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
KEYWORD
nonn,cons
AUTHOR
Iaroslav V. Blagouchine, May 10 2015
STATUS
approved