Fibonacci numbers and golden ratio: $\Phi = \lim \sqrt[n]{F_n}$ 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...
 A: 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
$$\sqrt[n]{F_n}\sim_\infty \alpha=:\Phi$$
A: 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.
A: 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 $R=\lim_{n\to\infty}R_n=\Phi$.
A: It is a standard result that $F_n = \frac{\phi^n - (-\phi)^{-n}}{\sqrt{5}}$.
Then $\lim_{n\rightarrow \infty} \sqrt[n]{F_n} = \lim_{n\rightarrow \infty} \sqrt[n]{\frac{\phi^n - (-\phi)^{-n}}{\sqrt{5}}} = \lim_{n\rightarrow \infty} \sqrt[n]{\frac{\phi^n}{\sqrt{5}}} = \lim_{n\rightarrow \infty} \frac{\phi}{\sqrt[n]{\sqrt{5}}} = \phi$ .
A: We know 
$$\lim_{n \to \infty} \frac{F_{n+1}}{F_n}=\Phi$$
by continuity of logarithm function, this implies
$$\lim_{n \to \infty} (log{F_{n+1}}-log{F_n})=log\Phi$$ 
By definition, $\forall{\epsilon \gt 0}, \exists{N\in {\mathbb N}}$ such that $n \ge N $ implies $|log{F_{n+1}}-log{F_n}-log\Phi|\lt \epsilon_1$
So by triangle inequality,
 $$|(\sum_{k=N}^{n}\frac{1}{n-N+1}(logF_{k+1}-logF_k))-log\Phi|\le \frac{1}{n-N+1}\sum_{k=N}^{n}|logF_{k+1}-logF_k-log\Phi|  \lt \epsilon_1
$$
Let $$\sum_{k=1}^{N-1} |logF_{k+1}-logF_k-log\Phi|=A$$
Then $$|\sum_{k=1}^{n}(logF_{k+1}-logF_k-log\Phi)| \le \sum_{k=1}^{n} |logF_{k+1}-logF_k-log\Phi|\;,$$
which can be split into
$$\sum_{k=1}^{N-1} |logF_{k+1}-logF_k-log\Phi|+\sum_{k=N}^{n} |logF_{k+1}-logF_k-log\Phi|$$
Hence, we have $$\frac{1}{n-N+1}(\sum_{k=1}^{N-1} |logF_{k+1}-logF_k-\Phi|+\sum_{k=N}^{n} |logF_{k+1}-logF_k-log\Phi|)\lt \frac{A}{n-N+1}+\epsilon_1$$
Now, fix $N=N_0$. Since A is also fixed, $A \gt 0, \forall{\epsilon_2 \gt 0}, \exists{N_1 \in{\mathbb N}}$ such that $n \ge N_1 \;\text{implies} \; (n-N_0+1)\epsilon_2\lt A, \;\text{so}\; \dfrac{A}{n-N_0+1}\lt \epsilon_2.$ 
Hence, $\forall{n \ge N_1},$
$$|(\sum_{k=1}^{n}\frac{1}{n}(logF_{k+1}-logF_k))-log\Phi|\lt|\sum_{k=1}^{n}\frac{1}{n-N_0+1}(logF_{k+1}-logF_k-log\Phi)|\lt \epsilon_1+\epsilon_2$$
$$|\frac{1}{n}logF_{n+1}-log\Phi|=|(\sum_{k=1}^{n}\frac{1}{n}(logF_{k+1}-logF_k))-log\Phi|\lt \varepsilon_1$$
Therefore, by definition, 
$$lim_{n \to \infty}\frac{1}{n}logF_{n+1}=lim_{n \to \infty}log\sqrt[n]{F_{n+1}}=log\Phi$$
By the continuity of the exponential function $e^x$, we get 
$$lim_{n\to \infty} \sqrt[n]{F_{n+1}}=\Phi$$
Hence proven. 
Remark: "Introduction to Calculus and Analysis I" by Richard Courant and Fritz John,page 114,SECTION 1.7, exercise $^*10$: "If $a_n \gt 0, \;\text{and} \; lim_{n\to \infty}\dfrac{a_{n+1}}{a_n}=L,\;\text{then}\;lim_{n\to \infty}\sqrt[n]{a_n}=L."$
