A115563 Decimal expansion of Sum_{n>=2} 1/(n*log(n)^2).

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

Offset

1,1

Comments

Sum_{n>1} 1/(n*log(n)^2) is a tiny bit greater than (zeta(2))^(3/2) = (Pi^2 / 6)^(3/2) = 2.109709908063657.... - Daniel Forgues, Mar 30 2012

From Bernard Schott, Oct 03 2021: (Start)

Theorem: Bertrand series Sum_{n>=2} 1/(n*log(n)^q) is convergent iff q > 1 (for q = 3, 4, 5 see respectively A145419, A145420, A145421).

As H(n) ~ log(n), compare with A347145. (End)

Links

Table of n, a(n) for n=1..141.

John V. Baxley, Euler's constant, Taylor's formula, and slowly converging series, Math. Mag. 65 (1992), 302-313.

Bart Braden, Calculating sums of infinite series, Am. Math. Monthly 99 (1992) 649-655.

David Broadhurst, Re: need help about 2 constants, primeforum, Mar 10 2006.

Pierre CAMI, David Broadhurst, Need help about 2 constants, digest of 3 messages in primeform Yahoo group, Mar 10, 2006. [Cached copy]

Rick Kreminski, Using Simpson's rule to approximate sums of infinite series, College Math. J. 28 (1997), 368-376.

R. J. Mathar, The series limit of sum_k 1/[k*log k *(log log k)^2], arXiv:0902.0789 [math.NA], 2009, App. A.

S.O.S. Math, Bertrand Series.

Eric Weisstein's World of Mathematics, Convergent Series

Wikipédia, Série de Bertrand (in French).

Example

2.10974280123689197447925719761655132638553198439474202264991560319281...

Mathematica

digits = 150; NSum[1/(n*Log[n]^2), {n, 2, Infinity}, NSumTerms -> 200000, WorkingPrecision -> digits + 5, Method -> {"EulerMaclaurin", Method -> {"NIntegrate", "MaxRecursion" -> 20}}] (* Vaclav Kotesovec, Mar 01 2016, after Jean-François Alcover *)
maxiter = 20; nn = 10000; alfa = 2; bas = Sum[1/(k*Log[k]^alfa), {k, 2, nn}] + 1/((alfa - 1)*Log[nn + 1/2]^(alfa - 1)); sub = 0; Do[sub = sub + 1/4^s/(2*s + 1)! * NSum[(D[1/(x*Log[x]^alfa), {x, 2 s}]) /. x -> k, {k, nn + 1, Infinity}, WorkingPrecision -> 120, NSumTerms -> 100000, PrecisionGoal -> 120, Method -> {"NIntegrate", "MaxRecursion" -> 100}]; Print[bas - sub], {s, 1, maxiter}] (* Vaclav Kotesovec, Jun 12 2022 *)

Crossrefs

Cf. A145419, A145420, A145421, A363632, A363633.

Cf. A137245, A257812. A097906 is a similar sum.

Cf. A118582, A347145.

Sequence in context: A158335 A111595 A021478 * A364068 A293881 A362308

Adjacent sequences: A115560 A115561 A115562 * A115564 A115565 A115566

Keyword

cons,nonn

Author

Pierre CAMI, Mar 11 2006

Extensions

Removed incorrect speculations about relations to A097906 - R. J. Mathar, Oct 14 2010

More terms from Robert G. Wilson v, Dec 12 2012

Corrected a(55) and beyond, Vaclav Kotesovec, Mar 01 2016

Status

approved