# How to prove that $\lim \limits_{n\rightarrow \infty} \frac{F_{n+1}}{F_n}=\frac{\sqrt{5}+1}{2}$

How would one prove that $$\lim_{n\rightarrow \infty} \frac{F_{n+1}}{F_n}=\frac{\sqrt{5}+1}{2}=\varphi$$

where $F_n$ is the nth Fibonacci number and $\varphi$ is the Golden Ratio?

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This is explained at the wikipedia page, en.wikipedia.org/wiki/Fibonacci_number –  Matthew Conroy Apr 15 '12 at 23:28
Where is it on the page? –  Argon Apr 15 '12 at 23:37
You will find this useful. –  Pedro Tamaroff Apr 15 '12 at 23:53
Read the ninth chapter of C. Stanley Ogilvy's magnificent book Excursions in Geometry. –  Michael Hardy Apr 15 '12 at 23:58
@Argon Your limit follows easily from the closed-form expression for $F_n$, the derivation of which is described in the section "Relation to the Golden Ratio". –  Matthew Conroy Apr 16 '12 at 1:34

$$F_{n+1}=F_n+F_{n-1}$$

$$F_{n+1}/F_n=1+F_{n-1}/F_n=1+1/(F_n/F_{n-1})$$

Call the limit $x$; then $$x=1+1/x$$

Take it from there.

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Thanks for this! –  Argon Apr 15 '12 at 23:42
@Gerry I think it is important to prove the limit exists, isn't it? i.e Considering even and odd subsequences and showing the sequence in general is bounded. –  Pedro Tamaroff Apr 15 '12 at 23:47
@PeterT.off, on principle, I leave details to the reader. –  Gerry Myerson Apr 15 '12 at 23:58
@GerryMyerson : You haven't up-voted the question yet. –  Michael Hardy Apr 15 '12 at 23:59
@Michael, I disagree with "if a question is worth answering, it's almost always worth up-voting," but this isn't the place to discuss it. It may be worth bringing up on meta (if you haven't already done so - I haven't checked recently). –  Gerry Myerson Apr 17 '12 at 0:27

Gerry's solution is quite elegant. One might take the less elegant route of first deriving the Binet formula:

$$F_n=\frac1{\sqrt{5}}(\phi^n-(-\phi)^{-n})$$

from which

$$\frac{F_{n+1}}{F_n}=\frac{\phi^{n+1}-(-\phi)^{-n-1}}{\phi^n-(-\phi)^{-n}}=\frac{\phi-\frac{\left(-\frac1{\phi}\right)^{n+1}}{\phi^n}}{1-\frac{\left(-\frac1{\phi}\right)^n}{\phi^n}}=\frac{\phi+\frac{(-1)^n}{\phi^{2n+1}}}{1-\frac{(-1)^n}{\phi^{2n}}}$$

$(-1)^n$ is a bounded sequence, while $\frac1{\phi^n}$ decays nicely to $0$, so... you can take it from there.

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If you know that the limit exists, you can proceed e.g. as in Gerry's answer.

There are probably many different ways to show that the limit exists. One of them uses Cassini identity $$F_{n+1}F_{n-1}-F_n^2=(-1)^n,$$ you can get $$\frac{F_{n+1}}{F_n}-\frac{F_n}{F_{n-1}}=(-1)^n\frac1{F_nF_{n-1}}.$$

So now you could use Leibniz test, you only have to show that $\lim\limits_{n\to\infty}\frac1{F_nF_{n-1}}=0$

(Proof of Cassini identity can be found on Wikipedia, on this site or elsewhere.)

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