{\displaystyle \ln 2\approx 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} .
^{\infty }{\frac {(-1)^{n+1}}{n}}=1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+{\frac {1}{5}}-{\frac {1}{6}}+\cdots .}
ln ⁡ 2 − 1 ^{\infty }{\frac {(-1)^{n}}{n(4n^{2}-1)}}=\ln 2-1} .
ln ⁡ 21 2 + 1 2 ∑ n1 ∞ ( − 1 ) n + 1 n ( n + 1 ) .
ln ⁡ 21 − ∑ n1 ∞ 1 2 n n ( n + 1 ) ^{\infty }{\frac {1}{2^{n}n(n+1)}}} .
{2}}+{\frac {1}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)}}.} ln ⁡ 25 8 + 1 2 ∑ n1 ∞ ( − 1 ) n + 1 n ( n + 1 ) ( n + 2 ) .
ln ⁡ 21 2 + 1 2 ∑ n1 ∞ ( − 1 ) n + 1 n ( n + 1 ) {2}}+{\frac {1}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)}}} .
{8}}+{\frac {1}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)}}.} ln ⁡ 22 3 + 3 4 ∑ n1 ∞ ( − 1 ) n + 1 n ( n + 1 ) ( n + 2 ) ( n + 3 ) .
ln ⁡ 22 3 + 3 4 ∑ n1 ∞ ( − 1 ) n + 1 n ( n + 1 ) ( n + 2 ) ( n + 3 ) {3}}+{\frac {3}{4}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)}}} .
{3}}+{\frac {3}{4}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)}}.}
2 ln ⁡ 2 − 3 2 ^{\infty }{\frac {(-1)^{n}}{n(9n^{2}-1)}}=2\ln 2-{\frac {3}{2}}} .
{192}}+{\frac {3}{2}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)(n+4)}}.} ln ⁡ 2661 960 + 15 4 ∑ n1 ∞ ( − 1 ) n + 1 n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) ( n + 5 ) .
ln ⁡ 2661 960 + 15 4 ∑ n1 ∞ ( − 1 ) n + 1 n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) ( n + 5 ) {960}}+{\frac {15}{4}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)(n+4)(n+5)}}} .
{960}}+{\frac {15}{4}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)(n+4)(n+5)}}.} ln ⁡ 22 3 ( 1 + 2 4 3 − 4 + 2 8 3 − 8 + 2 12 3 − 12 + … ) .
ln ⁡ 22 3 + 3 4 ∑ n1 ∞ ( − 1 ) n + 1 n ( n + 1 ) ( n + 2 ) ( n + 3 ) {3}}+{\frac {3}{4}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n(n+1)(n+2)(n+3)}}} .
ln ⁡ 21 − ∑ n1 ∞ 1 2 n n ( n + 1 ) .
ln ⁡ 21 − ∑ n1 ∞ 1 2 n n ( n + 1 ) ^{\infty }{\frac {1}{2^{n}n(n+1)}}} .
^{\infty }{\frac {1}{2^{n}n(n+1)}}.}
ln ⁡ 2 − 1 ^{\infty }{\frac {(-1)^{n}}{n(4n^{2}-1)}}=\ln 2-1} .
ln ⁡ 21 2 + 2 ∑ n1 ∞ 1 2 n n ( n + 1 ) ( n + 2 ) .
ln ⁡ 21 − ∑ n1 ∞ 1 2 n n ( n + 1 ) ^{\infty }{\frac {1}{2^{n}n(n+1)}}} .
{2}}+2\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)}}.} ln ⁡ 25 6 − 6 ∑ n1 ∞ 1 2 n n ( n + 1 ) ( n + 2 ) ( n + 3 ) .
ln ⁡ 25 6 − 6 ∑ n1 ∞ 1 2 n n ( n + 1 ) ( n + 2 ) ( n + 3 ) {6}}-6\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)(n+3)}}}
{6}}-6\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)(n+3)}}.} ln ⁡ 27 12 + 24 ∑ n1 ∞ 1 2 n n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) .
ln ⁡ 25 6 − 6 ∑ n1 ∞ 1 2 n n ( n + 1 ) ( n + 2 ) ( n + 3 ) {6}}-6\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)(n+3)}}}
{12}}+24\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)(n+3)(n+4)}}.}
ln ⁡ 2 − 1 ^{\infty }{\frac {(-1)^{n}}{n(4n^{2}-1)}}=\ln 2-1} .
ln ⁡ 247 60 − 120 ∑ n1 ∞ 1 2 n n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) ( n + 5 ) .
ln ⁡ 247 60 − 120 ∑ n1 ∞ 1 2 n n ( n + 1 ) ( n + 2 ) ( n + 3 ) ( n + 4 ) ( n + 5 ) {60}}-120\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)(n+3)(n+4)(n+5)}}} .
{60}}-120\sum _{n=1}^{\infty }{\frac {1}{2^{n}n(n+1)(n+2)(n+3)(n+4)(n+5)}}.}
ln ⁡ 2 − 1 ^{\infty }{\frac {(-1)^{n}}{n(4n^{2}-1)}}=\ln 2-1} .
^{\infty }{\frac {(-1)^{n}}{n(9n^{2}-1)}}=2\ln 2-{\frac {3}{2}}.} ∑ n1 ∞ 1 4 n 2 − 2 nln ⁡ 2. ^{\infty }{\frac {1}{4n^{2}-2n}}=\ln 2.} ∑ n1 ∞ 2 ( − 1 ) n + 1 ( 2 n − 1 ) + 1 8 n 2 − 4 nln ⁡ 2. ^{\infty }{\frac {2(-1)^{n+1}(2n-1)+1}{8n^{2}-4n}}=\ln 2.} ∑ n0 ∞ ( − 1 ) n 3 n + 1ln ⁡ 2 3 + π 3 3 .
2 ln ⁡ 2 3 ^{\infty }{\frac {(-1)^{n}}{(3n+1)(3n+2)}}={\frac {2\ln 2}{3}}} . ∑ n1 ∞ 1 ∑ k1 n k 218 − 24 ln ⁡ 2 ^{\infty }{\frac {1}{\sum _{k=1}^{n}k^{2}}}=18-24\ln 2} menggunakan lim N → ∞ ∑ nN 2 N 1 nln ⁡ 2 \sum _{n=N}^{2N}{\frac {1}{n}}=\ln 2} .
^{\infty }{\frac {(-1)^{n}}{3n+1}}={\frac {\ln 2}{3}}+{\frac {\pi }{3{\sqrt {3}}}}.} ∑ n0 ∞ ( − 1 ) n 3 n + 2− ln ⁡ 2 3 + π 3 3 .
∑ n0 ∞ ( − 1 ) n 3 n + 2− ln ⁡ 2 3 + π 3 3 ^{\infty }{\frac {(-1)^{n}}{3n+2}}=-{\frac {\ln 2}{3}}+{\frac {\pi }{3{\sqrt {3}}}}} .
^{\infty }{\frac {(-1)^{n}}{3n+2}}=-{\frac {\ln 2}{3}}+{\frac {\pi }{3{\sqrt {3}}}}.} ∑ n0 ∞ ( − 1 ) n ( 3 n + 1 ) ( 3 n + 2 )
∑ n0 ∞ ( − 1 ) n 3 n + 2− ln ⁡ 2 3 + π 3 3 ^{\infty }{\frac {(-1)^{n}}{3n+2}}=-{\frac {\ln 2}{3}}+{\frac {\pi }{3{\sqrt {3}}}}} .
^{\infty }{\frac {(-1)^{n}}{(3n+1)(3n+2)}}={\frac {2\ln 2}{3}}.} ∑ n1 ∞ 1 ∑ k1 n k 218 − 24 ln ⁡ 2 ^{\infty }{\frac {1}{\sum _{k=1}^{n}k^{2}}}=18-24\ln 2} using lim N → ∞ ∑ nN 2 N 1 nln ⁡ 2 \sum _{n=N}^{2N}{\frac {1}{n}}=\ln 2} ∑ n1 ∞ 1 4 n 2 − 3 nln ⁡ 2 + π 6 ^{\infty }{\frac {1}{4n^{2}-3n}}=\ln 2+{\frac {\pi }{6}}} (sums of the reciprocals of decagonal numbers)
2 ln ⁡ 2 3 ^{\infty }{\frac {(-1)^{n}}{(3n+1)(3n+2)}}={\frac {2\ln 2}{3}}} . ∑ n1 ∞ 1 ∑ k1 n k 218 − 24 ln ⁡ 2 ^{\infty }{\frac {1}{\sum _{k=1}^{n}k^{2}}}=18-24\ln 2} menggunakan lim N → ∞ ∑ nN 2 N 1 nln ⁡ 2 \sum _{n=N}^{2N}{\frac {1}{n}}=\ln 2} .
^{\infty }{\frac {1}{2^{n}}}[\zeta (n)-1]=\ln 2-{\frac {1}{2}}.} ∑ n1 ∞ 1 2 n + 1 [ ζ ( 2 n + 1 )
ln ⁡ 2 − 1 2 ^{\infty }{\frac {1}{2^{n}}}[\zeta (n)-1]=\ln 2-{\frac {1}{2}}} .
^{\infty }{\frac {1}{2n+1}}[\zeta (2n+1)-1]=1-\gamma -{\frac {\ln 2}{2}}.} ∑ n1 ∞ 1 2 2 n − 1 ( 2 n + 1 ) ζ ( 2 n )
− 1 ]1 − γ − ln ⁡ 2 2 ^{\infty }{\frac {1}{2n+1}}[\zeta (n)-1]=1-\gamma -{\frac {\ln 2}{2}}} .
1 − ln ⁡ 2. ^{\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.)
− 1 ]1 − γ − ln ⁡ 2 2 ^{\infty }{\frac {1}{2n+1}}[\zeta (n)-1]=1-\gamma -{\frac {\ln 2}{2}}} .
ln ⁡ 22 3 + 1 2 ∑ k1 ∞ ( 1 2 k + 1 4 k + 1 + 1 8 k + 4 + 1 16 k + 12 ) 1 16 k . {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.)
ln ⁡ 22 3 + 1 2 ∑ k1 ∞ ( 1 2 k + 1 4 k + 1 + 1 8 k + 4 + 1 16 k + 12 ) 1 16 k {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}}}} (Lihat lebih banyak mengenai wakilan tipe Bailey−Borwein−Plouffe (BBP).)
Applying the three general series for natural logarithm to 2 directly gives: ln ⁡ 2∑ n1 ∞ ( − 1 ) n − 1 n .
Menerapkan ketiga deret umum untuk logaritma natural ke 2 secara langsung memberikan: ln ⁡ 2∑ n1 ∞ ( − 1 ) n − 1 n ^{\infty }{\frac {(-1)^{n-1}}{n}}} .
^{\infty }{\frac {(-1)^{n-1}}{n}}.}
ln ⁡ 2 − 1 ^{\infty }{\frac {(-1)^{n}}{n(4n^{2}-1)}}=\ln 2-1} .
{3}}\sum _{k=0}^{\infty }{\frac {1}{9^{k}(2k+1)}}.} Applying them to 23 2 ⋅ 4 3 {2}}\cdot {\frac {4}{3}}} gives: ln ⁡ 2∑ n1 ∞ ( − 1 ) n − 1 2 n n + ∑ n1 ∞ ( − 1 ) n − 1 3 n n .
2 ln ⁡ 2 3 ^{\infty }{\frac {(-1)^{n}}{(3n+1)(3n+2)}}={\frac {2\ln 2}{3}}} . ∑ n1 ∞ 1 ∑ k1 n k 218 − 24 ln ⁡ 2 ^{\infty }{\frac {1}{\sum _{k=1}^{n}k^{2}}}=18-24\ln 2} menggunakan lim N → ∞ ∑ nN 2 N 1 nln ⁡ 2 \sum _{n=N}^{2N}{\frac {1}{n}}=\ln 2} .
^{\infty }{\frac {(-1)^{n-1}}{2^{n}n}}+\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{3^{n}n}}.}
ln ⁡ 2 − 1 ^{\infty }{\frac {(-1)^{n}}{n(4n^{2}-1)}}=\ln 2-1} .
ln ⁡ 2∑ n1 ∞ 1 3 n n + ∑ n1 ∞ 1 4 n n .
∑ n1 ∞ 1 2 2 n − 1 ( 2 n + 1 )
^{\infty }{\frac {1}{3^{n}n}}+\sum _{n=1}^{\infty }{\frac {1}{4^{n}n}}.}
ln ⁡ 2 − 1 ^{\infty }{\frac {(-1)^{n}}{n(4n^{2}-1)}}=\ln 2-1} .
{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 })^{2}} gives: ln ⁡ 22 ∑ n1 ∞ ( − 1 ) n − 1 ( 2 + 1 ) n n .
2 ln ⁡ 2 3 ^{\infty }{\frac {(-1)^{n}}{(3n+1)(3n+2)}}={\frac {2\ln 2}{3}}} . ∑ n1 ∞ 1 ∑ k1 n k 218 − 24 ln ⁡ 2 ^{\infty }{\frac {1}{\sum _{k=1}^{n}k^{2}}}=18-24\ln 2} menggunakan lim N → ∞ ∑ nN 2 N 1 nln ⁡ 2 \sum _{n=N}^{2N}{\frac {1}{n}}=\ln 2} .
^{\infty }{\frac {(-1)^{n-1}}{({\sqrt {2}}+1)^{n}n}}.}
ln ⁡ 2 − 1 ^{\infty }{\frac {(-1)^{n}}{n(4n^{2}-1)}}=\ln 2-1} .
^{\infty }{\frac {1}{(2+{\sqrt {2}})^{n}n}}.}
2 ln ⁡ 2 − 1 ^{\infty }{\frac {1}{n(4n^{2}-1)}}=2\ln 2-1} .
{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 {15}}\right)}^{7}\cdot {\left({\frac {81}{80}}\right)}^{3}\cdot {\left({\frac {25}{24}}\right)}^{5}} gives: ln ⁡ 27 ∑ n1 ∞ ( − 1 ) n − 1 15 n n + 3 ∑ n1 ∞ ( − 1 ) n − 1 80 n n + 5 ∑ n1 ∞ ( − 1 ) n − 1 24 n n .
2 ln ⁡ 2 3 ^{\infty }{\frac {(-1)^{n}}{(3n+1)(3n+2)}}={\frac {2\ln 2}{3}}} . ∑ n1 ∞ 1 ∑ k1 n k 218 − 24 ln ⁡ 2 ^{\infty }{\frac {1}{\sum _{k=1}^{n}k^{2}}}=18-24\ln 2} menggunakan lim N → ∞ ∑ nN 2 N 1 nln ⁡ 2 \sum _{n=N}^{2N}{\frac {1}{n}}=\ln 2} .
{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)}}.} The natural logarithm of 2 occurs frequently as the result of integration.
2 ln ⁡ 2 3 ^{\infty }{\frac {(-1)^{n}}{(3n+1)(3n+2)}}={\frac {2\ln 2}{3}}} . ∑ n1 ∞ 1 ∑ k1 n k 218 − 24 ln ⁡ 2 ^{\infty }{\frac {1}{\sum _{k=1}^{n}k^{2}}}=18-24\ln 2} menggunakan lim N → ∞ ∑ nN 2 N 1 nln ⁡ 2 \sum _{n=N}^{2N}{\frac {1}{n}}=\ln 2} .
− arccot ⁡ ( 55 ) + arccot ⁡ ( 14187 ) − ⋯ ) .
− arccot ⁡ ( 1 ) + arccot ⁡ ( 5 )
This generalized continued fraction: ln ⁡ 2[ 0 ; 1 , 2 , 3 , 1 , 5 , 2 3 , 7 , 1 2 , 9 , 2 5 , .
ln ⁡ 2[ 0 ; 1 , 2 , 3 , 1 , 5 , 2 3 , 7 , 1 2 , 9 , 2 5 , .
. . ] {3}},7,{\tfrac {1}{2}},9,{\tfrac {2}{5}},...,2k-1,{\frac {2}{k}},...\right]} , also expressible as ln ⁡ 21 1 + 1 2 + 1 3 + 2 2 + 2 5 + 3 2 + 3 7 + 4 2 + ⋱2 3 − 1 2 9 − 2 2 15 − 3 2 21 − ⋱ {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 }}}}}}}}}
{3}},7,{\tfrac {1}{2}},9,{\tfrac {2}{5}},...,2k-1,{\frac {2}{k}},...\right]} , dapat diekspresikan sebagai ln ⁡ 21 1 + 1 2 + 1 3 + 2 2 + 2 5 + 3 2 + 3 7 + 4 2 + ⋱2 3 − 1 2 9 − 2 2 15 − 3 2 21 − ⋱ {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 }}}}}}}}} Diberikan sebuah nilai dari ln ⁡ 2 , sebuah skema menghitung logaritma dari bilangan bulat lainnya adalah untuk mentabulasi logaritma dari bilangan prima dan di lapisan berikutnya, logaritma dari bilangan komposit c berdasarkan faktorisasinya c2 i 3 j 5 k 7 l ⋯ → ln ⁡ ( c )
i ln ⁡ ( 2 ) + j ln ⁡ ( 3 ) + k ln ⁡ ( 5 ) + l ln ⁡ ( 7 ) + ⋯ 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 In a third layer, the logarithms of rational numbers r⁠a/b⁠ are computed with ln(r)ln(a)
i ln ⁡ ( 2 ) + j ln ⁡ ( 3 ) + k ln ⁡ ( 5 ) + l ln ⁡ ( 7 ) + ⋯ 3^{j}5^{k}7^{l}\cdots \rightarrow \ln(c)=i\ln(2)+j\ln(3)+k\ln(5)+l\ln(7)+\cdots } Ini memakai DI lapisan ketiga, logaritma bilangan rasional ra b {b}}} dihitung dengan menggunakan ln ⁡ ( r )ln ⁡ ( a ) − ln ⁡ ( b ) , dan logaritma akar melalui ln ⁡ c n1 n ln ⁡ c ]{c}}={\frac {1}{n}}\ln c} .
{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 q2 represents ln p by ln 2 and a series of a parameter ⁠d/qt⁠ that one wishes to keep small for quick convergence.
ln ⁡ ( 1 + d q t )∑ m1 ∞ ( − 1 ) m + 1 ( d q t ) m m∑ n0 ∞ 2 2 n + 1 ( d 2 q t + d ) 2 n + 1 {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}} Memilih q2 mewakili ln ⁡ ( p ) oleh ln ⁡ 2 dan sebuah deret dari sebuah parameter d q t {q^{t}}}} yang ingin tetap kecil untuk konvergen cepat.
As of December 2018, it has been calculated to more digits than any other natural logarithm of a natural number, except that of 1.
Mulai Desember 2018, ini telah dihitung lebih banyak digit dari setiap logaritma natural dari sebuah bilangan asli, kecuali 1.
Archived from the original (PDF) on 2011-06-06.
Diarsipkan dari asli (PDF) tanggal 2011-06-06.
Gourévitch, Boris; Jesús Goyanes Guillera (2007).
Gourévitch, Boris; Guillera Goyanes, Jesús (2007).