# Fibonacci Sequence converge exercise

1. Sep 29, 2013

### Calabi_Yau

Let Fn denote the Fibonacci sequence.
un is the sequence given by: un= Fn+1/Fn. Show that mod(un - $\phi$) $\leq$$\frac{1}{\phi}$mod(un-1-$\phi$) and therefore mod(un - $\phi$) $\leq$ $\frac{1}{\phin-1}$[/itex]mod(u1-$\phi$) and then conclude un converges to $\phi$

I have tried with the identity $\phi$ = 1+ $\frac{1}{\phi}$ if anything came to light... And I tried dividing the mods but it got even more complicated.

I can prove from the seocnd equation that un converges to $\phi$ as n-1 converges to infinity and thus 1/+inf =0, the right side becomes zero and we get mod(un - $\phi$) $\leq$0 which is the definiton of convergence. But I can't get from the first equation to the second. I don't know how to pass from un-1 to u1 and the part of the $\phi$n-1. Can someone shed some light on this issue?

2. Sep 29, 2013

### pasmith

We have
$$|u_n - \phi| \leq \frac{1}{\phi} |u_{n-1} - \phi|$$
But since this is true for all n, we also have
$$|u_n - \phi| \leq \frac{1}{\phi} |u_{n-1} - \phi| \leq \frac{1}{\phi} \frac{1}{\phi} |u_{n-2} - \phi| = \frac{1}{\phi^2} |u_{n-2} - \phi|$$
and we can clearly keep going, picking up a factor of $\phi^{-1}$ each time, until we have a multiple of $|u_1 - \phi|$ on the right.

3. Sep 29, 2013

### Calabi_Yau

Hmm, I didn't see that pattern. That is the same as making n-1=p and then doing the same for up and up-1.

Nice, thank you :)