Recurrence relation/with limit











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










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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















up vote
2
down vote

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










share|cite|improve this question









New contributor




Nekarts is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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













up vote
2
down vote

favorite
1









up vote
2
down vote

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










share|cite|improve this question









New contributor




Nekarts is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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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edited Nov 27 at 12:37









Mostafa Ayaz

13.1k3735




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asked Nov 27 at 12:05









Nekarts

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Nekarts is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Check out our Code of Conduct.












  • 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










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






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    up vote
    5
    down vote













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



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



    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




























      up vote
      3
      down vote













      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





















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        3 Answers
        3






        active

        oldest

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        3 Answers
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        active

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        up vote
        1
        down vote



        accepted










        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

























          up vote
          1
          down vote



          accepted










          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























            up vote
            1
            down vote



            accepted







            up vote
            1
            down vote



            accepted






            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

            13.1k3735




            13.1k3735






















                up vote
                5
                down vote













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



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



                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

























                  up vote
                  5
                  down vote













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



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



                  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























                    up vote
                    5
                    down vote










                    up vote
                    5
                    down vote









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



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



                    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












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                    answered Nov 27 at 13:06









                    lab bhattacharjee

                    221k15154271




                    221k15154271






















                        up vote
                        3
                        down vote













                        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

























                          up vote
                          3
                          down vote













                          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























                            up vote
                            3
                            down vote










                            up vote
                            3
                            down vote









                            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












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                            answered Nov 27 at 12:12









                            José Carlos Santos

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