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%I #19 May 28 2015 08:21:47
%S 7,6,9,4,0,2,1,5,0,2,8,0,8,0,0,4,8,4,1,2,2,1,2,6,9,7,1,9,4,6,0,0,5,3,
%T 1,5,5,7,6,2,0,5,5,3,2,0,3,3,5,4,3,5,8,7,7,1,5,5,6,3,4,4,4,8,1,1,1,6,
%U 2,1,5,3,7,1,4,1,0,2,9,9,9,0,9,7,0,5,4,8,0,7,3,4,1,4,1,0,0,3,7,2,0,4,3,5,5,6,7,3,3
%N Decimal expansion of Sum_{n=1..infinity} (-1)^(n-1)/(n + log(n)).
%C This alternating series converges relatively slowly, but can be efficiently computed via an integral representation, which converges exponentially fast (see my formula below). I used this formula and PARI to compute 1000 digits of this series. Modern CAS are also able to evaluate it very quickly and to a high degree of accuracy.
%H Iaroslav V. Blagouchine, <a href="/A257964/b257964.txt">Table of n, a(n) for n = 0..1000</a>
%F Equals 1/2 + integral_{x=0..infinity} (x+arctan(x))/(sinh(Pi*x)*((1+1/2*log(1+x^2))^2 + (x+arctan(x))^2)).
%e 0.769402150280800484122126971946005315576205532033543...
%p evalf(sum((-1)^(n-1)/(n+ln(n)), n = 1..infinity), 120);
%p evalf(1/2+int((x+arctan(x))/(sinh(Pi*x)*((1+(1/2)*ln(1+x^2))^2+(x+arctan(x))^2)), x = 0..infinity), 120);
%t N[NSum[(-1)^(n-1)/(n+Log[n]), {n, 1, Infinity}, AccuracyGoal -> 120, Method -> "AlternatingSigns", WorkingPrecision -> 200],120]
%t N[1/2 + NIntegrate[(x+ArcTan[x])/(Sinh[Pi*x]*((1+1/2*Log[1+x^2])^2 + (x+ArcTan[x])^2)), {x, 0, Infinity}, AccuracyGoal -> 120, WorkingPrecision -> 200],120]
%o (PARI) default(realprecision, 120); sumalt(n=1, (-1)^(n-1)/(n+log(n)))
%o (PARI) allocatemem(50000000);
%o default(realprecision, 1200); 1/2 + intnum(x=0, 1000, (x+atan(x))/(sinh(Pi*x)*((1+0.5*log(1+x^2))^2 + (x+atan(x))^2)))
%Y Cf. A099769, A257837, A257812, A257898, A257960, A257972.
%K nonn,cons
%O 0,1
%A _Iaroslav V. Blagouchine_, May 14 2015
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