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Assume we are given any number a. Write it in the form $a = \frac{b}{c}$ (if rational, in the usual way, if irrational, use forms like $\frac{a}{1}$).

Construct a sequence $\frac{b}{c},\frac{c}{b+c},\frac{b+c}{b+2 c},\frac{b+2c}{2b+3c}...$

The nth term is $\frac{F_{n-1}*b+F_n*c}{F_n*b+F_{n+1}*c}$ where $F_n$ is the nth Fibonacci number.

I have noticed that no matter what the starting point $a$ is, the sequence always converges to $\phi-1$ , ie. $\frac{(\sqrt{5}+1)}{2}-1$ or $\frac{(\sqrt{5}-1)}{2}$.

I would like to prove this. If $a=1$, then the nth term is $F_{n+1}/F_{n+2}$ and the limit is $\phi-1$. But If $a \not= 1$ this seems harder to show. Especially if $a < 0$.

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Do you mean "numerator"? –  blue Feb 21 '13 at 20:02

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up vote 2 down vote accepted

Given $\displaystyle\frac{F_{n+1}}{F_n}\to\varphi$ and $\displaystyle\frac{F_{n-1}}{F_n}\to\varphi^{-1}$, your sequence converges as

$$\frac{1/F_n}{1/F_n}\cdot\frac{F_{n-1}b+F_nc}{F_nb+F_{n+1}c}=\frac{b\displaystyle\frac{F_{n-1}}{F_n}+c}{\displaystyle b+c\frac{F_{n+1}}{F_n}}\longrightarrow \frac{b\varphi^{-1}+c}{b+c\varphi}=\varphi^{-1}\frac{b+c\varphi}{b+c\varphi}=\varphi^{-1}.$$

Note that $\varphi^{-1}=\varphi-1$.

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Very good, thanks. I think the last number should be $\phi^{-1}$. But I got that. –  Valtteri Feb 21 '13 at 20:16

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