Concerning the limit of the numerator of a rational function












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I have a basic question about the limits of rational sequences:



Suppose we know that a sequence $(a_{n})$ converges to $0$ and that each $a_n$ is defined by $a_n := frac{b_n}{c_n}$ for two sequences $(b_{n})$ and $(c_{n})$. We further know that $(c_{n})$ goes to infinity, i.e



$$lim_{n rightarrow infty} c_n = infty.$$



Can we deduce from this that $lim_{n rightarrow infty} b_n = 0$ and if yes how?



Edit: I have an additional question: Are we at least able to deduce that $b_{(n)}$ converges?






sequences-and-series limits







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    0












    $begingroup$


    I have a basic question about the limits of rational sequences:



    Suppose we know that a sequence $(a_{n})$ converges to $0$ and that each $a_n$ is defined by $a_n := frac{b_n}{c_n}$ for two sequences $(b_{n})$ and $(c_{n})$. We further know that $(c_{n})$ goes to infinity, i.e



    $$lim_{n rightarrow infty} c_n = infty.$$



    Can we deduce from this that $lim_{n rightarrow infty} b_n = 0$ and if yes how?



    Edit: I have an additional question: Are we at least able to deduce that $b_{(n)}$ converges?






    sequences-and-series limits







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I have a basic question about the limits of rational sequences:



      Suppose we know that a sequence $(a_{n})$ converges to $0$ and that each $a_n$ is defined by $a_n := frac{b_n}{c_n}$ for two sequences $(b_{n})$ and $(c_{n})$. We further know that $(c_{n})$ goes to infinity, i.e



      $$lim_{n rightarrow infty} c_n = infty.$$



      Can we deduce from this that $lim_{n rightarrow infty} b_n = 0$ and if yes how?



      Edit: I have an additional question: Are we at least able to deduce that $b_{(n)}$ converges?






      sequences-and-series limits







      share|cite|improve this question











      $endgroup$




      I have a basic question about the limits of rational sequences:



      Suppose we know that a sequence $(a_{n})$ converges to $0$ and that each $a_n$ is defined by $a_n := frac{b_n}{c_n}$ for two sequences $(b_{n})$ and $(c_{n})$. We further know that $(c_{n})$ goes to infinity, i.e



      $$lim_{n rightarrow infty} c_n = infty.$$



      Can we deduce from this that $lim_{n rightarrow infty} b_n = 0$ and if yes how?



      Edit: I have an additional question: Are we at least able to deduce that $b_{(n)}$ converges?






      sequences-and-series limits




      sequences-and-series limits






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      share|cite|improve this question













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      edited Dec 17 '18 at 14:10







      3nondatur

















      asked Dec 17 '18 at 13:10









      3nondatur3nondatur

      399111




      399111






















          2 Answers
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          $begingroup$

          No.



          What about $$left( a_n right)=left( frac{1}{n}right)$$



          where $a_n=frac{1}{n}$,
          $ b_n=1$ and $ c_n=n$



          Edit-Edited to answer your additional question.



          Again No.



          What about $$left( a_n right)=left( frac{n}{n^2}right)$$



          where $a_n=frac{n}{n^2}$,
          $ b_n=n$ and $ c_n=n^2$






          share|cite|improve this answer











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            1












            $begingroup$

            No, that’s not true. For instance, for $b_n = n$ and $c_n = n^2$, we have



            $$lim_{n to infty}frac{b_n}{c_n} = 0$$



            while both sequences tend to $infty$. More generally, when both $b_n$ and $c_n$ are polynomials with $c_n$ having the higher degree, the limit tends to $0$, even though both polynomials may tend to $infty$.



            Edit: Also no, the example above once again shows this. Both polynomials clearly diverge to $infty$, but since the denominator has a greater degree, it diverges faster. Hence, the limit tends to $0$.






            share|cite|improve this answer











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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

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              active

              oldest

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              active

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              3












              $begingroup$

              No.



              What about $$left( a_n right)=left( frac{1}{n}right)$$



              where $a_n=frac{1}{n}$,
              $ b_n=1$ and $ c_n=n$



              Edit-Edited to answer your additional question.



              Again No.



              What about $$left( a_n right)=left( frac{n}{n^2}right)$$



              where $a_n=frac{n}{n^2}$,
              $ b_n=n$ and $ c_n=n^2$






              share|cite|improve this answer











              $endgroup$


















                3












                $begingroup$

                No.



                What about $$left( a_n right)=left( frac{1}{n}right)$$



                where $a_n=frac{1}{n}$,
                $ b_n=1$ and $ c_n=n$



                Edit-Edited to answer your additional question.



                Again No.



                What about $$left( a_n right)=left( frac{n}{n^2}right)$$



                where $a_n=frac{n}{n^2}$,
                $ b_n=n$ and $ c_n=n^2$






                share|cite|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  No.



                  What about $$left( a_n right)=left( frac{1}{n}right)$$



                  where $a_n=frac{1}{n}$,
                  $ b_n=1$ and $ c_n=n$



                  Edit-Edited to answer your additional question.



                  Again No.



                  What about $$left( a_n right)=left( frac{n}{n^2}right)$$



                  where $a_n=frac{n}{n^2}$,
                  $ b_n=n$ and $ c_n=n^2$






                  share|cite|improve this answer











                  $endgroup$



                  No.



                  What about $$left( a_n right)=left( frac{1}{n}right)$$



                  where $a_n=frac{1}{n}$,
                  $ b_n=1$ and $ c_n=n$



                  Edit-Edited to answer your additional question.



                  Again No.



                  What about $$left( a_n right)=left( frac{n}{n^2}right)$$



                  where $a_n=frac{n}{n^2}$,
                  $ b_n=n$ and $ c_n=n^2$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 17 '18 at 13:58

























                  answered Dec 17 '18 at 13:13









                  Rakesh BhattRakesh Bhatt

                  952214




                  952214























                      1












                      $begingroup$

                      No, that’s not true. For instance, for $b_n = n$ and $c_n = n^2$, we have



                      $$lim_{n to infty}frac{b_n}{c_n} = 0$$



                      while both sequences tend to $infty$. More generally, when both $b_n$ and $c_n$ are polynomials with $c_n$ having the higher degree, the limit tends to $0$, even though both polynomials may tend to $infty$.



                      Edit: Also no, the example above once again shows this. Both polynomials clearly diverge to $infty$, but since the denominator has a greater degree, it diverges faster. Hence, the limit tends to $0$.






                      share|cite|improve this answer











                      $endgroup$


















                        1












                        $begingroup$

                        No, that’s not true. For instance, for $b_n = n$ and $c_n = n^2$, we have



                        $$lim_{n to infty}frac{b_n}{c_n} = 0$$



                        while both sequences tend to $infty$. More generally, when both $b_n$ and $c_n$ are polynomials with $c_n$ having the higher degree, the limit tends to $0$, even though both polynomials may tend to $infty$.



                        Edit: Also no, the example above once again shows this. Both polynomials clearly diverge to $infty$, but since the denominator has a greater degree, it diverges faster. Hence, the limit tends to $0$.






                        share|cite|improve this answer











                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          No, that’s not true. For instance, for $b_n = n$ and $c_n = n^2$, we have



                          $$lim_{n to infty}frac{b_n}{c_n} = 0$$



                          while both sequences tend to $infty$. More generally, when both $b_n$ and $c_n$ are polynomials with $c_n$ having the higher degree, the limit tends to $0$, even though both polynomials may tend to $infty$.



                          Edit: Also no, the example above once again shows this. Both polynomials clearly diverge to $infty$, but since the denominator has a greater degree, it diverges faster. Hence, the limit tends to $0$.






                          share|cite|improve this answer











                          $endgroup$



                          No, that’s not true. For instance, for $b_n = n$ and $c_n = n^2$, we have



                          $$lim_{n to infty}frac{b_n}{c_n} = 0$$



                          while both sequences tend to $infty$. More generally, when both $b_n$ and $c_n$ are polynomials with $c_n$ having the higher degree, the limit tends to $0$, even though both polynomials may tend to $infty$.



                          Edit: Also no, the example above once again shows this. Both polynomials clearly diverge to $infty$, but since the denominator has a greater degree, it diverges faster. Hence, the limit tends to $0$.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Dec 17 '18 at 14:01

























                          answered Dec 17 '18 at 13:22









                          KM101KM101

                          6,0251525




                          6,0251525






























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