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:

$\ds \phi$

$:=$

$\ds \dfrac {1 + \sqrt 5} 2$

$\ds \hat \phi$

$:=$

$\ds \paren {1 - \phi}$

$\ds = \dfrac {1 - \sqrt 5} 2$

$\ds \alpha$

$:=$

$\ds \dfrac {\phi} {\hat \phi}$

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 top and bottom by $1 + \sqrt 5$

$\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 and subtracting $\phi \hat \phi^n$ 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 $\phi$ 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 top and bottom by $\dfrac 1 {\hat \phi^n }$

$\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.

$\blacksquare$

Proof 3

Let:

$\ds a_n$

$:=$

$\ds \dfrac {f_{n + 1} } {f_n}$

$\ds \map g x$

$:=$

$\ds 1 + \dfrac 1 x$

Then:

$\ds \map g {a_n}$

$=$

$\ds 1 + \dfrac {f_n} {f_{n + 1} }$

$\ds $

$=$

$\ds \dfrac {f_{n + 1} + f_n} {f_{n + 1} }$

$\ds $

$=$

$\ds \dfrac {f_{n + 2} } {f_{n + 1 } }$

Definition of Fibonacci Number

$\ds $

$=$

$\ds a_{n + 1}$

This sequence represents the iterated dynamical system $a_{n + 1} = \map g {a_n}$ with initial condition $a_1 = 1$.

We have that:

$\size {\map {g'} \phi} = \dfrac 1 {\phi^2} < 1$

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Hence the golden mean $\phi$ is an attracting fixed point of $g$.

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Additionally, since the interval $I := \closedint {\dfrac 3 2} 2$ is positively invariant under $g$

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and the derivative of $g$ is bounded above by $\dfrac 1 {\paren {3 / 2}^2} < 1$,

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then $g$ is a contraction mapping on $I$.

Thus, by the Banach Fixed-Point Theorem, $I$ is in the region of attraction of $\phi$.

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Although this article appears correct, it's inelegant. There has to be a better way of doing it.
In particular: Banach Fixed-Point Theorem only asserts the uniqueness of the fix point $\phi$, nothing about its attracting property.
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Because $a_2 = 2 \in I$ is in the orbit of the dynamical system, the iterations converge to $\phi$.

Therefore,

$\ds \lim_{n \mathop \to \infty} \dfrac {f_{n + 1} } {f_n} = \lim_{n \mathop \to \infty} a_n = \phi$

$\blacksquare$

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 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)
2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): golden section