Ratio of Consecutive Fibonacci Numbers/Proof 1
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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:
| \(\ds \lim_{n \mathop \to \infty} \frac {f_{n + 1} } {f_n}\) | \(=\) | \(\ds \lim_{n \mathop \to \infty}\ \phi + \dfrac {\sqrt 5} {\alpha^n - 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \phi + 0\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \phi\) |
$\blacksquare$