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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Although this article appears correct, it's inelegant. There has to be a better way of doing it.
In particular: Banach Fixed-Point Theorem only asserts the uniqueness of the fix point $\phi$, nothing about its attracting property.
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Because $a_2 = 2 \in I$ is in the orbit of the dynamical system, the iterations converge to $\phi$.

Therefore,

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

$\blacksquare$