OFFSET
0,1
COMMENTS
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.
LINKS
Iaroslav V. Blagouchine, Table of n, a(n) for n = 0..1000
FORMULA
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)).
EXAMPLE
0.769402150280800484122126971946005315576205532033543...
MAPLE
evalf(sum((-1)^(n-1)/(n+ln(n)), n = 1..infinity), 120);
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);
MATHEMATICA
N[NSum[(-1)^(n-1)/(n+Log[n]), {n, 1, Infinity}, AccuracyGoal -> 120, Method -> "AlternatingSigns", WorkingPrecision -> 200], 120]
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]
PROG
(PARI) default(realprecision, 120); sumalt(n=1, (-1)^(n-1)/(n+log(n)))
(PARI) allocatemem(50000000);
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)))
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Iaroslav V. Blagouchine, May 14 2015
STATUS
approved