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?
 A: $$F_{n+1}=F_n+F_{n-1}$$
$$F_{n+1}/F_n=1+F_{n-1}/F_n=1+1/(F_n/F_{n-1})$$
Call the limit $x$; then $$x=1+1/x$$ 
Take it from there. 
A: 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.) 
A: 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.
A: 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.
