Fibonacci Number less than Golden Section to Power less One

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Theorem

For all $n \in \N_{> 0}$:

$F_n \le \phi^{n - 1}$

where:

$F_n$ is the $n$th Fibonacci number
$\phi$ is the golden section: $\phi = \dfrac {1 + \sqrt 5} 2$


Proof

The proof proceeds by induction.

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

$F_n \le \phi^{n - 1}$


Basis for the Induction

$\map P 1$ is true, as this just says:

$F_1 = 1 = \phi^0 = \phi^{1 - 1}$

It is also necessary to demonstrate $\map P 2$ is true:

$F_2 = 1 \le \dfrac {1 + \sqrt 5} 2 = \phi = \phi^{2 - 1}$

This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $\map P k$ is true, for all $1 \le k \le n$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$F_k \le \phi^{k - 1}$


from which it is to be shown that:

$F_{k + 1} \le \phi^k$


Induction Step

This is the induction step:


\(\ds F_{k + 1}\) \(=\) \(\ds F_{k - 1} + F_k\) Definition of Fibonacci Numbers
\(\ds \) \(\le\) \(\ds \phi^{k - 2} + \phi^{k - 1}\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \phi^{k - 2} \paren {1 + \phi}\)
\(\ds \) \(=\) \(\ds \phi^{k - 2} \paren {\phi^2}\) Square of Golden Mean equals One plus Golden Mean
\(\ds \) \(=\) \(\ds \phi^k\)

So $\map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction.


Therefore:

$\forall n \in \N_{> 0}: F_n \le \phi^{n - 1}$

$\blacksquare$


Also see


Sources

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