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?

  • 7
    $\begingroup$ This is explained at the wikipedia page, en.wikipedia.org/wiki/Fibonacci_number $\endgroup$ Apr 15, 2012 at 23:28
  • $\begingroup$ Where is it on the page? $\endgroup$
    – Argon
    Apr 15, 2012 at 23:37
  • 1
    $\begingroup$ You will find this useful. $\endgroup$
    – Pedro
    Apr 15, 2012 at 23:53
  • $\begingroup$ Read the ninth chapter of C. Stanley Ogilvy's magnificent book Excursions in Geometry. $\endgroup$ Apr 15, 2012 at 23:58
  • 1
    $\begingroup$ @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". $\endgroup$ Apr 16, 2012 at 1:34

4 Answers 4




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

Take it from there.

  • 27
    $\begingroup$ @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. $\endgroup$
    – Pedro
    Apr 15, 2012 at 23:47
  • 1
    $\begingroup$ @PeterT.off, on principle, I leave details to the reader. $\endgroup$ Apr 15, 2012 at 23:58
  • 1
    $\begingroup$ 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. $\endgroup$ Apr 16, 2012 at 17:38
  • 4
    $\begingroup$ @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). $\endgroup$ Apr 17, 2012 at 0:27
  • 2
    $\begingroup$ 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.) $\endgroup$ Sep 19, 2017 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.)

  • $\begingroup$ How does the limit of 1/FnFn-1 = 0 imply that the limit of Fn+1/Fn exists? Please help ASAP. $\endgroup$
    – Max Li
    Sep 30, 2017 at 0:30
  • $\begingroup$ As mentioned in the post, all you have to do is to use Leibniz test for the alternating series $\sum (-1)^n \frac1{F_nF_{n-1}}$. $\endgroup$ Sep 30, 2017 at 0:59
  • $\begingroup$ Yes, that implies that the right side of the equation is equal to 0. but how does that imply that Fn+1/Fn exists? $\endgroup$
    – Max Li
    Sep 30, 2017 at 20:43
  • 2
    $\begingroup$ 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. $\endgroup$ Sep 30, 2017 at 23:27
  • $\begingroup$ Why is $$\frac{F_{n+1}}{F_n}-\frac{F_n}{F_{n-1}}=(-1)^n\frac1{F_nF_{n-1}}.$$ the following $$\frac{F_{n+1}}{F_n}=1+\sum^n_{k=2}\ldots $$ ? $\endgroup$
    – pls_halp
    Oct 2, 2017 at 19:04

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


from which


$(-1)^n$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.