A257960 -id:A257960 - cp's OEIS frontend

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

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

A257812 Decimal expansion of Sum_{n>=2} (-1)^n/(n*log(n)).

Original entry on oeis.org

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

Offset: 0

Iaroslav V. Blagouchine, May 10 2015

This alternating series converges quite slowly. However, it can be efficiently computed via its integral representation (see formula below), which converges exponentially quickly. This formula and PARI were used to compute 1000 digits.

			0.5264122465333104109306965014111314137217905978875585...

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

Cf. A099769, A115563, A257898, A257960, A257964, A257972, A257837.

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

NSum[(-1)^n/(n*Log[n]), {n, 2, Infinity}, AccuracyGoal -> 120, WorkingPrecision -> 200, Method -> AlternatingSigns]
1/(4*Log[2])+2*NIntegrate[(2*ArcTan[x]+x*Log[4+4*x^2])/((x^2+1)*Sinh[2*Pi*x]*(Log[4+4*x^2]^2+4*ArcTan[x]^2)), {x, 0,Infinity}, WorkingPrecision->120]

default(realprecision,120); sumalt(n=2, (-1)^n/(n*log(n))) \\ Vaclav Kotesovec, May 10 2015

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

Equals 1/(4*log(2)) + 2*Integral_{x=0..oo} (2*arctan(x)+x*log(4+4*x^2))/(sinh(2*Pi*x)*(log(4+4*x^2)^2+4*arctan(x)^2)*(x^2+1)) dx.

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)).

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)).

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

A351687 Decimal expansion of Sum_{n>=2} (-1)^n/log(n!).

Original entry on oeis.org

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

Offset: 1

Bernard Schott, May 05 2022

This series is convergent according to the alternating series test, while series Sum_{n>=2} 1/log(n!) -> infinity (link).

			1.0769010278586314719973748207332879382948126467776416116987...

Socratic Q&A, How do you test the Series Sum_{n>=2} 1/log(n!) for convergence?

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

Cf. A099871, A257898, A257960, A336741.

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

RealDigits[NSum[(-1)^k/Log[k!], {k, 2, Infinity}, WorkingPrecision -> 120, Method -> "AlternatingSigns"]][[1]] (* Amiram Eldar, May 05 2022 *)

sumalt(k=2, (-1)^k/log(k!)) \\ Vaclav Kotesovec, May 05 2022

Equals Sum_{k>=2} (-1)^k/log(k!).

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

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A257812 Decimal expansion of Sum_{n>=2} (-1)^n/(n*log(n)).

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

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

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

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

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