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A002162
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Decimal expansion of the natural logarithm of 2.
(Formerly M4074 N1689)
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197
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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, 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, 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
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refs;
listen;
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OFFSET
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0,1
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COMMENTS
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Newton calculated the first 16 terms of this sequence.
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REFERENCES
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G. Boros and V. H. Moll, Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals, Cambridge University Press, 2004.
S. R. Finch, Mathematical Constants, Cambridge, 2003, Sections 1.3.3 and 6.2.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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M. Kontsevich and D. Zagier, Periods, pp. 4-5.
Albert Stadler, Problem 3567, Crux Mathematicorum, Vol. 36 (Oct. 2010), p. 396; Oliver Geupel, Solution, Crux Mathematicorum, Vol. 37 (Oct. 2011), pp. 400-401.
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FORMULA
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log(2) = Sum_{k>=1} 1/(k*2^k) = Sum_{j>=1} (-1)^(j+1)/j.
log(2) = Integral_{t=0..1} dt/(1+t).
log(2) = (2/3) * (1 + Sum_{k>=1} 2/((4*k)^3-4*k)) (Ramanujan).
log(2) = 4*Sum_{k>=0} (3-2*sqrt(2))^(2*k+1)/(2*k+1) (Y. Luke). - R. J. Mathar, Jul 13 2006
log(2) = (1/4)*(3 - Sum_{n>=1} 1/(n*(n+1)*(2*n+1))).
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)
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
log(2) = 105*(Sum_{n>=1} 1/(2*n*(2*n+1)*(2*n+3)*(2*n+5)*(2*n+7)))-319/44100).
log(2) = 319/420 - (3/2)*Sum_{n>=1} 1/(6*n^2+39*n+63)). (End)
log(2) = lim_{s->1} zeta(s)*(1-1/2^(s-1)). - Mats Granvik, Jun 18 2013
log(2) = 2*Sum_{n>=1} 1/( n*A008288(n-1,n-1)*A008288(n,n) ), a result due to Burnside.
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.
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.
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)
log(2) = (2/3)*Sum_{k >= 0} 1/((2*k + 1)*9^k).
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)
Asymptotic expansions:
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);
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)
log(2) = lim_{n->oo} Sum_{k=1..n} sin(1/(n+k)). See Mathematical Reflections link. - Michel Marcus, Jan 07 2017
Equals Sum_{k>=2} zeta(k)/2^k.
Equals -Sum_{k>=2} log(1 - 1/k^2).
Equals Integral_{x=0..Pi/3} tan(x) dx. (End)
log(2) = Integral_{x=0..Pi/2} (sec(x) - tan(x)) dx. - Clark Kimberling, Jul 08 2020
log(2) = Integral_{x = 0..1} (x - 1)/log(x) dx (Boros and Moll, p. 97).
log(2) = (1/2)*Integral_{x = 0..1} (x + 2)*(x - 1)^2/log(x)^2 dx.
log(2) = (1/4)*Integral_{x = 0..1} (x^2 + 3*x + 4)*(x - 1)^3/log(x)^3 dx. (End)
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
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
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.
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, ...].
Let P(n,k) = n*(n + 1)*...*(n + k).
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)
log(2) = 5/8 + (1/8)*Sum_{k >= 1} (-1)^(k+1) * (2*k + 1)^2 / ( k*(k + 1) )^4
= 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
= 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
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)
log(2) = eta(1) is a period, where eta(x) is the Dirichlet eta function. - Andrea Pinos, Mar 19 2024
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EXAMPLE
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0.693147180559945309417232121458176568075500134360255254120680009493393...
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MATHEMATICA
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PROG
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(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
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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