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 - $\phi$) $\leq$ [itex]\frac{1}{\phi^n-1}[/itex][/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 - $\phi$) $\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 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 - $\phi$) $\leq$ [itex]\frac{1}{\phi^n-1}[/itex][/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 - $\phi$) $\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 u_n-1 to u₁ and the part of the $\phi$^n-1. Can someone shed some light on this issue?

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.

 

[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 :)