A079586 Decimal expansion of Sum_{k>=1} 1/F(k) where F(k) is the k-th Fibonacci number A000045(k).
3, 3, 5, 9, 8, 8, 5, 6, 6, 6, 2, 4, 3, 1, 7, 7, 5, 5, 3, 1, 7, 2, 0, 1, 1, 3, 0, 2, 9, 1, 8, 9, 2, 7, 1, 7, 9, 6, 8, 8, 9, 0, 5, 1, 3, 3, 7, 3, 1, 9, 6, 8, 4, 8, 6, 4, 9, 5, 5, 5, 3, 8, 1, 5, 3, 2, 5, 1, 3, 0, 3, 1, 8, 9, 9, 6, 6, 8, 3, 3, 8, 3, 6, 1, 5, 4, 1, 6, 2, 1, 6, 4, 5, 6, 7, 9, 0, 0, 8, 7, 2, 9, 7, 0, 4
Offset: 1
Examples
3.35988566624317755317201130291892717968890513373...
References
- Daniel Duverney, Number Theory, World Scientific, 2010, 5.22, pp.75-76.
- Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 358.
Links
- Kenny Lau, Table of n, a(n) for n = 1..10000 (First 1000 terms computed by Joerg Arndt)
- Richard André-Jeannin, Irrationalité de la somme des inverses de certaines suites récurrentes, Comptes Rendus de l'Académie des Sciences - Series I - Mathematics 308:19 (1989), pp. 539-541.
- Richard André-Jeannin, Problem H-450, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 29, No. 1 (1991), p. 89; Comparable, Solution to Problem H-450 by Paul S. Bruckman, ibid., Vol. 30, No. 2 (1992), p. 191-192.
- Richard André-Jeannin, Sequences of Integers Satisfying Recurrence Relations, The Fibonacci Quarterly, Vol. 29, No. 3 (1991), pp. 205-208;
- Joerg Arndt, On computing the generalized Lambert series, arXiv:1202.6525v3 [math.CA], (2012).
- Paul S. Bruckman, Problem B-602, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 25, No. 3 (1987), p. 279; Fibonacci Infinite Series, Solution to Problem B-602 by C. Georghiou, ibid., Vol. 26, No. 3 (1988), pp. 281-282.
- Peter Bundschuh and Keijo Väänänen, Arithmetical investigations of a certain infinite product, Compositio Mathematica, Vol. 91, No. 2 (1994), pp. 175-199.
- Daniel Duverney, Irrationalité de la somme des inverses de la suite Fibonacci, Elemente der Mathematik, Vol. 52, No. 1 (1997), pp. 31-36.
- William Gosper, Acceleration of Series, Artificial Intelligence Memo #304 (1974).
- W. E. Greig, Sums of Fibonacci reciprocals, The Fibonacci Quarterly, Vol. 15, No. 1 (1977), pp. 46-48.
- Peter Griffin, Acceleration of the Sum of Fibonacci Reciprocals, The Fibonacci Quarterly, Vol. 30, No. 2 (1992), pp. 179-181.
- Sarah H. Holliday and Takao Komatsu, On the sum of reciprocal generalized Fibonacci numbers, Integers 11A (2011), Article 11. Alternate link.
- A. F. Horadam, Elliptic functions and Lambert series in the summation of reciprocals in certain recurrence-generated sequences, The Fibonacci Quarterly, Vol. 26, No.2 (May-1988), pp. 98-114.
- Fredrik Johansson, The reciprocal Fibonacci constant.
- Paul Kinlaw, Michael Morris, and Samanthak Thiagarajan, Sums related to the Fibonacci sequence, Husson University (2021).
- Tapani Matala-Aho and Marc Prévost, Quantitative irrationality for sums of reciprocals of Fibonacci and Lucas numbers, Ramanujan J., Vol. 11 (2006), pp. 249-261.
- Z.W. M. Trzaska, Fibonacci Polynomials their Properties and Applications, Z. Anal. Anwend. 15 (1996), no. 3, pp. 729-746.
- Eric Weisstein's World of Mathematics, Reciprocal Fibonacci Constant.
- Wikipedia, Reciprocal Fibonacci constant.
Crossrefs
Programs
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Maple
Digits := 120: c := Pi/2 + I*arccsch(2): Jeannin := n -> sqrt(5/4)*add(I^(1-j)/sin(j*c), j = 1..n): evalf(Jeannin(1000)); # Peter Luschny, Nov 15 2023
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Mathematica
digits = 105; Sqrt[5]*NSum[(-1)^n/(GoldenRatio^(2*n + 1) - (-1)^n), {n, 0, Infinity}, WorkingPrecision -> digits, NSumTerms -> digits] // RealDigits[#, 10, digits] & // First (* Jean-François Alcover, Apr 09 2013 *) First@RealDigits[Sqrt[5]/4 ((Log[5] + 2 QPolyGamma[1, 1/GoldenRatio^4] - 4 QPolyGamma[1, 1/GoldenRatio^2])/(2 Log[GoldenRatio]) + EllipticTheta[2, 0, 1/GoldenRatio^2]^2), 10, 105] (* Vladimir Reshetnikov, Nov 18 2015 *) -
PARI
/* Fast computation without splitting into even and odd indices, see the Arndt reference */ lambert2(x, a, S)= { /* Return G(x,a) = Sum_{n>=1} a*x^n/(1-a*x^n) (generalized Lambert series) computed as Sum_{n=1..S} x^(n^2)*a^n*( 1/(1-x^n) + a*x^n/(1-a*x^n) ) As series in x correct up to order S^2. We also have G(x,a) = Sum_{n>=1} a^n*x^n/(1-x^n) */ return( sum(n=1,S, x^(n^2)*a^n*( 1/(1-x^n) + a*x^n/(1-a*x^n) ) ) ); } inv_fib_sum(p=1, q=1, S)= { /* Return Sum_{n>=1} 1/f(n) where f(0)=0, f(1)=1, f(n) = p*f(n-1) + q*f(n-1) computed using generalized Lambert series. Must have p^2+4*q > 0 */ my(al,be); \\ Note: the q here is -q in the Horadam paper. \\ The following numerical examples are for p=q=1: al=1/2*(p+sqrt(p^2+4*q)); \\ == +1.6180339887498... be=1/2*(p-sqrt(p^2+4*q)); \\ == -0.6180339887498... return( (al-be)*( 1/(al-1) + lambert2(be/al, 1/al, S) ) ); \\ == 3.3598856... } default(realprecision,100); S = 1000; /* (be/al)^S == -0.381966^S == -1.05856*10^418 << 10^-100 */ inv_fib_sum(1,1,S) /* 3.3598856... */ /* Joerg Arndt, Jan 30 2011 */ -
PARI
suminf(k=1, 1/(fibonacci(k))) \\ Michel Marcus, Feb 19 2019
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Sage
m=120; numerical_approx(sum(1/fibonacci(k) for k in (1..10*m)), digits=m) # G. C. Greubel, Feb 20 2019
Formula
Alternating series representation: 3 + Sum_{k >= 1} (-1)^(k+1)/(F(k)*F(k+1)*F(k+2)). - Peter Bala, Nov 30 2013
From Amiram Eldar, Oct 04 2020: (Start)
Equals sqrt(5) * Sum_{k>=0} (1/(phi^(2*k+1) - 1) - 2*phi^(2*k+1)/(phi^(4*(2*k+1)) - 1)), where phi is the golden ratio (A001622) (Greig, 1977).
Equals sqrt(5) * Sum_{k>=0} (-1)^k/(phi^(2*k+1) - (-1)^k) (Griffin, 1992).
From Gleb Koloskov, Sep 14 2021: (Start)
Equals 1 + c1*(c2 + 32*Integral_{x=0..infinity} f(x) dx),
phi = (1+sqrt(5))/2 = A001622,
f(x) = sin(x)*(4+cos(2*x))/((exp(Pi*x/log(phi))-1)*(2*cos(2*x)+3)*(7-2*cos(2*x))) (End)
From Amiram Eldar, Jan 27 2022: (Start)
Equals 3 + 2 * Sum_{k>=1} 1/(F(2*k-1)*F(2*k+1)*F(2*k+2)) (Bruckman, 1987).
Equals 2 + Sum_{k>=1} 1/A350901(k) (André-Jeannin, Problem H-450, 1991).
Equals sqrt(5/4)*Sum_{j>=1} i^(1-j)/sin(j*c) where c = Pi/2 + i*arccsch(2). - Peter Luschny, Nov 15 2023
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