
ln 2 = 7 ∑ n = 1 ∞ 1 16 n n + 3 ∑ n = 1 ∞ 1 81 n n + 5 ∑ n = 1 ∞ 1 25 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}}.}
ln 2 = 14 31 ∑ k = 0