# Fibonacci numbers and golden ratio

Let $\Phi$ be the golden ratio and $F_n$ be the usual Fibonacci numbers. How can I derive the following formula?

$$\Phi = \lim_{n\rightarrow \infty} \sqrt[n]{F_n}$$

I know the usual relation $$\Phi = \lim_{n\rightarrow \infty} \frac{F_{n+1}}{F_n} \quad ,$$ and Wikipedia tells me that $$\Phi^a = \lim_{n\rightarrow \infty} \frac{F_{n+a}}{F_n} \quad .$$

My first idea was to set $a = n$, which gives $$\Phi = \lim_{n\rightarrow \infty} \sqrt[n]\frac{F_{n+n}}{F_n} \quad ,$$

EDIT: We can also do $$\Phi = \lim_{n\rightarrow \infty} \sqrt[n]{\frac{F_{n+n}}{F_n}\frac{F_n}{F_n}} \quad ,$$ but I am totally stuck here...

-
One way following what you know. –  ABC Mar 27 at 13:01
The work you have done so far suggests that as n gets large $F_{2n}$ approaches $F_n^2$ –  kleineg Mar 27 at 13:05
If you are willing to use Wikipedia, the same page also has some formulas saying, more or less, $F_n \approx\phi^n/\sqrt{5}$. Therefore $\sqrt[n]{F_n} \approx \phi / \sqrt[(2n)]{5}$. –  Jeppe Stig Nielsen Mar 27 at 13:05
@kleinig: I cannot see this. Can you elaborate? One of my ideas was to multiply the term under the root by $F_n/F_n$ - when I then apply your hint, the result follows. –  mort Mar 27 at 13:12
@mort Sorry, that was less of a hint and more of an observation. When you set $a = n$ you got $\frac{F_{2n}}{F_n}$ in the nth root. And (once you prove it) we have the same equation with $F_n$ in the nth root. That suggests that $\frac{F_{2n}}{F_n} = F_n$ as n goes to infinity. –  kleineg Mar 27 at 13:27

It is a standard result that $F_n = \frac{\phi^n - (-\phi)^{-n}}{\sqrt{5}}$.

Then $\lim_{n\rightarrow \infty} \sqrt[n]{F_n} = \lim_{n\rightarrow \infty} \sqrt[n]{\frac{\phi^n - (-\phi)^{-n}}{\sqrt{5}}} = \lim_{n\rightarrow \infty} \sqrt[n]{\frac{\phi^n}{\sqrt{5}}} = \lim_{n\rightarrow \infty} \frac{\phi}{\sqrt[n]{\sqrt{5}}} = \phi$ .

-
Ok, I actually understand this! –  mort Mar 27 at 13:11

We have

$$F_n=\frac1{\sqrt5}\left(\underbrace{\frac{1+\sqrt5}{2}}_{=:\alpha}\right)^n-\frac1{\sqrt5}\left(\underbrace{\frac{1-\sqrt5}{2}}_{=:\beta}\right)^n$$ and since $|\beta|<|\alpha|$ then $$|\beta|^n=_\infty o(|\alpha|^n)$$ hence $$\sqrt[n]{F_n}\sim_\infty \alpha=:\Phi$$

-

The process is to look at the approximation that $F_n = \Phi^n/\sqrt{5}$. The n'th root of this is $\sqrt[n]{F_n} = \Phi / 5^{1/2n}$.

The denominator approaches unity.

-

Let $\Phi_n=\frac{F_{n+1}}{F_n}$ and $R_n=\sqrt[n]{F_n}$, then we have $$R_n^n=F_n=F_{n-1}\Phi_{n-1}=R_{n-1}^{n-1}\Phi_{n-1}$$ Take $n\to\infty$, $$R^n=R^{n-1}\Phi$$ Hence $R=\lim_{n\to\infty}R_n=\Phi$.

-