Fibonacci Sequence converge exercise

[Calabi_Yau]

Let F_n denote the Fibonacci sequence.
u_n is the sequence given by: u_n= F_n+1/F_n. Show that mod(u_n - \phi) \leq\frac{1}{\phi}mod(u_n-1-\phi) and therefore mod(u_n - \phi) \leq \frac{1}{\phi^n-1}[/itex]mod(u₁-\phi) and then conclude u_n 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 u_n converges to \phi as n-1 converges to infinity and thus 1/+inf =0, the right side becomes zero and we get mod(u_n - \phi) \leq0 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 u_n-1 to u₁ and the part of the \phi^n-1. Can someone shed some light on this issue?

 

[pasmith]

Let F_n denote the Fibonacci sequence.
u_n is the sequence given by: u_n= F_n+1/F_n. Show that mod(u_n - \phi) \leq\frac{1}{\phi}mod(u_n-1-\phi) and therefore mod(u_n - \phi) \leq \frac{1}{\phi^n-1}[/itex]mod(u₁-\phi) and then conclude u_n 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 u_n converges to \phi as n-1 converges to infinity and thus 1/+inf =0, the right side becomes zero and we get mod(u_n - \phi) \leq0 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 u_n-1 to u₁ and the part of the \phi^n-1. Can someone shed some light on this issue?

We have
<br /> |u_n - \phi| \leq \frac{1}{\phi} |u_{n-1} - \phi| <br />
But since this is true for all n, we also have
<br /> |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|<br />
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.

 

[Calabi_Yau]

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

Nice, thank you :)