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Consider the sequence $(a_n)$ with $a_n = F_{n+1}/F_n$ for $n \in \Bbb N$, where $F_n$ are the Fibonacci numbers. Show that this sequence converges to $\phi =(\sqrt{5}+1)/2$.

Can someone help me? The hint they give me is to find two bounding sequence, but I don't understand how this could help me

Thanks

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Hint. The $n$-th Fibbonacci number is given by,

$$F_n=\frac{\phi^n-(-\phi)^{-n}}{\sqrt{5}},$$

where $\phi=\frac{1+\sqrt{5}}{2}$ is the golden ratio.

Then,

$$\begin{align} \frac{F_{n+1}}{F_n} &=\frac{\phi^{n+1}-(-\phi)^{-n-1}}{\phi^n-(-\phi)^{-n}}\\ &=\frac{\phi^{n+1}-(-1)^{-n-1}\phi^{-n-1}}{\phi^n-(-1)^{-n}\phi^{-n}}\\ &=\frac{\phi-(-1)^{-n-1}\phi^{-2n-1}}{1-(-1)^{-n}\phi^{-2n}}. \end{align}$$

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You can prove the following facts, then conclude (bounding sequences appear after step 3):

  1. $\phi$ satisfies $\phi = 1 + \dfrac{1}{\phi}$,
  2. $a_{n+1} = 1 + \dfrac{1}{a_n}$ for all $n$,
  3. deduce from 1 and 2 that $|a_{n+1}-\phi| \leq \frac{1}{\phi}|a_n - \phi|$ for all $n$.
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  • $\begingroup$ Can you be more precise? I don't understand last point $\endgroup$ – geek4079 Aug 29 '14 at 13:15
  • $\begingroup$ @user3685627: done. Did you already prove the first two facts? $\endgroup$ – Siméon Aug 29 '14 at 13:18
  • $\begingroup$ Well the second point It's kinda of obvious cause Fibonacci is defined for n>=2 $\endgroup$ – geek4079 Aug 29 '14 at 13:22
  • $\begingroup$ @user3685627: this is not what "bounded sequence" mean. $\endgroup$ – Siméon Aug 29 '14 at 13:24
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    $\begingroup$ @user3685627: think, try, fail, think, try, fail,...,think, try, understand. $\endgroup$ – Siméon Aug 29 '14 at 13:33

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