Why is $\phi^x=\underset{n\to\infty}{\lim} \frac{F_n}{F_{n-x}}$?

Question

The function $F_n$ denotes the nth Fibonacci number and $\phi$ is the golden ratio $\frac{1+\sqrt{5}}{2}$. I found this while trying to create a fun math puzzle. Is there a name for this? Also, how do you prove it?

For anyone wondering, the puzzle was going to be something along the lines of:

Let $F_n$ be the Fibonacci Sequence for all integers $n$ and let $f(x)=\underset{n\to\infty}{\lim} \frac{F_n}{F_{n-x}}$. Find the exact value of $a$ if $f(x)=a^x$.

It's essentially been known since the time of Kepler. In modern terminology you would call it a consequence of Binet's formula. https://en.wikipedia.org/wiki/Fibonacci_number#Limit_of_consecutive_quotients — Sassatelli Giulio, Feb 12 '23 at 20:26
Sorry, I didn't intend for my saying that I found it to mean that I thought it didn't exist, just that I didn't find it on google or something. Thanks for fixing the question. — Dylan Levine, Feb 12 '23 at 20:44
Does this answer your question? Fibonacci numbers and golden ratio: $\Phi = \lim \sqrt[n]{F_n}$ — Jyrki Lahtonen, Feb 13 '23 at 10:16

score 1 · Accepted Answer · answered Feb 12 '23 at 20:29

1

For integer $m$ you have $\dfrac{F_{n+m}}{F_n}= \prod_{k=1}^m \dfrac{F_{n+k}}{F_{n+k-1}}\to \prod_{k=1}^m \phi $ as $n\to \infty$, using $\dfrac{F_{n+1}}{F_n}\to \phi$.

answered Feb 12 '23 at 20:29

Dosidis

1,255

Why is $\phi^x=\underset{n\to\infty}{\lim} \frac{F_n}{F_{n-x}}$?

1 Answers1