Thread: Fibonacci reciprocal sum

Hi.

On Wiki there is one particular formula (actually that last one in Reciprocal sums section) which I can not find out way to prove it.
sum(k=0,n)(1/F(2^k))=3-(F(2^n-1)/F(2^n))

I have tried to expand sum and take it to common denominator and I issued some problems.
Let's say we have n=4 so there is
1/F(1)+1/F(2)+1/F(4)+1/F(8)+1/F(16)
So common denominator is F(16) - that because of rule gcd(F(n),F(m))=F(gcd(n,m)) this means 16 is dividable by 8, 4, 2 and 1. So there is no problem. Now new equation is
(F(16)/F(1)+F(16)/F(2)+F(16)/F(4)+F(16)/F(8)+F(16)/F(16))/F(16)
So F(16)/F(16) is of course 1. By the chance I found out this rule
F(2^n)/F(2^(n-1))=F(2^(n-1)+1)+F(2^(n-1)-1)
By that rule F(16)/F(8) = F(9)+F(7)
Is there any rule like F(2^n)/F(2^m) = ... where m+1<n
For instance I tried to find out relationship between F(16)/F(4) = 329 and I really don't know how to express 329 as some sort of sum of Fibonacci numbers (to produce a rule).

Am I on the right way to prove given equation?
Is there any other property of Fibonacci numbers which I have to apply there? Is there some other thing to do, to prove this?

Thx
Regards Zmeda

Could you please re-post this using LaTeX format? It's really hard to read right now.

From the identity



take

n>0
so


dividing by we get



so

now applying the sum





now add in both sides

taking the limit and using that follows