Fibonacci golden ratio question without assuming convergence a priori 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?
 A: 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 $\phi$.
A: 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
A: Binet's formula gives an answer:
$F(n)=\frac{(\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2})^n}{\sqrt(5)}$
so
$\lim_{n_\to\infty}\frac{F(n+1)}{F(n)}=$
$\lim_{n_\to\infty}\frac{(\frac{1+\sqrt{5}}{2})^{n+1}-(\frac{1-\sqrt{5}}{2})^{n+1}}{(\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2})^n}=$
$\lim_{n_\to\infty}\frac{(\frac{1+\sqrt(5)}{2})(-3-\sqrt(5))^n-(\frac{1-\sqrt(5)}{2})}{(-3-\sqrt(5))^n-1}=$
$\lim_{n_\to\infty}\frac{1+\sqrt(5)}{2}=$
$\phi$
