A099769 - OEIS

%I #62 Sep 25 2021 18:12:11

%S 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,

%T 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,

%U 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

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

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

%C Mathematica program derived from method #3 in the reference (R. E. Shafer). - _Ryan Propper_, Sep 25 2006

%C 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

%D 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.

%H Iaroslav V. Blagouchine, <a href="/A099769/b099769.txt">Table of n, a(n) for n = 0..1000</a>

%H D. A. MacDonald, <a href="https://doi.org/10.1007/BF01733790">A note on the summation of slowly convergent alternating series</a>, BIT Numerical Mathematics, Vol. 36 (1996), pp. 766-774.

%H Mathematics Stack Exchange, <a href="https://math.stackexchange.com/questions/2806079/sum-n-2-infty-1n-frac1-log-n">Sum_{n=2..oo} (-1)^n/log(n) = ?</a>, 2018.

%H R. E. Shafer (proposer), <a href="http://www.jstor.org/stable/2031414">Problem 89-15</a>, SIAM Rev., Vol. 31, No. 3 (1989), p. 495; <a href="http://www.jstor.org/stable/2031625">Numerical Evaluation of a Slowly Convergent Series</a>, Solution to Problem 89-15 by Alan Gibbs, ibid., Vol. 32, No. 3 (1990), pp. 481-483.

%F 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

%F From _Amiram Eldar_, Jun 29 2021: (Start)

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

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

%F Both from Mathematics Stack Exchange. (End)

%e 0.9242998972229388559595701813595900537733193978869190...

%p evalf(sum((-1)^n/log(n), n=2..infinity),120); # _Vaclav Kotesovec_, May 11 2015

%t 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 *)

%t 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 *)

%t 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_ *)

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

%o (PARI) allocatemem(50000000);

%o   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

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

%K nonn,cons

%O 0,1

%A _N. J. A. Sloane_, Nov 11 2004

%E a(9)-a(17) from _Ryan Propper_, Sep 25 2006

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