Prove that the limit of two consecutive fibonacci numbers EXISTS.

Question

Using the definition of Fibonacci numbers, $F_n=F_{n-1}+F_{n-2}$, I can prove that the limit of $\frac{F_{n+1}}{F_n}$ as $n\to\infty$ is $\phi$ if we assume that the limit exists.

How can we prove that the limit does in fact exist?

Is there more than one method?

I do not think this is a duplicate. Please can someone explicitly show how to split the odd and even terms of this ratio sequence into two sequences- one monotonic increasing and one monotonic decreasing- and given that all ratios are between 1 and 2, show that the limit exists and we do not oscillate forever.

Most of the question has been answered.

I have shown that there are two subsequences- one increasing and bounded above by 2 and one decreasing, bounded below by 1. Using the fact that the limit exists, I can show it has value $\phi$. But how can I show the limit is the same for both subsequences?!

I would start with the fact that the limit is obviously bounded above by $2$ (can you see this?), and show it is monotonic. — preferred_anon, Feb 23 '14 at 21:12
Daniel Littlewood- thank you. The only problem is that the sequence of ratios oscillates to the limit, right? If I could say it was monotonic increasing and bounded above then I would know what to do, or monotonic decreasing and bounded below, but that isn't the case.
Andre Nicolas- Thanks for that answer. The Binet formula is quite incredible! I've seen that proof and would like to use that as a second solution. I would also really like to be able to show the limit exists using only real analysis if possible. — Guest, Feb 23 '14 at 21:17
@preferred_anon only monotonic considering even or odd terms. — qwr, Feb 15 '19 at 17:15

preferred_anon · Answer 1 · 2014-02-23T22:11:47.840

7

By Cassini's Identity: $$\left|\frac{F_{n+1}}{F_{n}}-\frac{F_{n}}{F_{n-1}}\right|=\left|\frac{F_{n+1}F_{n-1}-F_{n}^{2}}{F_{n}F_{n-1}}\right|=\left|\frac{(-1)^{n}}{F_{n}F_{n-1}}\right| \to 0$$

Proof of Cassini's Identity: $$F_{n+1}F_{n-1}-F_{n}^{2}\\ =(F_{n}+F_{n-1})F_{n-1}-F_{n}^{2}\\ =F_{n-1}F_{n}-F_{n}^{2}+F_{n-1}^{2}\\ =-(F_{n}(F_{n}-F_{n-1})-F_{n-1}^{2})\\ =-(F_{n}F_{n-2}-F_{n-1}^{2})$$ You can fill in the rest by induction.

edited Feb 23 '14 at 22:11

answered Feb 23 '14 at 21:26

preferred_anon

17,121

I like this a lot. I haven't heard of Cassini's Identity before. Could you please link to a proof or derivation of it? – Guest Feb 23 '14 at 21:45
I've sketched a proof in the answer - is this clear enough? – preferred_anon Feb 23 '14 at 22:13
Yes this is good, thank you!
What I really want to do is split the sequence of ratios into two sequences- a monotonic increasing one and a monotonic decreasing one and go from there. But I don't know how to do it...
– Guest Feb 23 '14 at 22:14
Simply replace $n$ by $2n$. Clearly when $n$ is even the difference is positive and when $n$ is odd the difference is negative! – preferred_anon Feb 23 '14 at 23:13
Yes, I did that. 2n and 2n+1 for even and odd respectively. But then what? – Guest Feb 23 '14 at 23:15
If the difference is positive it is increasing, if negative it is decreasing. – preferred_anon Feb 23 '14 at 23:16
Substitute into the Cassini's? Or into the general definition? Where do you substitute in? – Guest Feb 23 '14 at 23:19
Explicitly: $\frac{F_{2n+1}}{F_{2n}}-\frac{F_{2n}}{F_{2n-1}}=\frac{1}{F_{2n}F_{2n-1}}>0$ – preferred_anon Feb 23 '14 at 23:21
OK so you have really just shown that the sequence is oscillating, right? – Guest Feb 23 '14 at 23:25
Consider the two subsequences: If $a_{n}=F_{n+1}/F_{n}$, we have seen that $a_{2n}$ is increasing and bounded above (e.g. by 2), and $a_{2n+1}$ is decreasing and bounded below (e.g. by 1). Thus both have limits (and this limit is $\varphi$ in both cases). – preferred_anon Feb 23 '14 at 23:29
Sorry if I'm being stupid, but I do not understand where you have shown that $a_{2n}$ is increasing and $a_{2n+1}$ is decreasing – Guest Feb 23 '14 at 23:31
By the above argument, $a_{2n}-a_{2(n-1)}>0$ so $a_{2n}$ is increasing. Similarly for $a_{2n-1}$. – preferred_anon Feb 23 '14 at 23:33
GOT IT. Thanks- perfect answer – Guest Feb 24 '14 at 09:32
let us continue this discussion in chat – Guest Feb 24 '14 at 11:16

Ma Ming · Answer 2 · 2014-02-23T21:21:18.793

3

Hint: Consider $\frac{F_n}{F_{n-1}}=1+\frac{F_{n-2}}{F_{n-1}}$. Let $f_n=\frac{F_n}{F_{n-1}}$. We have $$f_n=1+\frac{1}{f_{n-1}}.$$

Let $f(x)=1+\frac{1}{x}$ for $1<x<2$. Show if $x<\phi$, then $x<f(x)<\phi$ (We also have $\phi<f(x)<x$ for $x>\phi$, but it is irrelevant to our concern.).

edited Feb 23 '14 at 21:21

answered Feb 23 '14 at 21:12

Ma Ming

7,482

Thank you for this answer. This is how I was able to show that the limit was $\phi$ if it exists, but how do you know $f_n$ exists and is a real number as $n \to \infty$? – Guest Feb 23 '14 at 21:19
Thanks for the edit. Would you be able to do the next step? Sorry- I am still unclear on how this would prove the existence of the limit given its oscillation? – Guest Feb 23 '14 at 21:44
@Guest If you showed my claim, then $f_1<f_2<f_3\cdots \le \phi$. – Ma Ming Feb 23 '14 at 22:57

Prove that the limit of two consecutive fibonacci numbers EXISTS.

2 Answers2