Fibonacci numbers and golden ratio: $\Phi = \lim \sqrt[n]{F_n}$

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

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

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.