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Fibonacci Sequence converge exercise

  1. Sep 29, 2013 #1
    Let Fn denote the Fibonacci sequence.
    un is the sequence given by: un= Fn+1/Fn. Show that mod(un - [itex]\phi[/itex]) [itex]\leq[/itex][itex]\frac{1}{\phi}[/itex]mod(un-1-[itex]\phi[/itex]) and therefore mod(un - [itex]\phi[/itex]) [itex]\leq[/itex] [itex]\frac{1}{\phin-1}[/itex][/itex]mod(u1-[itex]\phi[/itex]) and then conclude un 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 un 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(un - [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 un-1 to u1 and the part of the [itex]\phi[/itex]n-1. Can someone shed some light on this issue?
     
  2. jcsd
  3. Sep 29, 2013 #2

    pasmith

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    Homework Helper

    We have
    [tex]
    |u_n - \phi| \leq \frac{1}{\phi} |u_{n-1} - \phi|
    [/tex]
    But since this is true for all n, we also have
    [tex]
    |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|
    [/tex]
    and we can clearly keep going, picking up a factor of [itex]\phi^{-1}[/itex] each time, until we have a multiple of [itex]|u_1 - \phi|[/itex] on the right.
     
  4. Sep 29, 2013 #3
    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 :)
     
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