A344041 Decimal expansion of Sum_{k>=1} F(k)/(k*2^k), where F(k) is the k-th Fibonacci number (A000045).

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

Offset: 0

Author

Amiram Eldar, May 07 2021

Keywords

• nonn
• cons

Comments

This constant is a transcendental number (Adhikari et al., 2001).

A similar series is Sum_{k>=1} F(k)/2^k = 2.

The corresponding series with Lucas numbers (A000032) is Sum_{k>=1} L(k)/(k*2^k) = 2*log(2) (A016627).

In general, for m>=2, Sum_{k>=1} F(k)/(k*m^k) = log(1 - 2*sqrt(5)/(1 + sqrt(5) - 2*m)) / sqrt(5) and Sum_{k>=1} L(k)/(k*m^k) = log(m^2 / (m^2 - m - 1)). - Vaclav Kotesovec, May 08 2021

Examples

			0.86081788192800807777886646590121085084914136508057...
		

Links

• S. D. Adhikari, N. Saradha, T. N. Shorey and R. Tijdeman, Transcendental infinite sums, Indagationes Mathematicae, Vol. 12, No. 1 (2001), pp. 1-14.
• István Mező, Several Generating Functions for Second-Order Recurrence Sequences, Journal of Integer Sequences, Vol. 12 (2009), Article 09.3.7.
• Renzo Sprugnoli, Sums of reciprocals of the central binomial coefficients, INTEGERS 6 (2006) #A27.
• Index entries for transcendental numbers.

Crossrefs

Cf. A000032, A000045, A002163, A002390, A002457, A036289.

Programs

Mathematica

RealDigits[Sum[Fibonacci[n]/n/2^n, {n, 1, Infinity}], 10, 100][[1]]

PARI

suminf(k=1, fibonacci(k)/(k*2^k)) \\ Michel Marcus, May 07 2021

Formula

Equals Sum_{k>=0} (-1)^k/A002457(k).

Equals 4*log(phi)/sqrt(5) = 4*arcsinh(1/2)/sqrt(5) = arccosh(7/2)/sqrt(5) = 4*A002390/A002163.

Equals Integral_{x>=2} 1/(x^2 - x - 1) dx.