# Ratio of Consecutive Fibonacci Numbers/Proof 1

## 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:

$\phi := \dfrac {1 + \sqrt 5} 2$,
$\hat \phi := \paren {1 - \phi } = \dfrac {1 - \sqrt 5} 2$
$\alpha := \dfrac {\phi}{\hat \phi}$

Then:

 $\ds \alpha$ $=$ $\ds \dfrac {\phi} {\hat \phi}$ $\ds$ $=$ $\ds \dfrac {1 + \sqrt 5} {1 - \sqrt 5}$ $\ds$ $=$ $\ds \dfrac {1 + \sqrt 5} {1 - \sqrt 5} \paren {\dfrac {1 + \sqrt 5} {1 + \sqrt 5} }$ multiplying by $1$ $\ds$ $=$ $\ds \dfrac {6 + 2 \sqrt 5} {-4}$ $\ds$ $=$ $\ds - \dfrac {3 + \sqrt 5} 2$

Recall the Euler-Binet Formula:

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

Let $n \ge 1$.

It immediately follows that:

 $\ds \frac {f_{n + 1} } {f_n}$ $=$ $\ds \dfrac {\phi^{n + 1} - \hat \phi^{n + 1} } {\phi^n - \hat \phi^n}$ $\ds$ $=$ $\ds \dfrac {\paren {\phi^{n + 1} - \phi \hat \phi^n} + \paren {\phi \hat \phi^n - \hat \phi^{n + 1} } } {\phi^n - \hat \phi^n}$ adding $0$ to the numerator $\ds$ $=$ $\ds \dfrac {\phi \paren {\phi^n - \hat \phi^n} + \hat \phi^n \paren {\phi - \hat \phi } } {\phi^n - \hat \phi^n}$ factoring out a $\phi$ and $\hat \phi^n$ in the numerator $\ds$ $=$ $\ds \phi + \dfrac {\hat \phi^n \paren {\phi - \hat \phi} } {\phi^n - \hat \phi^n}$ $\ds$ $=$ $\ds \phi + \dfrac {\hat \phi^n \paren {\phi - \hat \phi} } {\phi^n - \hat \phi^n} \dfrac {\dfrac 1 {\hat \phi^n } } {\dfrac 1 {\hat \phi^n } }$ multiplying by $1$ $\ds$ $=$ $\ds \phi + \dfrac {\paren {\phi - \hat \phi} } {\dfrac {\phi^n} {\hat \phi^n} - 1}$ $\ds$ $=$ $\ds \phi + \dfrac {\sqrt 5} {\alpha^n - 1}$

From the definition of $\alpha$:

$\size \alpha > 1$

Therefore:

 $\ds \lim_{n \mathop \to \infty} \frac {f_{n + 1} } {f_n}$ $=$ $\ds \lim_{n \mathop \to \infty}\ \phi + \dfrac {\sqrt 5} {\alpha^n - 1}$ $\ds$ $=$ $\ds \phi + 0$ $\ds$ $=$ $\ds \phi$

$\blacksquare$