Prove $\frac{F(n+1)}{F(n)}$ converges to $\phi$ without assuming a priori that it converges
If I know it converges, then I know it converges to $\phi$
Since, if $\lim_{n\to\infty} \frac{F(n+1)}{F(n)} \to r$ then,
$\lim_{n\to\infty}\frac{F(n+1)}{F(n)} = \lim_{n\to\infty}\frac{F(n)+F(n-1)}{F(n)} = 1+\frac{1}{r} = r \to r=\phi$
But how do I know $\lim_{n\to\infty} \frac{F(n+1)}{F(n)}$ converges in the first place?
Let $\phi, \varphi$ be the positive & negative solutions, respectively, of $$x^2 = x + 1$$ So $$\phi + \varphi = 1$$ and $$\phi\varphi = -1$$ That is, $$\varphi = 1 - \phi = -\phi^{-1}$$ So $$\phi, \varphi = \frac{1\pm\sqrt5}2$$ That is, $$\phi\approx 1.6180339$$ and $$\varphi\approx -0.6180339$$ Also, $$\phi - \varphi = \sqrt5$$
Now, for any $n$, $$x^{n+1} = x^n + x^{n-1}$$ and so with a simple induction we can construct a Fibonacci sequence of powers of $x$, as this table shows:
| n | $F_n$ | $x^n$ | $x^n$ |
|---|---|---|---|
| 0 | 0 | $x^0$ | $0x +1$ |
| 1 | 1 | $x^1$ | $1x + 0$ |
| 2 | 1 | $x^2$ | $1x + 1$ |
| 3 | 2 | $x^3$ | $2x + 1$ |
| 4 | 3 | $x^4$ | $3x + 2$ |
| 5 | 5 | $x^5$ | $5x + 3$ |
| n | $F_n$ | $x^n$ | $F_nx + F_{n-1}$ |
Thus $$\phi^n = F_n\phi+ F_{n-1}$$ and $$\varphi^n = F_n\varphi+ F_{n-1}$$ We can easily use these results to derive the Binet formula, but we don't need that here.
$$\varphi^n= F_n\varphi+ F_{n-1}$$ $$(-1)^n\phi^{-n} = F_n(1 - \phi) + F_{n-1}$$ $$\frac{(-1)^n}{\phi^n} = F_{n+1} - \phi F_n$$ $$\frac{F_{n+1}}{F_n} = \phi + \frac{(-1)^n}{F_n\phi^n}$$
Clearly, the RHS converges to $\phi$.
you can easily show that it is bounded, as all terms $t_n$ are greater (or equal to) than 0, $t_n\ge0$, and we can also know that
$$t_{n+1}=t_n+t_{n1}\le t_n+t_n$$ as the next Fibonacci number is grater then the one before, from this we can show, assuming $t_n \neq0. $$ \frac{t_{n+1}}{t_{n}}=\frac{t_n+t_{n1}}{t_n}\le \frac{2t_n}{t_n}=2$$
so $\frac{t_{n+1}}{t_{n}} \le 2$ and as all the number are greater than $0$ so we get that $$0\le \frac{t_{n+1}}{t_{n}} \le 2$$ and as $t_{n+1}\ge t_n$ we know that it converges to a value and doesn't oscillate.
Hope that helps
Binet's formula gives an answer:
$F(n)=\frac{(\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2})^n}{\sqrt(5)}$
so
$\lim_{n_\to\infty}\frac{F(n+1)}{F(n)}=$
$\lim_{n_\to\infty}\frac{(\frac{1+\sqrt{5}}{2})^{n+1}-(\frac{1-\sqrt{5}}{2})^{n+1}}{(\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2})^n}=$
$\lim_{n_\to\infty}\frac{(\frac{1+\sqrt(5)}{2})(-3-\sqrt(5))^n-(\frac{1-\sqrt(5)}{2})}{(-3-\sqrt(5))^n-1}=$
$\lim_{n_\to\infty}\frac{1+\sqrt(5)}{2}=$
$\phi$
