Euler-Binet Formula/Corollary 2

Corollary to Euler-Binet Formula
For even $n$:
 * $F_n < \dfrac {\phi^n} {\sqrt 5}$

For odd $n$:
 * $F_n > \dfrac {\phi^n} {\sqrt 5}$

where:
 * $F_n$ denotes the $n$th Fibonacci number
 * $\phi$ denotes the golden mean.

Proof
From Euler-Binet Formula:


 * $F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5} = \dfrac {\phi^n} {\sqrt 5} - \dfrac {\hat \phi^n} {\sqrt 5}$

We have that:
 * $\hat \phi^n = -\dfrac 1 \phi$

and so:
 * $\hat \phi^n = \left({-1}\right)^n \dfrac 1 {\phi^n}$

$\phi > 0$ and so $\dfrac 1 {\phi^n} > 0$ for all $n$.

For even $n$:
 * $\left({-1}\right)^n = 1$

and so:
 * $\hat \phi^n > 0$

Thus:
 * $\dfrac {\phi^n} {\sqrt 5} - \dfrac {\hat \phi^n} {\sqrt 5} < \dfrac {\phi^n} {\sqrt 5}$

and so $F_n < \dfrac {\phi^n} {\sqrt 5}$.

For odd $n$:
 * $\left({-1}\right)^n = -1$

and so:
 * $\hat \phi^n < 0$

Thus:
 * $\dfrac {\phi^n} {\sqrt 5} - \dfrac {\hat \phi^n} {\sqrt 5} > \dfrac {\phi^n} {\sqrt 5}$

and so $F_n > \dfrac {\phi^n} {\sqrt 5}$.