Fibonacci Number greater than Golden Section to Power less Two

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Theorem

For all $n \in \N_{\ge 2}$:

$F_n \ge \phi^{n - 2}$

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 $n \in \N_{\ge 2}$, let $\map P n$ be the proposition:

$F_n \ge \phi^{n - 2}$


Basis for the Induction

$\map P 2$ is true, as this just says:

$F_2 = 1 = \phi^0 = \phi^{2 - 2}$

It is also necessary to demonstrate $\map P 3$ is true:

$F_3 = 2 \ge \dfrac {1 + \sqrt 5} 2 = \phi = \phi^{3 - 2}$

This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $\map P j$ is true for all $2 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$F_j \ge \phi^{j - 2}$ for $2 \le j \le k$


from which it is to be shown that:

$F_{k + 1} \ge \phi^{k - 1}$


Induction Step

This is the induction step:

As we have already shown $\map P 2$ and $\map P 3$, we just need to prove the result for $k \ge 3$.

\(\ds F_{k + 1}\) \(=\) \(\ds F_{k - 1} + F_k\) Definition of Fibonacci Numbers
\(\ds \) \(\ge\) \(\ds \phi^{k - 3} + \phi^{k - 2}\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \phi^{k - 3} \paren {1 + \phi}\)
\(\ds \) \(=\) \(\ds \phi^{k - 3} \paren {\phi^2}\) Square of Golden Mean equals One plus Golden Mean
\(\ds \) \(=\) \(\ds \phi^{k - 1}\)

So $\map P 2 \land \map P 3 \land \dots \land \map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction.


Therefore:

$\forall n \in \N_{\ge 2}: F_n \ge \phi^{n - 2}$

$\blacksquare$


Also see


Sources

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