OFFSET
0,1
COMMENTS
This alternating series converges quite slowly. However, it can be efficiently computed via its integral representation (see formula below), which converges exponentially fast. This formula and PARI was used to compute 1000 digits.
LINKS
Iaroslav V. Blagouchine, Table of n, a(n) for n = 0..1000
FORMULA
Equals 1/(4*log(2))+2*integral_{x=0..infinity} ((2*arctan(x)+x*log(4+4*x^2))/(sinh(2*Pi*x)*(log(4+4*x^2)^2+4*arctan(x)^2)*(x^2+1)).
EXAMPLE
0.5264122465333104109306965014111314137217905978875585...
MAPLE
evalf(sum((-1)^n/(n*log(n)), n=2..infinity), 120);
evalf(1/(4*log(2))+2*(Int((2*arctan(x)+x*log(4+4*x^2))/(sinh(2*Pi*x)*(log(4+4*x^2)^2+4*arctan(x)^2)*(x^2+1)), x=0..infinity)), 120);
MATHEMATICA
NSum[(-1)^n/(n*Log[n]), {n, 2, Infinity}, AccuracyGoal -> 120, WorkingPrecision -> 200, Method -> AlternatingSigns]
1/(4*Log[2])+2*NIntegrate[(2*ArcTan[x]+x*Log[4+4*x^2])/((x^2+1)*Sinh[2*Pi*x]*(Log[4+4*x^2]^2+4*ArcTan[x]^2)), {x, 0, Infinity}, WorkingPrecision->120]
PROG
(PARI) default(realprecision, 120); sumalt(n=2, (-1)^n/(n*log(n))) \\ Vaclav Kotesovec, May 10 2015
(PARI) allocatemem(50000000);
default(realprecision, 1200); 1/(4*log(2))+2*intnum(x=0, 1000, (2*atan(x)+x*log(4+4*x^2))/(sinh(2*Pi*x)*(log(4+4*x^2)^2+4*atan(x)^2)*(x^2+1)))
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Iaroslav V. Blagouchine, May 10 2015
STATUS
approved