A099769 Decimal expansion of Sum_{n >= 2} (-1)^n/log(n).

9, 2, 4, 2, 9, 9, 8, 9, 7, 2, 2, 2, 9, 3, 8, 8, 5, 5, 9, 5, 9, 5, 7, 0, 1, 8, 1, 3, 5, 9, 5, 9, 0, 0, 5, 3, 7, 7, 3, 3, 1, 9, 3, 9, 7, 8, 8, 6, 9, 1, 9, 0, 7, 4, 7, 7, 9, 6, 3, 0, 4, 3, 7, 2, 5, 0, 7, 0, 0, 5, 4, 1, 7, 1, 1, 4, 3, 4, 6, 8, 9, 7, 9, 8, 9, 9, 1, 3, 4, 7, 6, 6, 4, 9, 4, 6, 9, 1, 9, 5, 3, 5, 7, 4, 1, 4, 5, 2, 8

Offset: 0

Author

N. J. A. Sloane, Nov 11 2004

Keywords

• nonn
• cons

Comments

A slowly converging series. The reference (R. E. Shafer) gives several methods for evaluating the sum.

Mathematica program derived from method #3 in the reference (R. E. Shafer). - Ryan Propper, Sep 25 2006

This alternating slowly convergent series may also be efficiently computed via a rapidly convergent integral (see my formula below). I used this formula and PARI to compute 1000 digits of it. - Iaroslav V. Blagouchine, May 11 2015

Examples

			0.9242998972229388559595701813595900537733193978869190...
		

References

• C. C. Grosjean, An Euler-Maclaurin type asymptotic series expansion of the Sum_{n=2..oo} (-1)^n/ln(n), Simon Stevin, Vol. 65 (1991), pp. 31-55.

Links

• Iaroslav V. Blagouchine, Table of n, a(n) for n = 0..1000
• D. A. MacDonald, A note on the summation of slowly convergent alternating series, BIT Numerical Mathematics, Vol. 36 (1996), pp. 766-774.
• Mathematics Stack Exchange, Sum_{n=2..oo} (-1)^n/log(n) = ?, 2018.
• R. E. Shafer (proposer), Problem 89-15, SIAM Rev., Vol. 31, No. 3 (1989), p. 495; Numerical Evaluation of a Slowly Convergent Series, Solution to Problem 89-15 by Alan Gibbs, ibid., Vol. 32, No. 3 (1990), pp. 481-483.

Crossrefs

Cf. A257837, A257964, A257972, A257898, A257960, A257812.

Programs

Maple

evalf(sum((-1)^n/log(n), n=2..infinity),120); # Vaclav Kotesovec, May 11 2015

Mathematica

Do[X = 2*i; T = Table[Table[0, {X}], {X}]; For[n = 2, n <= X, n++, T[[n, 2]] = Sum[(-1)^k/Log[k], {k, 2, n}]]; For[k = 2, k <= X, k++, For[n = 2, n <= X - k + 1, n++, T[[n, k+1]] = T[[n+1, k-1]] + 1/(T[[n+1, k]] - T[[n, k]])]]; Print[N[T[[2, X]], 50]], {i, 50}] (* Ryan Propper, Sep 25 2006 *)
digits = 105; NSum[(-1)^n/Log[n], {n, 2, Infinity}, WorkingPrecision -> digits+10, Method -> "AlternatingSigns"] // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Feb 12 2013 *)
1/(2*Log[2])+8*NIntegrate[ArcTan[x]/((Log[4+4*x^2]^2+4*ArcTan[x]^2)*Sinh[2*Pi*x]), {x, 0, Infinity}, WorkingPrecision -> 109] // RealDigits // First (* Jean-François Alcover, May 12 2015, after Iaroslav V. Blagouchine *)

PARI

sumalt(n=2,(-1)^n/log(n)) \\ Herman Jamke (hermanjamke(AT)fastmail.fm), Apr 28 2007

PARI

allocatemem(50000000);
  default(realprecision, 1100);  1/(2*log(2)) + intnum(x=0,1000, 8*atan(x)/((log(4+4*x^2)^2+4*atan(x)^2)*sinh(2*Pi*x))) \\ Iaroslav V. Blagouchine, May 11 2015

Formula

Equals 1/(2*log(2)) + 8*Integral_{x=0..infinity} arctan(x)/((log(4+4*x^2)^2+4*arctan(x)^2)*sinh(2*Pi*x)) dx. - Iaroslav V. Blagouchine, May 11 2015

From Amiram Eldar, Jun 29 2021: (Start)

Equals Integral_{x>=0} (1 - (1-2^(1-x))*zeta(x)) dx.

Equals Integral_{x>=0} (1 + Li(x, -1)) dx, where Li(s, z) is the polylogarithm function.

Both from Mathematics Stack Exchange. (End)

Extensions

a(9)-a(17) from Ryan Propper, Sep 25 2006

a(18)-a(104) from Herman Jamke (hermanjamke(AT)fastmail.fm), Apr 28 2007