

A257972


Decimal expansion of Sum_{n=1..infinity} (1)^(n1)/(n  log(n)).


5



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
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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} (xarctan(x))/(sinh(Pi*x)*((11/2*log(1+x^2))^2 + (xarctan(x))^2)).


EXAMPLE

0.542666732570282754288850747639624748791420363763092...


MAPLE

evalf(sum((1)^(n1)/(nlog(n)), n = 1..infinity), 120);
evalf(1/2+Int((xarctan(x))/(sinh(Pi*x)*((1(1/2)*log(1+x^2))^2+(xarctan(x))^2)), x = 0..infinity), 120);


MATHEMATICA

N[NSum[(1)^(n1)/(nLog[n]), {n, 1, Infinity}, AccuracyGoal > 120, Method > "AlternatingSigns", WorkingPrecision > 200], 119]
N[1/2 + NIntegrate[(xArcTan[x])/(Sinh[Pi*x]*((11/2*Log[1+x^2])^2 + (xArcTan[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)^(n1)/(nlog(n)))
(PARI) allocatemem(50000000);
default(realprecision, 1200); 1/2 + intnum(x=0, 1, (xatan(x))/(sinh(Pi*x)*((10.5*log(1+x^2))^2 + (xatan(x))^2))) + intnum(x=1, 3, (xatan(x))/(sinh(Pi*x)*((10.5*log(1+x^2))^2 + (xatan(x))^2))) + intnum(x=3, 1000, (xatan(x))/(sinh(Pi*x)*((10.5*log(1+x^2))^2 + (xatan(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.dps = 110; mp.pretty = True
nsum(lambda n: (1)^(n1)/(nlog(n)), [1, inf], method='alternating') # Peter Luschny, May 17 2015


CROSSREFS

Cf. A099769, A257837, A257812, A257898, A257960, A257964.
Sequence in context: A260849 A246746 A180131 * A222307 A175838 A097960
Adjacent sequences: A257969 A257970 A257971 * A257973 A257974 A257975


KEYWORD

nonn,cons


AUTHOR

Iaroslav V. Blagouchine, May 15 2015


STATUS

approved



