# [seqfan] fast computation of sum(n=1, oo, 1/fibonacci(n))

Joerg Arndt arndt at jjj.de
Sat Nov 27 10:14:55 CET 2010

```Regarding seq A079586 "Decimal expansion of sum(k>=1,1/F(k)) ..."

Relation (4.17) on p.108 in
{Elliptic functions and Lambert series in the
summation of reciprocals in certain recurrence-generated sequences},
The Fibonacci Quarterly, vol.26, no.2, pp.98-114, (May-1988).
URL: http://www.fq.math.ca/26-2.html
gives an expression in terms of the 'generalized' Lambert series
L(a,x)=sum(n=1,N, a*x^n/(1-a*x^n))  \\ N==infty
The only missing morsel is a fast way to compute this one.
Now I observed the following identity for accelerated computation:
L(a,x)=sum(n=1,S, x^(n^2)*a^n*( 1/(1-x^n) + a*x^n/(1-a*x^n) ) )

The details:
let u_0 = 0, u_1=1, u_{n+2} = p * u_{n+1} - q * u_{n}
(where p^2>4*q), set
al=1/2*(p+sqrt(p^2-4*q))  \\ == +1.6180339887498...
be=1/2*(p-sqrt(p^2-4*q))  \\ == -0.6180339887498...
then
(al-be)*( 1/(al-1) + L(be/q, be^2/q) ) == 3.359885666...
(this is == (al-be)*( 1/(al-1) + L(1/al, be/al) ) )

(btw. p=3, q=+1, gives 1.53537... == sum(k=1,infty, 1/fib(2*n))
which is A153386)

The fast methods to computed the even (A153386) an odd (A153387) part
separately (in terms of Lambert and Theta) are described at
http://mathworld.wolfram.com/ReciprocalFibonacciConstant.html
(see the references).

A (very special) identity for sum(n=1, oo, 1/fibonacci(n))
was given by Gosper, see p.66 of
ftp://publications.ai.mit.edu/ai-publications/pdf/AIM-304.pdf

I didn't put comments into A079586 because I haven't checked
how far one needs to sum to get N digits
(shouldn't be rocket science, though).

```