Ratio of Consecutive Fibonacci Numbers

Theorem
For $n \in \N$, let $f_n$ be the $n^{\text{th}}$ Fibonacci number.

Denote by $\phi$ the golden mean (so $\phi = \dfrac {1 + \sqrt 5} 2$).

Then $\displaystyle \lim_{n \to \infty} \frac {f_{n + 1}} {f_n} = \phi$.

Proof
Denote $\phi = \dfrac {1 + \sqrt 5} 2$, $\hat \phi = \dfrac {1 - \sqrt 5} 2$ and $\alpha = \dfrac {\phi} {\hat \phi} = - \dfrac {3 + \sqrt 5} {2}$.

The Euler-Binet formula gives us that:
 * $f_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$

It immediately follows that we have (assume $n \geq 1$):

From the definition of $\alpha$, we observe that $|\alpha| > 1$. Therefore, we conclude:
 * $\displaystyle \lim_{n \to \infty} \frac {f_{n + 1}} {f_n} = \lim_{n \to \infty}\ \phi + \dfrac {\sqrt 5} {\alpha^n - 1} = \phi$