Let $F_{n+1}=F_{n-1}+F_{n-2}$ for $n \in \mathbb{N}$ with $n \geq 2$
$F_0:=0$ and $F_1:=1$.
How to compute
$\lim\limits_{n\to\infty}\frac{F_{n-1}}{F_{n+1}}$?
I tried to use Binet's formula:
$\lim\limits_{n\to\infty}\frac{F_{n-1}}{F_{n+1}}=\lim\limits_{n\to\infty}\frac{\frac{1}{\sqrt{5}}(\xi^{n-1}-\phi^{n-1})}{\frac{1}{\sqrt{5}}(\xi^{n+1}-\phi^{n+1})}=\lim\limits_{x\to\infty}\frac{\xi^{n-1}(1-\frac{\phi^{n-1}}{\xi^{n-1}})}{\xi^{n+1}(1-\frac{\phi^{n+1}}{\xi^{n+1}})}$
But I don't know what to do next.
I suppose $\xi^{n+1}(1-\frac{\phi^{n+1}}{\xi^{n+1}})=\xi$ but what about ${\xi^{n-1}(1-\frac{\phi^{n-1}}{\xi^{n-1}})}$?
Best Answer
$$F_{n+1}=F_n+F_{n-1}$$
$$\implies\dfrac{F_{n+1}}{F_n}=\dfrac{F_{n-1}}{F_n}+1$$
If $\lim_{n\to\infty}\dfrac{F_{n+1}}{F_n}=a,$ we have $$a=\dfrac1a+1\iff a^2-a-1=0, a=?$$
Finally $\lim_{n\to\infty}\dfrac{F_{n+1}}{F_{n-1}}=\lim_{n\to\infty}\dfrac{F_{n+1}}{F_n}\cdot\lim_{n\to\infty}\dfrac{F_n}{F_{n-1}}=a^2$