A085361 Decimal expansion of the number c = Sum_{n>=1} (zeta(n+1)-1)/n.
7, 8, 8, 5, 3, 0, 5, 6, 5, 9, 1, 1, 5, 0, 8, 9, 6, 1, 0, 6, 0, 2, 7, 6, 3, 2, 3, 4, 5, 4, 5, 5, 4, 6, 6, 6, 4, 7, 2, 7, 4, 9, 6, 6, 8, 2, 2, 3, 2, 8, 1, 6, 4, 9, 7, 5, 5, 1, 5, 6, 4, 0, 2, 3, 0, 1, 7, 8, 0, 6, 4, 3, 5, 6, 3, 3, 0, 1, 6, 2, 2, 8, 7, 4, 7, 1, 5, 9, 2, 1, 3, 3, 2, 2, 4, 3, 1, 9, 6, 7, 5, 6
Offset: 0
Examples
0.78853056591150896106027632345455466647274966822328164975515640230178...
References
- Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 1.8.1 Alternative representations [of real numbers], p. 62.
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Steven R. Finch, Mathematical Constants II, Encyclopedia of Mathematics and Its Applications, Cambridge University Press, Cambridge, 2018, p. 528 and 538.
- Sofia Kalpazidou, Khintchine's constant for Lüroth representation, Journal of Number Theory, Vol. 29, No. 2 (June 1988), pp. 196-205.
- Eric Weisstein's World of Mathematics, Alladi-Grinstead Constant.
- Eric Weisstein's World of Mathematics, Convergence Improvement.
Programs
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Magma
SetDefaultRealField(RealField(120)); L:=RiemannZeta(); (&+[(Evaluate(L,n+1)-1)/n: n in [1..1000]]); // G. C. Greubel, Nov 15 2018
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Maple
evalf(sum((Zeta(n+1)-1)/n, n=1..infinity), 120); # Vaclav Kotesovec, Dec 11 2015 evalf(Sum(-(-1)^k*Zeta(1, k), k = 2..infinity), 120); # Vaclav Kotesovec, Jun 18 2021
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Mathematica
Sum[(-1+Zeta[1+n])/n,{n,Infinity}] NSum[Log[k]/(k*(k+1)), {k, 1, Infinity}, WorkingPrecision -> 120, NSumTerms ->5000, Method -> {NIntegrate, MaxRecursion -> 100}] (* Vaclav Kotesovec, Dec 11 2015 *) -
PARI
suminf(n=1,(zeta(n+1)-1-2^(-n-1))/n)+log(2)/2 \\ Charles R Greathouse IV, Feb 20 2012
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PARI
sumalt(k=2, -(-1)^k * zeta'(k)) \\ Vaclav Kotesovec, Jun 17 2021
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Sage
import mpmath mpmath.mp.pretty=True; mpmath.mp.dps=108 #precision mpmath.nsum(lambda n: (-1+mpmath.zeta(1+n))/n, [1,mpmath.inf]) # Peter Luschny, Jul 14 2012
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Sage
numerical_approx(sum((zeta(k+1)-1)/k for k in [1..1000]), digits=120) # G. C. Greubel, Nov 15 2018
Formula
Equals Sum_{n>=2} log(n/(n-1))/n = Sum_{n>=1, k>=2} 1/(n*k^(n+1)). [From Mathworld links]
Equals -Sum_{k>=2} (-1)^k * zeta'(k). - Vaclav Kotesovec, Jun 17 2021
Equals log(A245254) = Sum_{k>=1} log(k)/(k*(k+1)). - Amiram Eldar, Jun 27 2021
Equals -log(A242624). - Amiram Eldar, Feb 06 2022
Comments