%I #31 Mar 01 2016 06:14:26
%S 5,2,6,4,1,2,2,4,6,5,3,3,3,1,0,4,1,0,9,3,0,6,9,6,5,0,1,4,1,1,1,3,1,4,
%T 1,3,7,2,1,7,9,0,5,9,7,8,8,7,5,5,8,5,4,0,7,4,6,9,9,5,7,0,0,8,3,3,7,8,
%U 3,2,2,3,1,3,0,2,0,8,4,4,6,9,8,4,6,3,6,2,2,7,2,9,7,3,4,6,1,5,1,7,8,8,7,6,4,9,5,5,8
%N Decimal expansion of Sum_{n=2..infinity} (-1)^n/(n*log(n)).
%C 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.
%H Iaroslav V. Blagouchine, <a href="/A257812/b257812.txt">Table of n, a(n) for n = 0..1000</a>
%F 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)).
%e 0.5264122465333104109306965014111314137217905978875585...
%p evalf(sum((-1)^n/(n*log(n)), n=2..infinity), 120);
%p 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);
%t NSum[(-1)^n/(n*Log[n]), {n, 2, Infinity}, AccuracyGoal -> 120, WorkingPrecision -> 200, Method -> AlternatingSigns]
%t 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]
%o (PARI) default(realprecision,120); sumalt(n=2, (-1)^n/(n*log(n))) \\ _Vaclav Kotesovec_, May 10 2015
%o (PARI) allocatemem(50000000);
%o 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)))
%Y Cf. A099769, A115563, A257898, A257960, A257964, A257972, A257837.
%K nonn,cons
%O 0,1
%A _Iaroslav V. Blagouchine_, May 10 2015