Ratio of Consecutive Fibonacci Numbers

From ProofWiki
Jump to navigation Jump to search

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

Retrieved from "https://proofwiki.org/w/index.php?title=Ratio_of_Consecutive_Fibonacci_Numbers&oldid=549015"