Ratio of Consecutive Fibonacci Numbers/Proof 3

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Theorem

For $n \in \N$, let $f_n$ be the $n$th Fibonacci number.


Then:

$\ds \lim_{n \mathop \to \infty} \frac {f_{n + 1} } {f_n} = \phi$

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


Proof

Let:

\(\ds a_n\) \(:=\) \(\ds \dfrac {f_{n + 1} } {f_n}\)
\(\ds \map g x\) \(:=\) \(\ds 1 + \dfrac 1 x\)

Then:

\(\ds \map g {a_n}\) \(=\) \(\ds 1 + \dfrac {f_n} {f_{n + 1} }\)
\(\ds \) \(=\) \(\ds \dfrac {f_{n + 1} + f_n} {f_{n + 1} }\)
\(\ds \) \(=\) \(\ds \dfrac {f_{n + 2} } {f_{n + 1 } }\) Definition of Fibonacci Number
\(\ds \) \(=\) \(\ds a_{n + 1}\)


This sequence represents the iterated dynamical system $a_{n + 1} = \map g {a_n}$ with initial condition $a_1 = 1$.


We have that:

$\size {\map {g'} \phi} = \dfrac 1 {\phi^2} < 1$


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Hence the golden mean $\phi$ is an attracting fixed point of $g$.


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Additionally, since the interval $I := \closedint {\dfrac 3 2} 2$ is positively invariant under $g$


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and the derivative of $g$ is bounded above by $\dfrac 1 {\paren {3 / 2}^2} < 1$,


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then $g$ is a contraction mapping on $I$.

Thus, by the Banach Fixed-Point Theorem, $I$ is in the region of attraction of $\phi$.


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Because $a_2 = 2 \in I$ is in the orbit of the dynamical system, the iterations converge to $\phi$.

Therefore,

$\ds \lim_{n \mathop \to \infty} \dfrac {f_{n + 1} } {f_n} = \lim_{n \mathop \to \infty} a_n = \phi$

$\blacksquare$

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