Suppose that {$a_n$} is the Fibonacci sequence, and $b_n=\frac{a_{n+1}}{a_n}$. Prove that {$b_n$} converges to the Golden Number. This is an exercise in my Analysis book that I find difficult. Suppose that {$a_n$} is the Fibonacci sequence, and $b_n=\frac{a_{n+1}}{a_n}$. Prove that {$b_n$} converges to the Golden Number.
First, of course, one has to prove that {$b_n$} does converge. I think since $a_n$ is increasing, and $a_{n+1}-a_n>a_n-a_{n-1}$, {$b_n$} is decreasing. Now since both the numerator and denominator are positive, {$b_n$} is bounded below by $0$. That means {$b_n$} must converge.
Now how to prove that {$b_n$} actually converges to $\frac{1+\sqrt5}{2}$ I don't know. I tried taking limits of $b_{n+1}$ and $b_n$ but that got messy and didn't seem to lead anywhere. 
How do I solve this?
Thanks in advance.
 A: First, from the definition of the sequence $\{b_n\}$:
$$b_{n+1}=\frac{a_{n+2}}{a_{n+1}}$$
But $\{a_n\}$ is the Fibonacci sequence, so $a_{n+2}=a_{n+1}+a_n$, then:
$$b_{n+1}=\frac{a_{n+1}+a_n}{a_{n+1}}=1+\frac{a_n}{a_{n+1}}=1+\frac{1}{b_n}$$
So, the equence $\{b_n\}$ satisfies:
$$b_{n+1}=\frac{b_n+1}{b_n}$$
Then, if the limit exists, let's say that it is $l$, it must satisfy the equation:
$$l=\frac{l+1}{l}$$
That's equivalent to:
$$l^2-l-1=0$$
Now solving this equation we find $l=\frac{1\pm \sqrt{5}}{2}$, but the sequence is bounded below by $0$ and $\frac{1-\sqrt{5}}{2}<0$ so $\lim b_n$ cannot be $\frac{1-\sqrt{5}}{2}$, then we conclude that:
$$l=\frac{1+\sqrt{5}}{2}$$ 
A: Using induction, we will prove
$$
F_{n+1}F_{n-1}-F_n^2=(-1)^n\tag{1}
$$
Note that $F_2F_0-F_1^2=-1$.
Assume that $F_nF_{n-2}-F_{n-1}^2=(-1)^{n-1}$. Then
$$
\begin{align}
F_{n+1}F_{n-1}-F_n^2
&=(F_n+F_{n-1})F_{n-1}-F_n^2\\
&=F_{n-1}^2-F_n(F_n-F_{n-1})\\
&=F_{n-1}^2-F_nF_{n-2}\\
&=-(-1)^{n-1}\\
&=(-1)^n\tag{2}
\end{align}
$$
Therefore, $(1)$ is true for $n\ge1$.

Dividing $(1)$ by $F_nF_{n-1}$ gives
$$
\frac{F_{n+1}}{F_n}-\frac{F_n}{F_{n-1}}=\frac{(-1)^n}{F_nF_{n-1}}\tag{3}
$$
Since $F_{n+1}\ge F_n$,
$$
\begin{align}
\frac{F_{n+2}F_{n+1}}{F_{n+1}F_n}
&=\frac{F_{n+2}}{F_n}\\
&=\frac{F_{n+1}}{F_n}+1\\[6pt]
&\ge2\tag{4}
\end{align}
$$
Therefore, for $n\ge1$,
$$
F_nF_{n+1}\ge2^{n-1}\tag{5}
$$
Thus, $(3)$ and $(5)$ show that
$$
\begin{align}
\lim_{n\to\infty}\frac{F_{n+1}}{F_n}
&=\frac{F_2}{F_1}+\lim_{n\to\infty}\sum_{k=2}^n\frac{(-1)^k}{F_kF_{k-1}}\\
&=1+\sum_{k=2}^\infty\frac{(-1)^k}{F_kF_{k-1}}\tag{6}
\end{align}
$$
converges.

The recursion for $F_n$ implies
$$
\begin{align}
0
&=\lim_{n\to\infty}\frac{F_{n+2}-F_{n+1}-F_n}{F_n}\\
&=\lim_{n\to\infty}\frac{F_{n+2}}{F_{n+1}}\frac{F_{n+1}}{F_n}-\lim_{n\to\infty}\frac{F_{n+1}}{F_n}-1\\
&=\left(\lim_{n\to\infty}\frac{F_{n+1}}{F_n}\right)^2-\lim_{n\to\infty}\frac{F_{n+1}}{F_n}-1\tag{7}
\end{align}
$$
Equation $(7)$ implies that
$$
\lim_{n\to\infty}\frac{F_{n+1}}{F_n}=\frac{1\pm\sqrt5}2\tag{8}
$$
Since $\frac{F_{n+1}}{F_n}\ge1$, $(8)$ implies
$$
\lim_{n\to\infty}\frac{F_{n+1}}{F_n}=\frac{1+\sqrt5}2\tag{9}
$$
