Proving $ \lim_{n\to\infty} \dfrac{ \Phi^{n+1} - (1 - \Phi)^{n+1}}{\Phi^{n} - (1 - \Phi)^n} = \Phi $

Question

$ \Phi = \frac{1 + \sqrt{5}}{2} $ is the golden ratio

I'm having hard time using proving that $$ \lim_{n\to\infty} \dfrac{ \Phi^{n+1} - (1 - \Phi)^{n+1}}{\Phi^{n} - (1 - \Phi)^n} = \Phi $$ dividing both the numerator and denominator by $ \Phi^{n} $ doesn't help, neither does $ \Phi^{n} - (1 - \Phi^{n}) = (2 \Phi +1 ) \sum\limits_{i=0}^{n-1} \Phi^i (1 - \Phi)^{n-1-i} $

Where is the trick?

Answer 1

Hint:

$\Phi-1=\dfrac{\sqrt5-1}2=\dfrac{5-1}{2(\sqrt5+1)}=\dfrac2{\sqrt5+1}<1$ and $>0$

$$\implies|1-\Phi|<1\text{ and }\left|\dfrac{1-\Phi}\Phi\right|<1$$

Divide the numerator and the denominator by $\Phi^n$

Answer 2

Use How do I prove Binet's Formula?

if $F(m)=\dfrac{\alpha^m-\beta^m}{\alpha-\beta}$ with $\alpha,\beta$ are the roots of $$t^2-t-1=0$$

we can prove $$F_{n+2}=F_{n+1}+F_n$$

$$\dfrac{F_{n+2}}{F_{n+1}}=1+\dfrac1{\dfrac{F_{n+1}}{F_{n}}}$$

If $\lim_{n\to\infty}\dfrac{F_{n+2}}{F_{n+1}}=r>0,$ $$r=1+\dfrac1r\iff r^2-r-1=0, r=?$$