Recurrence relation/with limit

Question

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 ${\xi^{n-1}(1-\frac{\phi^{n-1}}{\xi^{n-1}})}$?

Answer 1

$$F_{n+1}=F_n+F_{n-1}$$

$$\implies\dfrac{F_{n+1}}{F_n}=\dfrac{F_{n-1}}{F_n}+1$$

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 $\lim_{n\to\infty}\dfrac{F_{n+1}}{F_{n-1}}=\lim_{n\to\infty}\dfrac{F_{n+1}}{F_n}\cdot\lim_{n\to\infty}\dfrac{F_n}{F_{n-1}}=a^2$

Answer 2

After$$\lim_{n\to\infty}\frac{F_{n-1}}{F_{n+1}}=\lim_{n\to\infty}\frac{\frac{1}{\sqrt{5}}\left(\xi^{n-1}-\phi^{n-1}\right)}{\frac{1}{\sqrt{5}}\left(\xi^{n+1}-\phi^{n+1}\right)},$$you should have obtained$$\lim_{n\to\infty}\frac{\xi^{n-1}\left(1-\frac{\phi^{n-1}}{\xi^{n-1}}\right)}{\xi^{n+1}\left(1-\frac{\phi^{n+1}}{\xi^{n+1}}\right)},$$which is equal to$$\frac1{\xi^2}\lim_{n\to\infty}\frac{1-\frac{\phi^{n-1}}{\xi^{n-1}}}{1-\frac{\phi^{n+1}}{\xi^{n+1}}}=\frac1{\xi^2}.$$

Answer 3

We have$$\lim\limits_{n\to\infty}\frac{\xi^{n-1}(1-\frac{\phi^{n-1}}{\xi^{n-1}})}{\xi^{n+1}(1-\frac{\phi^{n+1}}{\xi^{n+1}})}={1\over \xi ^2}\lim_{n\to \infty}{1-\left({\phi\over \xi}\right)^{n-1}\over 1-\left({\phi\over \xi}\right)^{n+1}}$$where $${\phi\over \xi}={{1-\sqrt 5\over 2}\over {1+\sqrt 5\over 2}}={1-\sqrt 5\over 1+\sqrt 5}$$therefore $-1<{\phi \over \xi}<0$and we obtain $$\lim_{n\to \infty}{1-\left({\phi\over \xi}\right)^{n-1}\over 1-\left({\phi\over \xi}\right)^{n+1}}={1\over \xi ^2}={3-\sqrt 5\over 2}$$

Answer 4

(Note that the sequence is ill-posed as you haven't specified $F_2$. Nevertheless, we may proceed regardless of the value.)

Let's suppose that $F_n=\alpha^n$. Then the recurrence relation tells us that $\alpha^{n+1}=\alpha^{n-1}+\alpha^{n-2}$, so that $\alpha^3-\alpha-1=0$. Because the recurrence is linear with three degrees of freedom, all solutions will be of the form $F_n=c_1\alpha_1+c_2\alpha_2+c_3\alpha_3$, where $c_i$ are constants that depend on $F_0$, $F_1$, and $F_2$, and $\alpha_i$ are the roots of $\alpha^3-\alpha-1=0$.

Luckily, this equation has a unique root with magnitude greater than 1, which is approximately 1.32. Call this $\alpha_*$. Then $$\lim_{n\to\infty}\frac{F_{n-1}}{F_{n+1}}=\lim_{n\to\infty}\frac{F_{n-1}}{F_n}\lim_{n\to\infty}\frac{F_n}{F_{n+1}}=\alpha_*^{-2}\approx.57.$$