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%I M4074 N1689 #231 Apr 10 2024 09:59:21
%S 6,9,3,1,4,7,1,8,0,5,5,9,9,4,5,3,0,9,4,1,7,2,3,2,1,2,1,4,5,8,1,7,6,5,
%T 6,8,0,7,5,5,0,0,1,3,4,3,6,0,2,5,5,2,5,4,1,2,0,6,8,0,0,0,9,4,9,3,3,9,
%U 3,6,2,1,9,6,9,6,9,4,7,1,5,6,0,5,8,6,3,3,2,6,9,9,6,4,1,8,6,8,7
%N Decimal expansion of the natural logarithm of 2.
%C Newton calculated the first 16 terms of this sequence.
%C Area bounded by y = tan x, y = cot x, y = 0. - _Clark Kimberling_, Jun 26 2020
%D G. Boros and V. H. Moll, Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals, Cambridge University Press, 2004.
%D S. R. Finch, Mathematical Constants, Cambridge, 2003, Sections 1.3.3 and 6.2.
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H Harry J. Smith, <a href="/A002162/b002162.txt">Table of n, a(n) for n = 0..20000</a>
%H D. H. Bailey and J. M. Borwein, <a href="http://www.ams.org/notices/200505/fea-borwein.pdf">Experimental Mathematics: Examples, Methods and Implications</a>, Notices of the AMS, May 2005, Volume 52, Issue 5.
%H Peter Bala, <a href="/A002117/a002117.pdf">New series for old functions</a>
%H J. M. Borwein, P. B. Borwein and K. Dilcher, <a href="http://www.jstor.org/stable/2324715">Pi, Euler numbers and asymptotic expansions</a>, Amer. Math. Monthly, 96 (1989), 681-687.
%H Paul Cooijmans, <a href="http://web.archive.org/web/20050302174449/http://members.chello.nl/p.cooijmans/gliaweb/tests/odds.html">Odds</a>.
%H X. Gourdon and P. Sebah, <a href="http://numbers.computation.free.fr/Constants/Log2/log2.html">The logarithm constant:log(2)</a>
%H M. Kontsevich and D. Zagier, <a href="http://www.ihes.fr/~maxim/TEXTS/Periods.pdf">Periods</a>, pp. 4-5.
%H Mathematical Reflections, <a href="https://www.awesomemath.org/wp-pdf-files/math-reflections/mr-2016-04/mr_3_2016_solutions.pdf">Solution to Problem U376</a>, Issue 4, 2016, p 17.
%H Isaac Newton, <a href="http://www.archive.org/details/methodoffluxions00newt">The method of fluxions and infinite series; with its application to the geometry of curve-lines</a>, 1736; see p. 96.
%H Michael Penn, <a href="https://www.youtube.com/watch?v=DVn7DL0lxBo">an alternating floor sum.</a>, YouTube video, 2020.
%H Michael Penn, <a href="https://www.youtube.com/watch?v=aQBdKtOgkeg">A wonderful limit from the 2011 Virginia Tech Regional Math Competition</a>, YouTube video, 2022.
%H Simon Plouffe, <a href="https://web.archive.org/web/20150911212957/http://www.worldwideschool.org/library/books/sci/math/MiscellaneousMathematicalConstants/chap58.html">log(2), natural logarithm of 2 to 2000 places</a>
%H S. Ramanujan, <a href="http://www.imsc.res.in/~rao/ramanujan/CamUnivCpapers/question/q260.htm">Question 260</a>, J. Ind. Math. Soc., III, p. 43.
%H Albert Stadler, <a href="https://cms.math.ca/wp-content/uploads/crux-pdfs/CRUXv36n6.pdf">Problem 3567</a>, Crux Mathematicorum, Vol. 36 (Oct. 2010), p. 396; Oliver Geupel, <a href="https://smc.math.ca/wp-content/uploads/crux-pdfs/CRUXv37n6.pdf">Solution</a>, Crux Mathematicorum, Vol. 37 (Oct. 2011), pp. 400-401.
%H D. W. Sweeney, <a href="http://dx.doi.org/10.1090/S0025-5718-1963-0160308-X">On the computation of Euler's constant</a>, Math. Comp., 17 (1963), 170-178.
%H Horace S. Uhler, <a href="https://doi.org/10.1073/pnas.26.3.205">Recalculation and extension of the modulus and of the logarithms of 2, 3, 5, 7 and 17</a>, Proc. Nat. Acad. Sci. U. S. A. 26, (1940). 205-212.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/NaturalLogarithmof2.html">Natural Logarithm of 2</a>, <a href="http://mathworld.wolfram.com/Masser-GramainConstant.html">Masser-Gramain Constant</a>, <a href="http://mathworld.wolfram.com/LogarithmicIntegral.html">Logarithmic Integral</a>, <a href="https://mathworld.wolfram.com/DirichletEtaFunction.html">Dirichlet Eta Function</a>.
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Natural_logarithm_of_2">Natural logarithm of 2</a>.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Period_(algebraic_geometry)">Period (algebraic geometry)</a>.
%H <a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>
%F log(2) = Sum_{k>=1} 1/(k*2^k) = Sum_{j>=1} (-1)^(j+1)/j.
%F log(2) = Integral_{t=0..1} dt/(1+t).
%F log(2) = (2/3) * (1 + Sum_{k>=1} 2/((4*k)^3-4*k)) (Ramanujan).
%F log(2) = 4*Sum_{k>=0} (3-2*sqrt(2))^(2*k+1)/(2*k+1) (Y. Luke). - _R. J. Mathar_, Jul 13 2006
%F log(2) = 1 - (1/2)*Sum_{k>=1} 1/(k*(2*k+1)). - _Jaume Oliver Lafont_, Jan 06 2009, Jan 08 2009
%F log(2) = 4*Sum_{k>=0} 1/((4*k+1)*(4*k+2)*(4*k+3)). - _Jaume Oliver Lafont_, Jan 08 2009
%F log(2) = 7/12 + 24*Sum_{k>=1} 1/(A052787(k+4)*A000079(k)). - _R. J. Mathar_, Jan 23 2009
%F From _Alexander R. Povolotsky_, Jul 04 2009: (Start)
%F log(2) = (1/4)*(3 - Sum_{n>=1} 1/(n*(n+1)*(2*n+1))).
%F log(2) = (230166911/9240 - Sum_{k>=1} (1/2)^k*(11/k + 10/(k+1) + 9/(k+2) + 8/(k+3) + 7/(k+4) + 6/(k+5) - 6/(k+7) - 7/(k+8) - 8/(k+9) - 9/(k+10) - 10/(k+11)))/35917. (End)
%F log(2) = A052882/A000670. - _Mats Granvik_, Aug 10 2009
%F From log(1-x-x^2) at x=1/2, log(2) = (1/2)*Sum_{k>=1} L(k)/(k*2^k), where L(n) is the n-th Lucas number (A000032). - _Jaume Oliver Lafont_, Oct 24 2009
%F log(2) = Sum_{k>=1} 1/(cos(k*Pi/3)*k*2^k) (cf. A176900). - _Jaume Oliver Lafont_, Apr 29 2010
%F log(2) = (Sum_{n>=1} 1/(n^2*(n+1)^2*(2*n+1)) + 11)/16. - _Alexander R. Povolotsky_, Jan 13 2011
%F log(2) = (Sum_{n>=1} (2*n+1)/(Sum_{k=1..n} k^2)^2))+396)/576. - _Alexander R. Povolotsky_, Jan 14 2011
%F From _Alexander R. Povolotsky_, Dec 16 2008: (Start)
%F log(2) = 105*(Sum_{n>=1} 1/(2*n*(2*n+1)*(2*n+3)*(2*n+5)*(2*n+7)))-319/44100).
%F log(2) = 319/420 - (3/2)*Sum_{n>=1} 1/(6*n^2+39*n+63)). (End)
%F log(2) = Sum_{k>=1} A191907(2,k)/k. - _Mats Granvik_, Jun 19 2011
%F log(2) = Integral_{x=0..oo} 1/(1 + e^x) dx. - _Jean-François Alcover_, Mar 21 2013
%F log(2) = lim_{s->1} zeta(s)*(1-1/2^(s-1)). - _Mats Granvik_, Jun 18 2013
%F From _Peter Bala_, Dec 10 2013: (Start)
%F log(2) = 2*Sum_{n>=1} 1/( n*A008288(n-1,n-1)*A008288(n,n) ), a result due to Burnside.
%F log(2) = (1/3)*Sum_{n >= 0} (5*n+4)/( (3*n+1)*(3*n+2)*C(3*n,n) )*(1/2)^n = (1/12)*Sum_{n >= 0} (28*n+17)/( (3*n+1)*(3*n+2)*C(3*n,n) )*(-1/4)^n.
%F log(2) = (3/16)*Sum_{n >= 0} (14*n+11)/( (4*n+1)*(4*n+3)*C(4*n,2*n) )*(1/4)^n = (1/12)*Sum_{n >= 0} (34*n+25)/( (4*n+1)*(4*n+3)*C(4*n,2*n) )*(-1/18)^n. For more series of this type see the Bala link.
%F See A142979 for series acceleration formulas for log(2) obtained from the Mercator series log(2) = Sum_{n >= 1} (-1)^(n+1)/n. See A142992 for series for log(2) related to the root lattice C_n. (End)
%F log(2) = lim_{n->oo} Sum_{k=2^n..2^(n+1)-1} 1/k. - _Richard R. Forberg_, Aug 16 2014
%F From _Peter Bala_, Feb 03: (Start)
%F log(2) = (2/3)*Sum_{k >= 0} 1/((2*k + 1)*9^k).
%F Define a pair of integer sequences A(n) = 9^n*(2*n + 1)!/n! and B(n) = A(n)*Sum_{k = 0..n} 1/((2*k + 1)*9^k). Both satisfy the same second-order recurrence equation u(n) = (40*n + 16)*u(n-1) - 36*(2*n - 1)^2*u(n-2). From this observation we obtain the continued fraction expansion log(2) = (2/3)*(1 + 2/(54 - 36*3^2/(96 - 36*5^2/(136 - ... - 36*(2*n - 1)^2/((40*n + 16) - ... ))))). Cf. A002391, A073000 and A105531 for similar expansions. (End)
%F log(2) = Sum_{n>=1} (Zeta(2*n)-1)/n. - _Vaclav Kotesovec_, Dec 11 2015
%F From _Peter Bala_, Oct 30 2016: (Start)
%F Asymptotic expansions:
%F for N even, log(2) - Sum_{k = 1..N/2} (-1)^(k-1)/k ~ (-1)^(N/2)*(1/N - 1/N^2 + 2/N^4 - 16/N^6 + 272/N^8 - ...), where the sequence of unsigned coefficients [1, 1, 2, 16, 272, ...] is A000182 with an extra initial term of 1. See Borwein et al., Theorem 1 (b);
%F for N odd, log(2) - Sum_{k = 1..(N-1)/2} (-1)^(k-1)/k ~ (-1)^((N-1)/2)*(1/N - 1/N^3 + 5/N^5 - 61/N^7 + 1385/N^9 - ...), by Borwein et al., Lemma 2 with f(x) := 1/(x + 1/2), h := 1/2 and then set x = (N - 1)/2, where the sequence of unsigned coefficients [1, 1, 5, 61, 1385, ...] is A000364. (End)
%F log(2) = lim_{n->oo} Sum_{k=1..n} sin(1/(n+k)). See Mathematical Reflections link. - _Michel Marcus_, Jan 07 2017
%F log(2) = Sum_{n>=1} (A006519(n) / ( (1+2^A006519(n)) * A000265(n) * (1 + A000265(n))). - _Nicolas Nagel_, Mar 19 2018
%F From _Amiram Eldar_, Jul 02 2020: (Start)
%F Equals Sum_{k>=2} zeta(k)/2^k.
%F Equals -Sum_{k>=2} log(1 - 1/k^2).
%F Equals Sum_{k>=1} 1/A002939(k).
%F Equals Integral_{x=0..Pi/3} tan(x) dx. (End)
%F log(2) = Integral_{x=0..Pi/2} (sec(x) - tan(x)) dx. - _Clark Kimberling_, Jul 08 2020
%F From _Peter Bala_, Nov 14 2020: (Start)
%F log(2) = Integral_{x = 0..1} (x - 1)/log(x) dx (Boros and Moll, p. 97).
%F log(2) = (1/2)*Integral_{x = 0..1} (x + 2)*(x - 1)^2/log(x)^2 dx.
%F log(2) = (1/4)*Integral_{x = 0..1} (x^2 + 3*x + 4)*(x - 1)^3/log(x)^3 dx. (End)
%F log(2) = 2*arcsinh(sqrt(2)/4) = 2*sqrt(2)*Sum_{n >= 0} (-1)^n*C(2*n,n)/ ((8*n+4)*32^n) = 3*Sum_{n >= 0} (-1)^n/((8*n+4)*(2^n)*C(2*n,n)). - _Peter Bala_, Jan 14 2022
%F log(2) = Integral_{x=0..oo} ( e^(-x) * (1-e^(-2x)) * (1-e^(-4x)) * (1-e^(-6x)) ) / ( x * (1-e^(-14x)) ) dx (see Crux Mathematicorum link). - _Bernard Schott_, Jul 11 2022
%F From _Peter Bala_, Oct 22 2023: (Start)
%F log(2) = 23/32 + 2!^3/16 * Sum_{n >= 1} (-1)^n * (n + 1)/(n*(n + 1)*(n + 2))^2 = 707/1024 - 4!^3/(16^2 * 2!^2) * Sum_{n >= 1} (-1)^n * (n + 2)/(n*(n + 1)*(n + 2)*(n + 3)*(n + 4))^2 = 42611/61440 + 6!^3/(16^3 * 3!^2) * Sum_{n >= 1} (-1)^n * (n + 3)/(n*(n + 1)*(n + 2)*(n + 3)*(n + 4)*(n + 5)*(n + 6))^2.
%F More generally, it appears that for k >= 0, log(2) = c(k) + (2*k)!^3/(16^k * k!^2) * Sum_{n >= 1} (-1)^(n+k+1) * (n + k)/(n*(n + 1)*...*(n + 2*k))^2 , where c(k) is a rational approximation to log(2). The first few values of c(k) are [0, 23/32, 707/1024, 42611/61440, 38154331/55050240, 76317139/110100480, 26863086823/38755368960, ...].
%F Let P(n,k) = n*(n + 1)*...*(n + k).
%F Conjecture: for k >= 0 and r even with r - 1 <= k, the series Sum_{n >= 1} (-1)^n * (d/dn)^r (P(n,k)) / (P(n,k)^2 = A(r,k)*log(2) + B(r,k), where A(r,k) and B(r,k) are both rational numbers. (End)
%F From _Peter Bala_, Nov 13 2023: (Start)
%F log(2) = 5/8 + (1/8)*Sum_{k >= 1} (-1)^(k+1) * (2*k + 1)^2 / ( k*(k + 1) )^4
%F = 257/384 + (3!^5/2^9)*Sum_{k >= 1} (-1)^(k+1) * (2*k + 1)*(2*k + 3)^2*(2*k + 5) / ( k*(k + 1)*(k + 2)*(k + 3) )^4
%F = 267515/393216 + (5!^5/2^19)*Sum_{k >= 1} (-1)^(k+1) * (2*k + 1)*(2*k + 3)*(2*k + 5)^2*(2*k + 7)*(2*k + 9) / ( k*(k + 1)*(k + 2)*(k + 3)*(k + 4)*(k + 5) )^4
%F log(2) = 3/4 - 1/128 * Sum_{k >= 0} (-1/16)^k * (10*k + 12)*binomial(2*k+2,k+1)/ ((k + 1)*(2*k + 3)). The terms of the series are O(1/(k^(3/2)*4^n)). (End)
%F log(2) = eta(1) is a period, where eta(x) is the Dirichlet eta function. - _Andrea Pinos_, Mar 19 2024
%e 0.693147180559945309417232121458176568075500134360255254120680009493393...
%t RealDigits[N[Log[2],200]][[1]] (* _Vladimir Joseph Stephan Orlovsky_, Feb 21 2011 *)
%t RealDigits[Log[2],10,120][[1]] (* _Harvey P. Dale_, Jan 25 2024 *)
%o (PARI) { default(realprecision, 20080); x=10*log(2); for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b002162.txt", n, " ", d)); } \\ _Harry J. Smith_, Apr 21 2009
%Y Cf. A016730 (continued fraction), A002939, A008288, A142979, A142992.
%K nonn,cons
%O 0,1
%A _N. J. A. Sloane_
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