
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^n1)/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^(n1))=F(2^(n1)+1)+F(2^(n1)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 repost this using LaTeX format? It's really hard to read right now.

From the identity
$\displaystyle F(t+p)F(p1)=F(p)F(t+p1)+(1)^p F(t)$
take
$\displaystyle p=t=2^n$ n>0
so
$\displaystyle F(2^{n+1})F(2^n1)F(2^n)F(2^{n+1}1)= F(2^n)$
dividing by $\displaystyle f(2^n).f(2^{n+1}) $ we get
$\displaystyle \frac{F(2^n1)}{F(2^n)}\frac{F(2^{n+1}1)}{F(2^{n+1})}= \frac{1}{F(2^{n+1})}$
so
$\displaystyle  \Delta \frac{F(2^k1)}{F(2^k)}=\frac{1}{F(2^{k+1})}$
now applying the sum $\displaystyle \sum\limits^{n1}_{k=1} $
$\displaystyle \sum\limits^{n1}_{k=1}\frac{1}{F(2^{k+1})}= \frac{F(2^k1)}{F(2^k)}\bigg^{n}_{1}= \frac{F(2^{n}1)}{F(2^{n})} + \frac{F(1)}{F(2)}=1\frac{F(2^{n}1)}{F(2^{n})}$
$\displaystyle \sum\limits^{n}_{k=2}\frac{1}{F(2^{k})}=1\frac{F(2^{n}1)}{F(2^{n})}$
now add $\displaystyle \frac{1}{F(1)} +\frac{1}{F(2)}=2$ in both sides
$\displaystyle \sum\limits^{n}_{k=0}\frac{1}{F(2^{k})}=3\frac{F(2^{n}1)}{F(2^{n})}$
taking the limit and using that $\displaystyle \lim \frac{F(n+1)}{F(n)} =\frac{1+\sqrt{5}}{2}$ follows
$\displaystyle \sum\limits^{\infty}_{k=0}\frac{1}{F(2^{k})}=\frac {7\sqrt{5}}{2}$