A214369 Decimal expansion of Sum_{n>=1} 1/(3^n-1).
6, 8, 2, 1, 5, 3, 5, 0, 2, 6, 0, 5, 2, 3, 8, 0, 6, 6, 7, 6, 1, 2, 6, 3, 1, 8, 6, 2, 2, 6, 6, 2, 4, 0, 0, 9, 6, 4, 9, 1, 9, 0, 2, 4, 8, 3, 2, 6, 9, 0, 3, 4, 1, 9, 2, 2, 8, 2, 5, 7, 8, 4, 7, 1, 3, 6, 7, 7, 1, 8, 3, 4, 7, 7, 4, 1, 7, 8, 7, 3, 2, 9, 0, 0, 9, 6, 2, 1, 2, 6, 9, 0, 3, 0, 4, 5, 3, 3, 1, 3, 7, 5, 0, 3, 2
Offset: 0
Examples
Equals 0.6821535026052380667...
Links
- G. C. Greubel, Table of n, a(n) for n = 0..10000
Programs
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Maple
evalf(sum(1/(3^k-1), k=1..infinity), 120); # Vaclav Kotesovec, Oct 18 2014 # second program with faster converging series evalf( add( (1/3)^(n^2)*(1 + 2/(3^n - 1)), n = 1..14 ), 105); # Peter Bala, Jan 30 2022
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Mathematica
RealDigits[ NSum[1/(3^n - 1), {n, 1, Infinity}, WorkingPrecision -> 110, NSumTerms -> 100], 10, 105] // First (* or *) 1 - (Log[2] + QPolyGamma[0, 1, 1/3])/Log[3] // RealDigits[#, 10, 105]& // First (* Jean-François Alcover, Jun 05 2013 *) x = 1/3; RealDigits[ Sum[ DivisorSigma[0, k] x^k, {k, 1000}], 10, 105][[1]] (* Robert G. Wilson v, Oct 12 2014 after an observation and the formula of Amarnath Murthy, see A073668 *) -
PARI
suminf(n=1, 1/(3^n-1)) \\ Michel Marcus, Mar 11 2017
Formula
Equals Sum_{n>=1} 1/A024023(n).
Equals Sum_{k>=1} d(k)/3^k, where d(k) is the number of divisors of k (A000005). - Amiram Eldar, May 17 2020
Extensions
More terms from Jean-François Alcover, Feb 12 2013