1. ## 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

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

3. From the identity

$F(t+p)F(p-1)=F(p)F(t+p-1)+(-1)^p F(t)$

take

$p=t=2^n$ n>0
so

$F(2^{n+1})F(2^n-1)-F(2^n)F(2^{n+1}-1)= F(2^n)$
dividing by $f(2^n).f(2^{n+1})$ we get

$\frac{F(2^n-1)}{F(2^n)}-\frac{F(2^{n+1}-1)}{F(2^{n+1})}= \frac{1}{F(2^{n+1})}$

so
$- \Delta \frac{F(2^k-1)}{F(2^k)}=\frac{1}{F(2^{k+1})}$
now applying the sum $\sum\limits^{n-1}_{k=1}$

$\sum\limits^{n-1}_{k=1}\frac{1}{F(2^{k+1})}=- \frac{F(2^k-1)}{F(2^k)}\bigg|^{n}_{1}=- \frac{F(2^{n}-1)}{F(2^{n})} + \frac{F(1)}{F(2)}=1-\frac{F(2^{n}-1)}{F(2^{n})}$

$\sum\limits^{n}_{k=2}\frac{1}{F(2^{k})}=1-\frac{F(2^{n}-1)}{F(2^{n})}$

now add $\frac{1}{F(1)} +\frac{1}{F(2)}=2$ in both sides
$\sum\limits^{n}_{k=0}\frac{1}{F(2^{k})}=3-\frac{F(2^{n}-1)}{F(2^{n})}$
taking the limit and using that $\lim \frac{F(n+1)}{F(n)} =\frac{1+\sqrt{5}}{2}$ follows
$\sum\limits^{\infty}_{k=0}\frac{1}{F(2^{k})}=\frac {7-\sqrt{5}}{2}$