A257837 Decimal expansion of Sum_{n>=2} (-1)^n/log(2*n-1).

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

Offset

0,1

Comments

This alternating series is a particular case of the Fatou series sin(alpha*n)/log(n) with alpha=Pi/2 and converges very slowly. However, it can be efficiently computed via its integral representation (see my formula below), which converges exponentially fast. I used this formula and PARI to compute 1000 digits of this series.

Links

Iaroslav V. Blagouchine, Table of n, a(n) for n = 0..1000

Formula

Equals 1/(2*log(3))+6*Integral_{x=0..infinity} arctan(x)/((log(9+9*x^2)^2+4*arctan(x)^2)*sinh(3*Pi*x/2)).

Example

0.5639914398242359108584254635830512736968995545268548...

Maple

evalf(sum((-1)^n/log(2*n-1), n=2..infinity), 120);
evalf(1/(2*log(3))+6*(Int(arctan(x)/((log(9+9*x^2)^2+4*arctan(x)^2)*sinh(3*Pi*x/2)), x=0..infinity)), 120);

Mathematica

NSum[(-1)^n/Log[2*n-1], {n, 2, Infinity}, AccuracyGoal -> 120, WorkingPrecision -> 200, Method -> "AlternatingSigns"]
1/(2*Log[3])+6*NIntegrate[ArcTan[x]/((Log[9+9*x^2]^2+4*ArcTan[x]^2)*Sinh[3*Pi*x/2]), {x, 0, Infinity}, WorkingPrecision->120] (* Mathematica 5.1 evaluates correctly only first 17 digits. In later versions, all digits are correct. *)

Prog

(PARI) default(realprecision, 120); sumalt(n=2, (-1)^n/log(2*n-1))
(PARI) allocatemem(50000000);
default(realprecision, 1200); 1/(2*log(3))+intnum(x=0, 1000, 6*atan(x)/((log(9+9*x^2)^2+4*atan(x)^2)*sinh(3*Pi*x/2)))

Crossrefs

Cf. A099769, A257898, A257960, A257964, A257972, A257812.

Sequence in context: A153614 A328905 A195709 * A099038 A064206 A087197

Adjacent sequences: A257834 A257835 A257836 * A257838 A257839 A257840

Keyword

nonn,cons

Author

Iaroslav V. Blagouchine, May 10 2015

Status

approved