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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7This is explained at the wikipedia page, http://en.wikipedia.org/wiki/Fibonacci_number – Matthew Conroy Apr 15 '12 at 23:28
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Where is it on the page? – Argon Apr 15 '12 at 23:37
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1You will find this useful. – Pedro Apr 15 '12 at 23:53
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Read the ninth chapter of C. Stanley Ogilvy's magnificent book Excursions in Geometry. – Michael Hardy Apr 15 '12 at 23:58
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Will do, thanks again. I will need to look up this book in the library. Thanks! – Argon Apr 15 '12 at 23:59
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1@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 01:34
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@Matthew, yes, but I reckon that's overkill, when it comes so fast from the recurrence. Still, the more methods a person knows, the better. – Gerry Myerson Apr 16 '12 at 09:28
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@Gerry I wasn't disparaging your answer - it is excellent. – Matthew Conroy Apr 16 '12 at 16:32
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$$F_{n+1}=F_n+F_{n-1}$$
$$F_{n+1}/F_n=1+F_{n-1}/F_n=1+1/(F_n/F_{n-1})$$
Call the limit $x$; then $$x=1+1/x$$
Take it from there.
Gerry Myerson
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27@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 Apr 15 '12 at 23:47
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@Michael How do you know who upvoted and who didn't? Or is the only upvote yours? – Pedro Apr 16 '12 at 00:00
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@MichaelHardy, perhaps you would like to start a meta thread (or direct me to an already existing thread) on the circumstances under which one is obligated to upvote a question. – Gerry Myerson Apr 16 '12 at 00:56
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@GerryMyerson I think Michael was making a joke, alluding to the fact that many users upvote questions that they have answered to attract more views and hence potential upvotes to their answer. – Ragib Zaman Apr 16 '12 at 07:40
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1The 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
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4@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 00:27
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@GerryMyerson : But this is the place to discuss what has happened in this particular instance. – Michael Hardy Apr 17 '12 at 01:33
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@Michael, I disagree with that, too. Try http://meta.math.stackexchange.com/questions/3315/why-should-i-upvote-questions – Gerry Myerson Apr 17 '12 at 04:07
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2I 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 04:18
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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.)
Martin Sleziak
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How does the limit of 1/FnFn-1 = 0 imply that the limit of Fn+1/Fn exists? Please help ASAP. – Jack Pan Sep 30 '17 at 00:30
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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 00:59
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Yes, that implies that the right side of the equation is equal to 0. but how does that imply that Fn+1/Fn exists? – Jack Pan Sep 30 '17 at 20:43
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2Based 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
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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 02 '17 at 19:04
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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.
J. M. ain't a mathematician
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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.
Rebecca Bonham
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