OFFSET
0,1
COMMENTS
This alternating series converges quite slowly, but 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. 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
0.542666732570282754288850747639624748791420363763092...
MAPLE
evalf(sum((-1)^(n-1)/(n-log(n)), n = 1..infinity), 120);
evalf(1/2+Int((x-arctan(x))/(sinh(Pi*x)*((1-(1/2)*log(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], 119]
N[1/2 + NIntegrate[(x-ArcTan[x])/(Sinh[Pi*x]*((1-1/2*Log[1+x^2])^2 + (x-ArcTan[x])^2)), {x, 0, 1, Infinity}, AccuracyGoal -> 120, WorkingPrecision -> 200], 119] (* The integrand reaches a local maximum near x=1.02, so for better numerical accuracy, split the interval of integration into two or three parts. *)
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, 1, (x-atan(x))/(sinh(Pi*x)*((1-0.5*log(1+x^2))^2 + (x-atan(x))^2))) + intnum(x=1, 3, (x-atan(x))/(sinh(Pi*x)*((1-0.5*log(1+x^2))^2 + (x-atan(x))^2))) + intnum(x=3, 1000, (x-atan(x))/(sinh(Pi*x)*((1-0.5*log(1+x^2))^2 + (x-atan(x))^2))) /* The integrand reaches a local maximum near x=1.02, so for better numerical accuracy, split the interval of integration into two or three parts. */
(Sage)
from mpmath import mp, nsum, inf
mp.dps = 110; mp.pretty = True
nsum(lambda n: (-1)^(n-1)/(n-log(n)), [1, inf], method='alternating') # Peter Luschny, May 17 2015
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Iaroslav V. Blagouchine, May 15 2015
STATUS
approved