A257972 Decimal expansion of Sum_{n=1..infinity} (-1)^(n-1)/(n - log(n)).

5, 4, 2, 6, 6, 6, 7, 3, 2, 5, 7, 0, 2, 8, 2, 7, 5, 4, 2, 8, 8, 8, 5, 0, 7, 4, 7, 6, 3, 9, 6, 2, 4, 7, 4, 8, 7, 9, 1, 4, 2, 0, 3, 6, 3, 7, 6, 3, 0, 9, 2, 7, 2, 0, 0, 9, 5, 0, 7, 8, 6, 6, 0, 1, 3, 8, 1, 0, 1, 1, 7, 9, 9, 6, 4, 3, 2, 3, 3, 3, 6, 7, 3, 6, 3, 9, 8, 3, 4, 5, 7, 0, 2, 2, 3, 6, 5, 4, 2, 0, 4, 8, 2, 8, 6, 3, 8, 5, 5

Offset: 0

Author

Iaroslav V. Blagouchine, May 15 2015

Keywords

• nonn
• cons

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.

Examples

			0.542666732570282754288850747639624748791420363763092...
		

Links

• Iaroslav V. Blagouchine, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A099769, A257837, A257812, A257898, A257960, A257964.

Programs

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. *)

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

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)).