Recurrence relation/with limit












1














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






recurrence-relations







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  • How do you prove existence of $limlimits_{ntoinfty}frac{F_{n-1}}{F_{n+1}}$.
    – Yadati Kiran
    Nov 27 at 12:07






  • 2




    Are you sure it's $$F_{n+1} = F_{n-1} + F_{n-2}$$ and not $$F_n=F_{n-1}+F_{n-2}?$$
    – 5xum
    Nov 27 at 12:08






  • 5




    What is $F_2{}$?
    – Arthur
    Nov 27 at 12:11






  • 3




    In that case, how do you evaluate $displaystyle F_{2}$ ?.
    – Felix Marin
    Nov 27 at 17:21






  • 1




    Binet's formula is only valid if $F_n=F_{n-1}+F_{n-2}$.
    – Teepeemm
    Nov 27 at 18:07
















1














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






recurrence-relations







share|cite|improve this question
























  • How do you prove existence of $limlimits_{ntoinfty}frac{F_{n-1}}{F_{n+1}}$.
    – Yadati Kiran
    Nov 27 at 12:07






  • 2




    Are you sure it's $$F_{n+1} = F_{n-1} + F_{n-2}$$ and not $$F_n=F_{n-1}+F_{n-2}?$$
    – 5xum
    Nov 27 at 12:08






  • 5




    What is $F_2{}$?
    – Arthur
    Nov 27 at 12:11






  • 3




    In that case, how do you evaluate $displaystyle F_{2}$ ?.
    – Felix Marin
    Nov 27 at 17:21






  • 1




    Binet's formula is only valid if $F_n=F_{n-1}+F_{n-2}$.
    – Teepeemm
    Nov 27 at 18:07














1












1








1


1





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






recurrence-relations







share|cite|improve this 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



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



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






recurrence-relations




recurrence-relations






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share|cite|improve this question













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edited Nov 27 at 12:37









Mostafa Ayaz

13.7k3836




13.7k3836










asked Nov 27 at 12:05









Nekarts

234




234












  • How do you prove existence of $limlimits_{ntoinfty}frac{F_{n-1}}{F_{n+1}}$.
    – Yadati Kiran
    Nov 27 at 12:07






  • 2




    Are you sure it's $$F_{n+1} = F_{n-1} + F_{n-2}$$ and not $$F_n=F_{n-1}+F_{n-2}?$$
    – 5xum
    Nov 27 at 12:08






  • 5




    What is $F_2{}$?
    – Arthur
    Nov 27 at 12:11






  • 3




    In that case, how do you evaluate $displaystyle F_{2}$ ?.
    – Felix Marin
    Nov 27 at 17:21






  • 1




    Binet's formula is only valid if $F_n=F_{n-1}+F_{n-2}$.
    – Teepeemm
    Nov 27 at 18:07


















  • How do you prove existence of $limlimits_{ntoinfty}frac{F_{n-1}}{F_{n+1}}$.
    – Yadati Kiran
    Nov 27 at 12:07






  • 2




    Are you sure it's $$F_{n+1} = F_{n-1} + F_{n-2}$$ and not $$F_n=F_{n-1}+F_{n-2}?$$
    – 5xum
    Nov 27 at 12:08






  • 5




    What is $F_2{}$?
    – Arthur
    Nov 27 at 12:11






  • 3




    In that case, how do you evaluate $displaystyle F_{2}$ ?.
    – Felix Marin
    Nov 27 at 17:21






  • 1




    Binet's formula is only valid if $F_n=F_{n-1}+F_{n-2}$.
    – Teepeemm
    Nov 27 at 18:07
















How do you prove existence of $limlimits_{ntoinfty}frac{F_{n-1}}{F_{n+1}}$.
– Yadati Kiran
Nov 27 at 12:07




How do you prove existence of $limlimits_{ntoinfty}frac{F_{n-1}}{F_{n+1}}$.
– Yadati Kiran
Nov 27 at 12:07




2




2




Are you sure it's $$F_{n+1} = F_{n-1} + F_{n-2}$$ and not $$F_n=F_{n-1}+F_{n-2}?$$
– 5xum
Nov 27 at 12:08




Are you sure it's $$F_{n+1} = F_{n-1} + F_{n-2}$$ and not $$F_n=F_{n-1}+F_{n-2}?$$
– 5xum
Nov 27 at 12:08




5




5




What is $F_2{}$?
– Arthur
Nov 27 at 12:11




What is $F_2{}$?
– Arthur
Nov 27 at 12:11




3




3




In that case, how do you evaluate $displaystyle F_{2}$ ?.
– Felix Marin
Nov 27 at 17:21




In that case, how do you evaluate $displaystyle F_{2}$ ?.
– Felix Marin
Nov 27 at 17:21




1




1




Binet's formula is only valid if $F_n=F_{n-1}+F_{n-2}$.
– Teepeemm
Nov 27 at 18:07




Binet's formula is only valid if $F_n=F_{n-1}+F_{n-2}$.
– Teepeemm
Nov 27 at 18:07










4 Answers
4






active

oldest

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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 $$lim_{nto infty}{1-left({phiover xi}right)^{n-1}over 1-left({phiover xi}right)^{n+1}}={1over xi ^2}={3-sqrt 5over 2}$$






share|cite|improve this answer





















  • Whether intentionally or not, OP is not using the Fibonacci sequence, so Binet's formula is not valid.
    – Teepeemm
    Dec 3 at 16:20










  • It is a Fibonacci sequence. $0,1,1,2,3,5,8,cdots$
    – Mostafa Ayaz
    Dec 3 at 16:22










  • Not if $F_2=15$. Double check the comments on the original post.
    – Teepeemm
    Dec 3 at 16:23



















5














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



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



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



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






share|cite|improve this answer





















  • OP specified that $F_{n+1}=F_{n-1}+F_{n-2}$.
    – Teepeemm
    Dec 3 at 16:21



















3














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$$frac1{xi^2}lim_{ntoinfty}frac{1-frac{phi^{n-1}}{xi^{n-1}}}{1-frac{phi^{n+1}}{xi^{n+1}}}=frac1{xi^2}.$$






share|cite|improve this answer





















  • Whether intentionally or not, OP is not using the Fibonacci sequence, so Binet's formula is not valid.
    – Teepeemm
    Dec 3 at 16:20



















0














(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 $F_n=c_1alpha_1+c_2alpha_2+c_3alpha_3$, 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
$$lim_{ntoinfty}frac{F_{n-1}}{F_{n+1}}=lim_{ntoinfty}frac{F_{n-1}}{F_n}lim_{ntoinfty}frac{F_n}{F_{n+1}}=alpha_*^{-2}approx.57.$$






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    4 Answers
    4






    active

    oldest

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    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    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 $$lim_{nto infty}{1-left({phiover xi}right)^{n-1}over 1-left({phiover xi}right)^{n+1}}={1over xi ^2}={3-sqrt 5over 2}$$






    share|cite|improve this answer





















    • Whether intentionally or not, OP is not using the Fibonacci sequence, so Binet's formula is not valid.
      – Teepeemm
      Dec 3 at 16:20










    • It is a Fibonacci sequence. $0,1,1,2,3,5,8,cdots$
      – Mostafa Ayaz
      Dec 3 at 16:22










    • Not if $F_2=15$. Double check the comments on the original post.
      – Teepeemm
      Dec 3 at 16:23
















    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 $$lim_{nto infty}{1-left({phiover xi}right)^{n-1}over 1-left({phiover xi}right)^{n+1}}={1over xi ^2}={3-sqrt 5over 2}$$






    share|cite|improve this answer





















    • Whether intentionally or not, OP is not using the Fibonacci sequence, so Binet's formula is not valid.
      – Teepeemm
      Dec 3 at 16:20










    • It is a Fibonacci sequence. $0,1,1,2,3,5,8,cdots$
      – Mostafa Ayaz
      Dec 3 at 16:22










    • Not if $F_2=15$. Double check the comments on the original post.
      – Teepeemm
      Dec 3 at 16:23














    1












    1








    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 $$lim_{nto infty}{1-left({phiover xi}right)^{n-1}over 1-left({phiover xi}right)^{n+1}}={1over xi ^2}={3-sqrt 5over 2}$$






    share|cite|improve this answer












    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 $$lim_{nto infty}{1-left({phiover xi}right)^{n-1}over 1-left({phiover xi}right)^{n+1}}={1over xi ^2}={3-sqrt 5over 2}$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 27 at 12:34









    Mostafa Ayaz

    13.7k3836




    13.7k3836












    • Whether intentionally or not, OP is not using the Fibonacci sequence, so Binet's formula is not valid.
      – Teepeemm
      Dec 3 at 16:20










    • It is a Fibonacci sequence. $0,1,1,2,3,5,8,cdots$
      – Mostafa Ayaz
      Dec 3 at 16:22










    • Not if $F_2=15$. Double check the comments on the original post.
      – Teepeemm
      Dec 3 at 16:23


















    • Whether intentionally or not, OP is not using the Fibonacci sequence, so Binet's formula is not valid.
      – Teepeemm
      Dec 3 at 16:20










    • It is a Fibonacci sequence. $0,1,1,2,3,5,8,cdots$
      – Mostafa Ayaz
      Dec 3 at 16:22










    • Not if $F_2=15$. Double check the comments on the original post.
      – Teepeemm
      Dec 3 at 16:23
















    Whether intentionally or not, OP is not using the Fibonacci sequence, so Binet's formula is not valid.
    – Teepeemm
    Dec 3 at 16:20




    Whether intentionally or not, OP is not using the Fibonacci sequence, so Binet's formula is not valid.
    – Teepeemm
    Dec 3 at 16:20












    It is a Fibonacci sequence. $0,1,1,2,3,5,8,cdots$
    – Mostafa Ayaz
    Dec 3 at 16:22




    It is a Fibonacci sequence. $0,1,1,2,3,5,8,cdots$
    – Mostafa Ayaz
    Dec 3 at 16:22












    Not if $F_2=15$. Double check the comments on the original post.
    – Teepeemm
    Dec 3 at 16:23




    Not if $F_2=15$. Double check the comments on the original post.
    – Teepeemm
    Dec 3 at 16:23











    5














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



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



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



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






    share|cite|improve this answer





















    • OP specified that $F_{n+1}=F_{n-1}+F_{n-2}$.
      – Teepeemm
      Dec 3 at 16:21
















    5














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



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



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



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






    share|cite|improve this answer





















    • OP specified that $F_{n+1}=F_{n-1}+F_{n-2}$.
      – Teepeemm
      Dec 3 at 16:21














    5












    5








    5






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



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



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



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






    share|cite|improve this answer












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



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



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



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







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 27 at 13:06









    lab bhattacharjee

    223k15156274




    223k15156274












    • OP specified that $F_{n+1}=F_{n-1}+F_{n-2}$.
      – Teepeemm
      Dec 3 at 16:21


















    • OP specified that $F_{n+1}=F_{n-1}+F_{n-2}$.
      – Teepeemm
      Dec 3 at 16:21
















    OP specified that $F_{n+1}=F_{n-1}+F_{n-2}$.
    – Teepeemm
    Dec 3 at 16:21




    OP specified that $F_{n+1}=F_{n-1}+F_{n-2}$.
    – Teepeemm
    Dec 3 at 16:21











    3














    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$$frac1{xi^2}lim_{ntoinfty}frac{1-frac{phi^{n-1}}{xi^{n-1}}}{1-frac{phi^{n+1}}{xi^{n+1}}}=frac1{xi^2}.$$






    share|cite|improve this answer





















    • Whether intentionally or not, OP is not using the Fibonacci sequence, so Binet's formula is not valid.
      – Teepeemm
      Dec 3 at 16:20
















    3














    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$$frac1{xi^2}lim_{ntoinfty}frac{1-frac{phi^{n-1}}{xi^{n-1}}}{1-frac{phi^{n+1}}{xi^{n+1}}}=frac1{xi^2}.$$






    share|cite|improve this answer





















    • Whether intentionally or not, OP is not using the Fibonacci sequence, so Binet's formula is not valid.
      – Teepeemm
      Dec 3 at 16:20














    3












    3








    3






    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$$frac1{xi^2}lim_{ntoinfty}frac{1-frac{phi^{n-1}}{xi^{n-1}}}{1-frac{phi^{n+1}}{xi^{n+1}}}=frac1{xi^2}.$$






    share|cite|improve this answer












    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$$frac1{xi^2}lim_{ntoinfty}frac{1-frac{phi^{n-1}}{xi^{n-1}}}{1-frac{phi^{n+1}}{xi^{n+1}}}=frac1{xi^2}.$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 27 at 12:12









    José Carlos Santos

    150k22120221




    150k22120221












    • Whether intentionally or not, OP is not using the Fibonacci sequence, so Binet's formula is not valid.
      – Teepeemm
      Dec 3 at 16:20


















    • Whether intentionally or not, OP is not using the Fibonacci sequence, so Binet's formula is not valid.
      – Teepeemm
      Dec 3 at 16:20
















    Whether intentionally or not, OP is not using the Fibonacci sequence, so Binet's formula is not valid.
    – Teepeemm
    Dec 3 at 16:20




    Whether intentionally or not, OP is not using the Fibonacci sequence, so Binet's formula is not valid.
    – Teepeemm
    Dec 3 at 16:20











    0














    (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 $F_n=c_1alpha_1+c_2alpha_2+c_3alpha_3$, 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
    $$lim_{ntoinfty}frac{F_{n-1}}{F_{n+1}}=lim_{ntoinfty}frac{F_{n-1}}{F_n}lim_{ntoinfty}frac{F_n}{F_{n+1}}=alpha_*^{-2}approx.57.$$






    share|cite|improve this answer


























      0














      (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 $F_n=c_1alpha_1+c_2alpha_2+c_3alpha_3$, 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
      $$lim_{ntoinfty}frac{F_{n-1}}{F_{n+1}}=lim_{ntoinfty}frac{F_{n-1}}{F_n}lim_{ntoinfty}frac{F_n}{F_{n+1}}=alpha_*^{-2}approx.57.$$






      share|cite|improve this answer
























        0












        0








        0






        (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 $F_n=c_1alpha_1+c_2alpha_2+c_3alpha_3$, 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
        $$lim_{ntoinfty}frac{F_{n-1}}{F_{n+1}}=lim_{ntoinfty}frac{F_{n-1}}{F_n}lim_{ntoinfty}frac{F_n}{F_{n+1}}=alpha_*^{-2}approx.57.$$






        share|cite|improve this answer












        (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 $F_n=c_1alpha_1+c_2alpha_2+c_3alpha_3$, 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
        $$lim_{ntoinfty}frac{F_{n-1}}{F_{n+1}}=lim_{ntoinfty}frac{F_{n-1}}{F_n}lim_{ntoinfty}frac{F_n}{F_{n+1}}=alpha_*^{-2}approx.57.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 at 16:46









        Teepeemm

        69259




        69259






























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