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 $n \geq 1$):

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac {f_{n + 1} } {f_n}\)	\(=\)	\(\displaystyle \dfrac {\phi^{n + 1} - \hat \phi^{n + 1} } {\phi^n - \hat \phi^n}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \dfrac {\left(\phi^{n + 1} - \phi \hat \phi^n\right) + \left(\phi \hat \phi^n - \hat \phi^{n + 1}\right)} {\phi^n - \hat \phi^n}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \phi + \dfrac {\hat \phi^n\left(\phi - \hat \phi\right)} {\phi^n - \hat \phi^n}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \phi + \dfrac {\sqrt 5} {\alpha^n - 1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

From the definition of $\alpha$, we observe that $|\alpha| > 1$. Therefore, we conclude:

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

$\blacksquare$