Cfrtjryk

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

$limlimits_{ntoinfty}frac{F_{n-1}}{F_{n+1}}$?

I tried to use Binet's formula:

$limlimits_{ntoinfty}frac{F_{n-1}}{F_{n+1}}=limlimits_{ntoinfty}frac{frac{1}{sqrt{5}}(xi^{n-1}-phi^{n-1})}{frac{1}{sqrt{5}}(xi^{n+1}-phi^{n+1})}=limlimits_{xtoinfty}frac{xi^{n-1}(1-frac{phi^{n-1}}{xi^{n-1}})}{xi^{n+1}(1-frac{phi^{n+1}}{xi^{n+1}})}$

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

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

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

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

If $lim_{ntoinfty}dfrac{F_{n+1}}{F_n}=a,$ we have $$a=dfrac1a+1iff a^2-a-1=0, a=?$$

Finally $lim_{ntoinfty}dfrac{F_{n+1}}{F_{n-1}}=lim_{ntoinfty}dfrac{F_{n+1}}{F_n}cdotlim_{ntoinfty}dfrac{F_n}{F_{n-1}}=a^2$

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