A257812 -id:A257812 - cp's OEIS frontend

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

Pierre CAMI, Mar 11 2006

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)

			2.10974280123689197447925719761655132638553198439474202264991560319281...

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).

Cf. A145419, A145420, A145421, A363632, A363633.
Cf. A137245, A257812. A097906 is a similar sum.
Cf. A118582, A347145.

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 *)

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

A099769 Decimal expansion of Sum_{n >= 2} (-1)^n/log(n).

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane, Nov 11 2004

A slowly converging series. The reference (R. E. Shafer) gives several methods for evaluating the sum.
Mathematica program derived from method #3 in the reference (R. E. Shafer). - Ryan Propper, Sep 25 2006
This alternating slowly convergent series may also be efficiently computed via a rapidly convergent integral (see my formula below). I used this formula and PARI to compute 1000 digits of it. - Iaroslav V. Blagouchine, May 11 2015

			0.9242998972229388559595701813595900537733193978869190...

C. C. Grosjean, An Euler-Maclaurin type asymptotic series expansion of the Sum_{n=2..oo} (-1)^n/ln(n), Simon Stevin, Vol. 65 (1991), pp. 31-55.

Iaroslav V. Blagouchine, Table of n, a(n) for n = 0..1000
D. A. MacDonald, A note on the summation of slowly convergent alternating series, BIT Numerical Mathematics, Vol. 36 (1996), pp. 766-774.
Mathematics Stack Exchange, Sum_{n=2..oo} (-1)^n/log(n) = ?, 2018.
R. E. Shafer (proposer), Problem 89-15, SIAM Rev., Vol. 31, No. 3 (1989), p. 495; Numerical Evaluation of a Slowly Convergent Series, Solution to Problem 89-15 by Alan Gibbs, ibid., Vol. 32, No. 3 (1990), pp. 481-483.

Cf. A257837, A257964, A257972, A257898, A257960, A257812.

evalf(sum((-1)^n/log(n), n=2..infinity),120); # Vaclav Kotesovec, May 11 2015

Do[X = 2*i; T = Table[Table[0, {X}], {X}]; For[n = 2, n <= X, n++, T[[n, 2]] = Sum[(-1)^k/Log[k], {k, 2, n}]]; For[k = 2, k <= X, k++, For[n = 2, n <= X - k + 1, n++, T[[n, k+1]] = T[[n+1, k-1]] + 1/(T[[n+1, k]] - T[[n, k]])]]; Print[N[T[[2, X]], 50]], {i, 50}] (* Ryan Propper, Sep 25 2006 *)
digits = 105; NSum[(-1)^n/Log[n], {n, 2, Infinity}, WorkingPrecision -> digits+10, Method -> "AlternatingSigns"] // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Feb 12 2013 *)
1/(2*Log[2])+8*NIntegrate[ArcTan[x]/((Log[4+4*x^2]^2+4*ArcTan[x]^2)*Sinh[2*Pi*x]), {x, 0, Infinity}, WorkingPrecision -> 109] // RealDigits // First (* Jean-François Alcover, May 12 2015, after Iaroslav V. Blagouchine *)

sumalt(n=2,(-1)^n/log(n)) \\ Herman Jamke (hermanjamke(AT)fastmail.fm), Apr 28 2007

allocatemem(50000000);
  default(realprecision, 1100);  1/(2*log(2)) + intnum(x=0,1000, 8*atan(x)/((log(4+4*x^2)^2+4*atan(x)^2)*sinh(2*Pi*x))) \\ Iaroslav V. Blagouchine, May 11 2015

Equals 1/(2*log(2)) + 8*Integral_{x=0..infinity} arctan(x)/((log(4+4*x^2)^2+4*arctan(x)^2)*sinh(2*Pi*x)) dx. - Iaroslav V. Blagouchine, May 11 2015
From Amiram Eldar, Jun 29 2021: (Start)
Equals Integral_{x>=0} (1 - (1-2^(1-x))*zeta(x)) dx.
Equals Integral_{x>=0} (1 + Li(x, -1)) dx, where Li(s, z) is the polylogarithm function.
Both from Mathematics Stack Exchange. (End)

a(9)-a(17) from Ryan Propper, Sep 25 2006

a(18)-a(104) from Herman Jamke (hermanjamke(AT)fastmail.fm), Apr 28 2007

A257837 Decimal expansion of Sum_{n>=2} (-1)^n/log(2*n-1).

Original entry on oeis.org

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

Offset: 0

Iaroslav V. Blagouchine, May 10 2015

This alternating series is a particular case of the Fatou series sin(alpha*n)/log(n) with alpha=Pi/2 and converges very slowly. However, it 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.

			0.5639914398242359108584254635830512736968995545268548...

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

Cf. A099769, A257898, A257960, A257964, A257972, A257812.

evalf(sum((-1)^n/log(2*n-1), n=2..infinity), 120);
evalf(1/(2*log(3))+6*(Int(arctan(x)/((log(9+9*x^2)^2+4*arctan(x)^2)*sinh(3*Pi*x/2)), x=0..infinity)), 120);

NSum[(-1)^n/Log[2*n-1], {n, 2, Infinity}, AccuracyGoal -> 120, WorkingPrecision -> 200, Method -> "AlternatingSigns"]
1/(2*Log[3])+6*NIntegrate[ArcTan[x]/((Log[9+9*x^2]^2+4*ArcTan[x]^2)*Sinh[3*Pi*x/2]), {x, 0,Infinity}, WorkingPrecision->120] (* Mathematica 5.1 evaluates correctly only first 17 digits. In later versions, all digits are correct. *)

default(realprecision,120); sumalt(n=2, (-1)^n/log(2*n-1))

allocatemem(50000000);
default(realprecision,1200); 1/(2*log(3))+intnum(x=0,1000,6*atan(x)/((log(9+9*x^2)^2+4*atan(x)^2)*sinh(3*Pi*x/2)))

Equals 1/(2*log(3))+6*Integral_{x=0..infinity} arctan(x)/((log(9+9*x^2)^2+4*arctan(x)^2)*sinh(3*Pi*x/2)).

A257898 Decimal expansion of Sum_{n=2..infinity} (-1)^n/log(log(n)), negated.

Original entry on oeis.org

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

Offset: 2

Iaroslav V. Blagouchine, May 12 2015

A very slowly convergent series, converging in virtue of Leibniz's rule.

			-11.47796801398707591150778896756796191665188684328765...

Eric Weisstein's World of Mathematics, Leibniz Criterion
Wikipedia, Alternating series test

Cf. A099769, A257837, A257812.

evalf(sum((-1)^n/log(log(n)), n = 2..infinity), 120);

NSum[(-1)^n/Log[Log[n]], {n, 2, Infinity}, AccuracyGoal -> 120, Method -> "AlternatingSigns", WorkingPrecision -> 200]

default(realprecision, 120); sumalt(n=2, (-1)^n/log(log(n)))

A257960 Decimal expansion of Sum_{n>=3} (-1)^n/log(log(log(n))).

Original entry on oeis.org

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

Offset: 3

Iaroslav V. Blagouchine, May 14 2015

An extremely slowly convergent series, converging in virtue of Leibniz's rule.

			277.8674989684568172306449945790310149069364211466765...

Eric Weisstein's World of Mathematics, Leibniz Criterion.
Wikipedia, Alternating series test.

Cf. A099769, A257837, A257812, A257898.

evalf(sum((-1)^n/log(log(log(n))), n = 3..infinity), 120); # Maple 12.0 computes this expression with no problems, but later versions of Maple may have some problems with it.

N[NSum[(-1)^n/Log[Log[Log[n]]], {n, 3, Infinity}, AccuracyGoal -> 500, Method -> "AlternatingSigns", WorkingPrecision -> 1000], 119] (* Mathematica needs higher precision than usual to compute this series *)

default(realprecision, 200); precision(sumalt(n=3, (-1)^n/log(log(log(n)))), 120) /* PARI needs higher precision than usual to compute this series */

A257964 Decimal expansion of Sum_{n=1..infinity} (-1)^(n-1)/(n + log(n)).

Original entry on oeis.org

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

Offset: 0

Iaroslav V. Blagouchine, May 14 2015

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.

			0.769402150280800484122126971946005315576205532033543...

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

Cf. A099769, A257837, A257812, A257898, A257960, A257972.

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);

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]

default(realprecision, 120); sumalt(n=1, (-1)^(n-1)/(n+log(n)))

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)))

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)).

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

Original entry on oeis.org

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

Iaroslav V. Blagouchine, May 15 2015

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.

			0.542666732570282754288850747639624748791420363763092...

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

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

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);

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. *)

default(realprecision, 120); sumalt(n=1, (-1)^(n-1)/(n-log(n)))

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. */

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

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)).

A168218 Decimal expansion of the Sum_{k=2..infinity} 1/(k^2*log(k)).

Original entry on oeis.org

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

Offset: 0

R. J. Mathar, Nov 20 2009

Also the value of the integral of the fractional part of the Riemann zeta function from 2 to infinity. - Alexander Bock, Apr 01 2014

			equals 1/(4*log(2))+1/(9*log(3))+1/(16*log(4))+ .... + = 0.605521788882600447699549005207240447303238898..

R. P. Boas, Jr, Partial sums of Infinite Series, and How They Grow, Am. Math. Monthly 84 (4) (1977) 237-235 [MR0440240].

Cf. A099769, A115563, A257812.

(* Computation needs a few minutes for 105 digits *) digits = 105; NSum[ 1/(n^2*Log[n]), {n, 2, Infinity}, NSumTerms -> 500000, WorkingPrecision -> digits + 5, Method -> {"EulerMaclaurin", Method -> {"NIntegrate", "MaxRecursion" -> 12}}] // RealDigits[#, 10, digits] & // First (* Jean-François Alcover, Feb 11 2013 *)
RealDigits[NIntegrate[Zeta[x] - 1, {x, 2, Infinity}, WorkingPrecision->110], 10, 100] (* Alexander Bock, Apr 01 2014 *)

intnum(x=2,[oo,log(2)],zeta(x)-1) \\ Charles R Greathouse IV, Apr 01 2014

suminf(k=2,1/k^2/log(k)) \\ Charles R Greathouse IV, Apr 01 2014

Equals Integral_{x>=2} (zeta(x) - 1) dx.

More terms from Jean-François Alcover, Feb 11 2013

A354295 Decimal expansion of Sum_{k>=1} (-1)^k * log(k) / (k+1).

Original entry on oeis.org

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

Offset: 0

Vaclav Kotesovec, May 23 2022

			0.096614934732226887013571828677661550056338319496835535473156803861323...

A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev, Integrals and Series, Vol. 1 (Overseas Publishers Association, Amsterdam, 1986), p. 746, section 5.5.1, formula 4.

Cf. A091812, A257812.

evalf(Sum((-1)^k*log(k)/(k + 1), k = 1..infinity), 105);

N[NSum[(-1)^k*Log[k]/(k+1), {k, 1, Infinity}, WorkingPrecision -> 120, Method -> "AlternatingSigns"], 105]

PARI
```
sumalt(k=1, (-1)^k*log(k)/(k+1))
```

A331421 Decimal expansion of Sum_{n >= 2} (-1)^n/log_2(n).

Original entry on oeis.org

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

Offset: 0

Daniel Hoyt, Jan 16 2020

A slowly converging series.

			0.64067586775192729126561661179177777...

Daniel Hoyt, Python 3 program using series acceleration.

Cf. A002162, A099769, A257837, A257964, A257972, A257898, A257960, A257812.

sumalt(k=2, (-1)^k/log(k))*log(2) \\ Edited by M. F. Hasler, Jan 31 2020

Equals A002162 * A099769. - M. F. Hasler, Jan 31 2020

cp's OEIS Frontend

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

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A099769 Decimal expansion of Sum_{n >= 2} (-1)^n/log(n).

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A257837 Decimal expansion of Sum_{n>=2} (-1)^n/log(2*n-1).

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A257898 Decimal expansion of Sum_{n=2..infinity} (-1)^n/log(log(n)), negated.

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A257960 Decimal expansion of Sum_{n>=3} (-1)^n/log(log(log(n))).

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A257964 Decimal expansion of Sum_{n=1..infinity} (-1)^(n-1)/(n + log(n)).

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Maple

Mathematica

PARI

PARI

Formula