A099769 -id:A099769 - cp's OEIS frontend

A091812 Decimal expansion of Sum_{k>=1} (-1)^k*log(k)/k.

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

Offset: 0

Benoit Cloitre, Mar 07 2004

Equal to the derivative eta'(1) of the Dirichlet eta function eta(s) = Sum_{k>=1} (-1)^(k-1)/k^s = (1 - 2^(1-s))*zeta(s) at s = 1. - Jonathan Sondow, Dec 28 2011

			0.15986890374243097175694787032491657049622202375645874267082452963965...

Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 2.21, p. 168.
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 3.

Henri Cohen, Fernando Rodriguez Villegas, and Don Zagier, Convergence Acceleration of Alternating Series, Exp. Math. 9 (1) (2000) 3-12.
Eric Weisstein's World of Mathematics, Dirichlet Eta Function.
Wikipedia, Dirichlet eta function.

Cf. A001620, A099769, A265162, A354295.

gamma*log(2)-log(2)^2/2 ; evalf(%) ; # R. J. Mathar, Jun 10 2024

RealDigits[EulerGamma*Log[2] - Log[2]^2/2, 10, 100][[1]] (* Amiram Eldar, Sep 12 2022 *)
RealDigits[Limit[Derivative[1][DirichletEta][x], x -> 1], 10, 110][[1]] (* Eric W. Weisstein, Jan 08 2024 *)

Euler*log(2)-log(2)^2/2 \\ Charles R Greathouse IV, Mar 28 2012

Equals gamma*log(2) - log(2)^2/2.

Equals -Sum_{k>=1} psi(k)/(k*2^k), where psi(x) is the digamma function. - Amiram Eldar, Sep 12 2022

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)).

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

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

A345928 Decimal expansion of Integral_{x>=0} (zeta(x)-1) dx (negated).

Original entry on oeis.org

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

Offset: 0

Amiram Eldar, Jun 29 2021

Robinson (1980) conjectured and Newman and Widder (1981) proved that this integral is equal to the limit given in the Formula section.

			-0.2432383428909807554150591354654623071783049...

Murray S. Klamkin (ed.), Problems in Applied Mathematics: Selections from SIAM Review, Philadelphia, PA: Society for Industrial and Applied Mathematics, 1990, pp. 281-282.

H. P. Robinson, A Conjectured Limit, Problem 80-7, SIAM Review, Vol. 22, No. 2 (1980), p. 229; D. J. Newman and D. V. Widder, Solution, ibid., Vol. 23, No. 2 (1981), pp. 256-257.

Cf. A099769, A168218.

Equals lim_{n->oo} (Sum_{k=2..n} 1/log(k) - Integral_{x=0..n} (1/log(x)) dx).

cp's OEIS Frontend

A091812 Decimal expansion of Sum_{k>=1} (-1)^k*log(k)/k.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI