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%I #42 Jul 03 2015 00:01:33
%S 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,
%T 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,
%U 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
%N Decimal expansion of Sum_{n>=2} (-1)^n/log(2*n-1).
%C 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.
%H Iaroslav V. Blagouchine, <a href="/A257837/b257837.txt">Table of n, a(n) for n = 0..1000</a>
%F 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)).
%e 0.5639914398242359108584254635830512736968995545268548...
%p evalf(sum((-1)^n/log(2*n-1), n=2..infinity), 120);
%p 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);
%t NSum[(-1)^n/Log[2*n-1], {n, 2, Infinity}, AccuracyGoal -> 120, WorkingPrecision -> 200, Method -> "AlternatingSigns"]
%t 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. *)
%o (PARI) default(realprecision,120); sumalt(n=2, (-1)^n/log(2*n-1))
%o (PARI) allocatemem(50000000);
%o 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)))
%Y Cf. A099769, A257898, A257960, A257964, A257972, A257812.
%K nonn,cons
%O 0,1
%A _Iaroslav V. Blagouchine_, May 10 2015
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