Proof the golden ratio with the limit of Fibonacci sequence Let $F_n=F_{n-1}+F_{n-2}$ the Fibonacci numbers, and $\phi=\frac{1+\sqrt5}{2}$
The exercise asks me to prove that:  $\lim\limits_{n \to \infty}\frac{F_{n+1}}{F_n}=\phi$.
Sorry as can be proceed??
 A: Let $R_n=\frac{F_{n+1}}{F_n}$. Since:
$$ F_n^2-F_{n-1}F_{n+1} = (-1)^n\tag{1}$$
(it is easy to prove by induction) we have that:
$$ \left|R_{n+1}-R_n\right|=\frac{1}{F_n F_{n+1}}, \tag{2}$$
hence $\{R_n\}_{n\in\mathbb{N}}$ is a Cauchy sequence and $R_n$ converges to some $L>1$ that satisfies $L=1+\frac{1}{L}$,
since $R_n>1$ and $R_{n+1}=1+\frac{1}{R_n}$.
A: Find $\alpha,\beta$ such that $F_0=\alpha+\beta$ and $F_1=\alpha\phi+\beta(1-\phi)$ and show by induction that 
$$F_n=\alpha\phi^n+\beta(1-\phi)^n.$$
Then compute the desired limit (and use that $|1-\phi|<1$).
A: $$F_{n+1}=F_{n}+F_{n-1} \Rightarrow \frac{F_{n+1}}{F_n}=1+\frac{F_{n-1}}{F_n}$$
Let 
$$x_n:= \frac{F_{n+1}}{F_n}$$
Then 
$$x_n=1+\frac{1}{x_{n-1}}$$
You can now prove that $1 \leq x_n \leq 2$ and by induction that
$$ x_1 \leq x_3 \leq x_5 \leq .. \leq x_{2n+1} \leq x_{2n} \leq x_{2n-2} \leq .. \leq x_2$$
From here you get that $x_{2n-1}$ and $x_{2n}$ converge, and you can use their definitions to get their limits. You prove that both limits are the same, which yields the desired result.
