# Ratio of Consecutive Fibonacci Numbers

## 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 1

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:

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

\(\ds \) | \(=\) | \(\ds \phi + 0\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \phi\) |

$\blacksquare$

## Proof 2

From Continued Fraction Expansion of Golden Mean: Successive Convergents, the $n$th convergent of the continued fraction expansion of $\phi$ is:

- $C_n = \dfrac {f_{n + 1} } {f_n}$

The result follows from Continued Fraction Expansion of Irrational Number Converges to Number Itself.

$\blacksquare$

## Historical Note

This result appears first to have been published by Simon Jacob in $1560$, but the work it appeared in was not well-known and subsequent mathematicians appear to have been unaware of it.

This result was observed by Johannes Kepler, but not established rigorously by him:

*It is so arranged that the two lesser terms of a progressive series added together constitute the third ... and so on to infinity, as the same proportion continues unbroken. It is impossible to provide a perfect example in round numbers. However ... Let the smalllest numbers be $1$ and $1$, which you must imagine as unequal. Add them, and the sum will be $2$; add this to $1$, result $3$; add $2$ to this, and get $5$; add $3$, get $8$ ... as $5$ is to $8$, so $8$ is to $13$, approximately, and as $8$ is to $13$, so $13$ is to $21$, approximately.*

However, it was Robert Simson in $1753$ who first explicitly stated that the Ratio of Consecutive Fibonacci Numbers tend to a limit.

## Sources

- 1986: David Wells:
*Curious and Interesting Numbers*... (previous) ... (next): $5$ - 1997: Donald E. Knuth:
*The Art of Computer Programming: Volume 1: Fundamental Algorithms*(3rd ed.) ... (previous) ... (next): $\S 1.2.8$: Fibonacci Numbers - 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $5$ - 1998: David Nelson:
*The Penguin Dictionary of Mathematics*(2nd ed.) ... (previous) ... (next):**Fibonacci sequence**(Fibonacci, 1202) - 1998: David Nelson:
*The Penguin Dictionary of Mathematics*(2nd ed.) ... (previous) ... (next):**golden section** - 2008: David Nelson:
*The Penguin Dictionary of Mathematics*(4th ed.) ... (previous) ... (next):**Fibonacci sequence**(Fibonacci, 1202)