Proving that ratio of two consecutive Fibonacci numbers to converges to golden ratio by induction $$\varphi = \frac{1 + \sqrt{5}}{2}$$
We want to prove that ratio of two consecutive Fibonacci numbers approaches $\varphi$ by induction and also utilizing Newton-Raphson method for approximating $\sqrt{5}$ as a rational number with relatively prime numerators and denominators.
Let us first define the Fibonacci Sequence and then write down what we want to prove using the symbolic notation.
$$\phi_1,\phi_2 = 1$$
$$\phi_{n+2} = \phi_{n+1} + \phi_{n}$$
$$1 \le n$$

$$n \to \infty \Rightarrow \frac{\phi_{n+1}}{\phi_{n}} \to \varphi$$

First we need rational approximations for the irrational number $\sqrt{5}$, so we can connect it up to integers:
$$x_{n} = \frac{a_{n}}{b_{n}}$$
$$x_{1} = \frac{2}{1} = \frac{a_1}{b_1}$$
Where the limit as $n$ goes to infinity is $\sqrt{5}$. So according to the Newton-Raphson method we can write that, 
$$x_{n+1} = x_n - \frac{F(x_n)}{F'(x_n)} = \frac{x_n^2 + 5}{2x_n}$$
Substituting the $\frac{a_n}{b_n}$ and $\frac{a_{n+1}}{b_{n+1}}$ respectively to the places of $x_n$ and $x_{n+1}$  we will get the following rational number with integer numerator and denominator,
$$\frac{a_{n+1}}{b_{n+1}} = \frac{a_n^2 + 5b_n^2}{2a_nb_n}$$
So that
$$a_{n+1} = a_n^2 + 5b_n^2$$
$$b_{n+1} = 2a_nb_n$$
Let us define
$$\varphi_n = \frac{1 + \frac{a_n}{b_n}}{2} = \frac{a_n + b_n}{2b_n}$$
$$n \to \infty \Rightarrow \varphi_{n} \to \varphi$$
For $n = 1$ (the first case):
$$\varphi_1 = \frac{1 + \frac{2}{1}}{2} = \frac{3}{2}$$
Numerator and denominator are two consecutive Fibonacci numbers, respectively the $\phi_4$ and $\phi_3$. Now here goes our induction hypothesis for $1 \le n$:
if numerator and denominator of $\varphi_n$ are two consecutive Fibonacci numbers, respectively $\phi_{3\cdot2^{n - 1} + 1}$ and $\phi_{3\cdot2^{n - 1}}$ (as the $\phi_{4}$ and $\phi_{3}$ are), then the numerator and denominator of $\varphi_{n+1}$ will be again two consecutive Fibonacci numbers, respectively $\phi_{3\cdot2^{n} + 1}$ and $\phi_{3\cdot2^{n}}$.
$$\varphi_1 = \frac{3}{2} = \frac{\phi_4}{\phi_3} = \frac{\phi_{3\cdot2^{0} + 1}}{\phi_{3\cdot2^{0}}}$$
$$\varphi_2 = \frac{13}{8} = \frac{\phi_7}{\phi_6} = \frac{\phi_{3\cdot2^{1} + 1}}{\phi_{3\cdot2^{1}}}$$
$$\varphi_3 = \frac{233}{144} = \frac{\phi_{13}}{\phi_{12}} = \frac{\phi_{3\cdot2^{2} + 1}}{\phi_{3\cdot2^{2}}}$$
$$...$$
If we can prove that, we will be proved that ratio of two consecutive Fibonacci numbers approaches $\varphi$, of course if i have no mistake to here.
Here is our assumption:
$$\varphi_n = \frac{a_n + b_n}{2b_n} = \frac{\phi_{3\cdot2^{n - 1} + 1}}{\phi_{3\cdot2^{n - 1}}}$$
And here is what we want to prove (i rewritten and manipulated the statement  using previous definitions of $a_{n+1}$ and $b_{n+1}$):
$$\varphi_{n+1} = \frac{(a_n + b_n)^2 + (2b_n)^2 }{(2b_n)^2} = \frac{\phi_{3\cdot2^{n} + 1}}{\phi_{3\cdot2^{n}}}$$
$$\varphi_{n+1} = \varphi_{n}^2 + 1 = \frac{\phi_{3\cdot2^{n} + 1}}{\phi_{3\cdot2^{n}}}$$
It gets really cumbersome for me after this point. Anyone could please give me hints or point out my mistakes if there is?
 A: This is just a suggestion:
From $F_{n+1}=F_n+F_{n-1}$, define the ratio as
$r_{n+1}=\frac{F_{n+1}}{F_n}=1+\frac{F_{n-1}}{F_n}=1+\frac{1}{r_n}$
There are a few thing too do now, for instance to show that $r_n$ converges. Once that is done, the limit should satisfy
$$x=1+\frac{1}{x}$$
which has the golden mean as solution.
Also notice that
$$
\begin{bmatrix}
F_{n+1}\\
F_n
\end{bmatrix} =\begin{bmatrix}
1 & 1 \\
1 & 0
\end{bmatrix} \begin{bmatrix}
F_n\\
F_{n-1}
\end{bmatrix}
$$
And iterating n times you get
$$
\begin{bmatrix}
F_{n+1}\\
F_n
\end{bmatrix} =\begin{bmatrix}
1 & 1 \\
1 & 0
\end{bmatrix}^n \begin{bmatrix}
F_1\\
F_0
\end{bmatrix}
$$
Using the diagonal decomposition of the matrix involve you get again something related to the golden mean.
A: With $\phi_n:=\dfrac{F_{n+1}}{F_n}$,
$$F_{n+2}=F_{n+1}+F_n\iff \phi_{n+1}=1+\frac1{\phi_{n}}.$$
Then if the sequence of $\phi_n$ converges, it converges to a root of
$$p=1+\frac1p.$$
As all terms are positive, convergence must be to the positive root.

As one can observe, the $\phi_n$ are alternating around $\phi$ and getting closer and closer. We can show that
$$|\phi_{n+1}-\phi|<|\phi_n-\phi|.$$
Indeed
$$|\phi_{n+1}-\phi|=\left|1+\frac1{\phi_{n}}-\phi\right|=\left|\frac1{\phi_{n}}-\frac1\phi\right|=\frac{|\phi_n-\phi|}{\phi_{n}\phi}<|\phi_n-\phi|.$$
As $\phi_n>1$, then the distance to $\phi$ is at least divided by $\phi$ (in fact by nearly $\phi^2$) on every iteration, and linear convergence is guaranteed.
