Sum of reciprocals of Fibonacci series

By Maria Jose Leon-Sotelo Esteban (T3819) on Sunday, December 24, 2000 - 09:39 am :

We need some information about this sum:

1+1+1/2+1/3+1/5+1/8+1/13+1/21+1/34+...+1/F(n) for n=1 to infinity, where F(n) are the terms of a Fibonacci series.

Thank you. Maria Jose.

By Michael Doré (Md285) on Sunday, January 7, 2001 - 03:00 pm :

Maria,

I think there's no closed form for this. I have tried to calculate it using the formula:

$begin{displaymath}frac{1}{sqrt{5}}Bigl(frac{1+sqrt{5}}{2}Bigr)^2 - frac{1}{sqrt{5}}Bigl(frac{1-sqrt{5}}{2}Bigr)^2 end{displaymath}$

but with no luck at all. Numerically the value is about 3.359885666243177. If you plug this into the inverse symbolic calculator it only returns "1/Fibonacci(n)" which probably means that there is no simpler way to write it.

It is easy to prove that the series converges. The nth Fibonacci number is asymptotic to (the golden ratio)ⁿ . The golden ratio > 1, so the ratio between consecutive terms in the 1/F_n sequence converges to a positive limit which is lower than 1. Therefore the infinite sum converges.

Yours,

Michael