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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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    @Gerry I wasn't disparaging your answer - it is excellent.2012-04-16

3 Answers 3

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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.

  • 1
    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](https://math.meta.stackexchange.com/q/10919). (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?](https://meta.stackexchange.com/q/509) (Posting mainly for the benefit of other users who see this post and are interested in the issue discussed in the above comments.)2017-09-19
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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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    @pls_halp Try induction or [telescoping series](https://en.wikipedia.org/wiki/Telescoping_series).2017-10-02
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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.