# 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?

• 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.

• @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
• The only up-vote at that point was mine (and still is now). "Obligation" is not what I had in mind. I just thought that if a question is worth answering, it's almost always worth up-voting. This question is worth having for future reference by others who come to this site. – Michael Hardy Apr 16 '12 at 17:38
• @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
• I will add to the above discussion between Gerry Myerson and @Michael Hardy that there is now this thread on meta: About not upvoted, answered questions. (It is from 2013, so it did not exist at the time of the above exchange.) There is also a related post on meta.SE: Why don't people upvote questions they answer? (Posting mainly for the benefit of other users who see this post and are interested in the issue discussed in the above comments.) – Martin Sleziak Sep 19 '17 at 4:18

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.)

• How does the limit of 1/FnFn-1 = 0 imply that the limit of Fn+1/Fn exists? Please help ASAP. – Max Li Sep 30 '17 at 0:30
• As mentioned in the post, all you have to do is to use Leibniz test for the alternating series $\sum (-1)^n \frac1{F_nF_{n-1}}$. – Martin Sleziak Sep 30 '17 at 0:59
• Yes, that implies that the right side of the equation is equal to 0. but how does that imply that Fn+1/Fn exists? – Max Li Sep 30 '17 at 20:43
• Based on the above equality, you can notice that $\frac{F_{n+1}}{F_n} = 1 + \sum\limits_{k=2}^n (-1)^k \frac1{F_kF_{k-1}}$. So the limit exists if and only if the series $\sum\limits_{k=2}^\infty (-1)^k \frac1{F_kF_{k-1}}$ converges. – Martin Sleziak Sep 30 '17 at 23:27
• Why is $$\frac{F_{n+1}}{F_n}-\frac{F_n}{F_{n-1}}=(-1)^n\frac1{F_nF_{n-1}}.$$ the following $$\frac{F_{n+1}}{F_n}=1+\sum^n_{k=2}\ldots$$ ? – pls_halp Oct 2 '17 at 19:04

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.

Several useful and cogent answers have already been given to this question, both here and in at least one duplicate. However, I haven't seen anyone mention the most elegant, concise and enlightening argument I could find when I was wondering about the same thing, i.e. this blog post by Carl McTague. I'm just linking it here in hopes that others may enjoy it.