Prove that $lim_{nto infty} a_n =0$












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Let $(a_n) _{nge 1}$ be a sequence of real numbers so that $lim_{ntoinfty} (a_n^2-a_{n-1}a_{n+1})=0$ and $lim_{ntoinfty} (a_{n+1}+a_{n-1})=0$. Prove that $lim_{nto infty} a_n =0$.
I observed that $a_n$ must be bounded because otherwise the second relation doesn't hold, but I can't see how to continue.






real-analysis sequences-and-series







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    $begingroup$


    Let $(a_n) _{nge 1}$ be a sequence of real numbers so that $lim_{ntoinfty} (a_n^2-a_{n-1}a_{n+1})=0$ and $lim_{ntoinfty} (a_{n+1}+a_{n-1})=0$. Prove that $lim_{nto infty} a_n =0$.
    I observed that $a_n$ must be bounded because otherwise the second relation doesn't hold, but I can't see how to continue.






    real-analysis sequences-and-series







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let $(a_n) _{nge 1}$ be a sequence of real numbers so that $lim_{ntoinfty} (a_n^2-a_{n-1}a_{n+1})=0$ and $lim_{ntoinfty} (a_{n+1}+a_{n-1})=0$. Prove that $lim_{nto infty} a_n =0$.
      I observed that $a_n$ must be bounded because otherwise the second relation doesn't hold, but I can't see how to continue.






      real-analysis sequences-and-series







      share|cite|improve this question









      $endgroup$




      Let $(a_n) _{nge 1}$ be a sequence of real numbers so that $lim_{ntoinfty} (a_n^2-a_{n-1}a_{n+1})=0$ and $lim_{ntoinfty} (a_{n+1}+a_{n-1})=0$. Prove that $lim_{nto infty} a_n =0$.
      I observed that $a_n$ must be bounded because otherwise the second relation doesn't hold, but I can't see how to continue.






      real-analysis sequences-and-series




      real-analysis sequences-and-series






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      asked Jan 21 at 19:08









      JustAnAmateurJustAnAmateur

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          $begingroup$

          $(a_{n+1} + a_{n-1}) to 0$ implies
          $$(a_{n+1} + a_{n-1})^2 = a_{n+1}^2 + 2 a_{n+1} a_{n-1} + a_{n-1}^2 to 0.$$
          Combining this with the other assumption shows that
          $$a_{n+1}^2 + 2 a_n^2 + a_{n-1}^2 to 0$$
          from which you may conclude $a_n to 0$.






          share|cite|improve this answer









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            $begingroup$

            $(a_{n+1} + a_{n-1}) to 0$ implies
            $$(a_{n+1} + a_{n-1})^2 = a_{n+1}^2 + 2 a_{n+1} a_{n-1} + a_{n-1}^2 to 0.$$
            Combining this with the other assumption shows that
            $$a_{n+1}^2 + 2 a_n^2 + a_{n-1}^2 to 0$$
            from which you may conclude $a_n to 0$.






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              $(a_{n+1} + a_{n-1}) to 0$ implies
              $$(a_{n+1} + a_{n-1})^2 = a_{n+1}^2 + 2 a_{n+1} a_{n-1} + a_{n-1}^2 to 0.$$
              Combining this with the other assumption shows that
              $$a_{n+1}^2 + 2 a_n^2 + a_{n-1}^2 to 0$$
              from which you may conclude $a_n to 0$.






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                $(a_{n+1} + a_{n-1}) to 0$ implies
                $$(a_{n+1} + a_{n-1})^2 = a_{n+1}^2 + 2 a_{n+1} a_{n-1} + a_{n-1}^2 to 0.$$
                Combining this with the other assumption shows that
                $$a_{n+1}^2 + 2 a_n^2 + a_{n-1}^2 to 0$$
                from which you may conclude $a_n to 0$.






                share|cite|improve this answer









                $endgroup$



                $(a_{n+1} + a_{n-1}) to 0$ implies
                $$(a_{n+1} + a_{n-1})^2 = a_{n+1}^2 + 2 a_{n+1} a_{n-1} + a_{n-1}^2 to 0.$$
                Combining this with the other assumption shows that
                $$a_{n+1}^2 + 2 a_n^2 + a_{n-1}^2 to 0$$
                from which you may conclude $a_n to 0$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 21 at 19:23









                angryavianangryavian

                42k23381




                42k23381






























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