- #1

- 35

- 1

_{n}denote the Fibonacci sequence.

u

_{n}is the sequence given by: u

_{n}= F

_{n+1}/F

_{n}. Show that mod(u

_{n}- [itex]\phi[/itex]) [itex]\leq[/itex][itex]\frac{1}{\phi}[/itex]mod(u

_{n-1}-[itex]\phi[/itex]) and therefore mod(u

_{n}- [itex]\phi[/itex]) [itex]\leq[/itex] [itex]\frac{1}{\phi

^{n-1}}[/itex][/itex]mod(u

_{1}-[itex]\phi[/itex]) and then conclude u

_{n}converges to [itex]\phi[/itex]

I have tried with the identity [itex]\phi[/itex] = 1+ [itex]\frac{1}{\phi}[/itex] 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 [itex]\phi[/itex] as n-1 converges to infinity and thus 1/+inf =0, the right side becomes zero and we get mod(u

_{n}- [itex]\phi[/itex]) [itex]\leq[/itex]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 u

_{1}and the part of the [itex]\phi[/itex]

^{n-1}. Can someone shed some light on this issue?