if $|a|<1$ so $lim_{nto infty}na^n=0$.












3












$begingroup$


Prove that if $|a|<1$ ($a$ is real) so $lim_{nto infty}na^n=0$.



I know that I need to use squeeze theory (because I have $-1<a<1$) but I dont see how.



thanks






sequences-and-series limits







share|cite|improve this question









$endgroup$

















    3












    $begingroup$


    Prove that if $|a|<1$ ($a$ is real) so $lim_{nto infty}na^n=0$.



    I know that I need to use squeeze theory (because I have $-1<a<1$) but I dont see how.



    thanks






    sequences-and-series limits







    share|cite|improve this question









    $endgroup$















      3












      3








      3


      2



      $begingroup$


      Prove that if $|a|<1$ ($a$ is real) so $lim_{nto infty}na^n=0$.



      I know that I need to use squeeze theory (because I have $-1<a<1$) but I dont see how.



      thanks






      sequences-and-series limits







      share|cite|improve this question









      $endgroup$




      Prove that if $|a|<1$ ($a$ is real) so $lim_{nto infty}na^n=0$.



      I know that I need to use squeeze theory (because I have $-1<a<1$) but I dont see how.



      thanks






      sequences-and-series limits




      sequences-and-series limits






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Apr 5 '15 at 12:03









      StabiloStabilo

      734512




      734512






















          4 Answers
          4






          active

          oldest

          votes


















          5












          $begingroup$

          If you want to do it by hand: since $|a| < 1$, there exists $N$ such that $frac{n+1}{n}|a| < q < 1, forall n geq N$. Hence $n|a|^n = frac{n}{n-1}|a|frac{n-1}{n-2}|a|dotsfrac{N+1}{N}|a| cdot N|a|^N < q^{n-N} cdot N|a|^N, , forall ngeq N.$
          The latter converges to zero and you get your squeezing sequence(s).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This is, as asked by the OP, the simplest, most basic answer here imo, with only basic stuff about sequences. +1
            $endgroup$
            – Timbuc
            Apr 5 '15 at 12:31



















          2












          $begingroup$

          If $|a| < 1,$ we claim that the series $sum_{1}^{infty}na^{n}$ converges, so that
          $na^{n} to 0$. It suffices to show that the series is absolutely convergent. But, since
          $$frac{(n+1)|a|^{n+1}}{n|a|^{n}} = (1+frac{1}{n})|a| to |a| < 1,$$ by the ratio test the series converges. Thus we have
          $$lim_{n to infty}na^{n} = 0.$$






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            You can use the relationship of sequences limits and function limits, just let $0<a<1$ for convenient, and study $xa^x,xrightarrowinfty$ you can use L.Hospital theorem, $lim frac{x}{a^{-x}}=lim frac{1}{-a^{-x}}frac{1}{lna}=0 $






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              just sequences, thanks
              $endgroup$
              – Stabilo
              Apr 5 '15 at 12:19










            • $begingroup$
              Oh, that's ok... in other conditions where you can use L.Hospital, don't forget it and it's so powerful.
              $endgroup$
              – Hans
              Apr 5 '15 at 12:23



















            0












            $begingroup$

            Here a very elementary way:





            • $|a| < 1 stackrel{a neq 0}{Rightarrow} |a| = frac{1}{1+p}$ for a $p > 0$.

            • $(1+p)^n = 1 + np + color{blue}{frac{n(n-1)}{2}p^2} + cdots + p^n color{blue}{>frac{n(n-1)}{2}p^2}$


            So, it follows
            $$0 leq |na^n| = n|a|^n = frac{n}{(1+p)^n} < frac{n}{color{blue}{frac{n(n-1)}{2}p^2}} = frac{2}{(n-1)p^2} stackrel{n to infty}{longrightarrow} 0$$






            share|cite|improve this answer









            $endgroup$













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              4 Answers
              4






              active

              oldest

              votes








              4 Answers
              4






              active

              oldest

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              active

              oldest

              votes






              active

              oldest

              votes









              5












              $begingroup$

              If you want to do it by hand: since $|a| < 1$, there exists $N$ such that $frac{n+1}{n}|a| < q < 1, forall n geq N$. Hence $n|a|^n = frac{n}{n-1}|a|frac{n-1}{n-2}|a|dotsfrac{N+1}{N}|a| cdot N|a|^N < q^{n-N} cdot N|a|^N, , forall ngeq N.$
              The latter converges to zero and you get your squeezing sequence(s).






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                This is, as asked by the OP, the simplest, most basic answer here imo, with only basic stuff about sequences. +1
                $endgroup$
                – Timbuc
                Apr 5 '15 at 12:31
















              5












              $begingroup$

              If you want to do it by hand: since $|a| < 1$, there exists $N$ such that $frac{n+1}{n}|a| < q < 1, forall n geq N$. Hence $n|a|^n = frac{n}{n-1}|a|frac{n-1}{n-2}|a|dotsfrac{N+1}{N}|a| cdot N|a|^N < q^{n-N} cdot N|a|^N, , forall ngeq N.$
              The latter converges to zero and you get your squeezing sequence(s).






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                This is, as asked by the OP, the simplest, most basic answer here imo, with only basic stuff about sequences. +1
                $endgroup$
                – Timbuc
                Apr 5 '15 at 12:31














              5












              5








              5





              $begingroup$

              If you want to do it by hand: since $|a| < 1$, there exists $N$ such that $frac{n+1}{n}|a| < q < 1, forall n geq N$. Hence $n|a|^n = frac{n}{n-1}|a|frac{n-1}{n-2}|a|dotsfrac{N+1}{N}|a| cdot N|a|^N < q^{n-N} cdot N|a|^N, , forall ngeq N.$
              The latter converges to zero and you get your squeezing sequence(s).






              share|cite|improve this answer









              $endgroup$



              If you want to do it by hand: since $|a| < 1$, there exists $N$ such that $frac{n+1}{n}|a| < q < 1, forall n geq N$. Hence $n|a|^n = frac{n}{n-1}|a|frac{n-1}{n-2}|a|dotsfrac{N+1}{N}|a| cdot N|a|^N < q^{n-N} cdot N|a|^N, , forall ngeq N.$
              The latter converges to zero and you get your squeezing sequence(s).







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Apr 5 '15 at 12:19









              user130065user130065

              1362




              1362












              • $begingroup$
                This is, as asked by the OP, the simplest, most basic answer here imo, with only basic stuff about sequences. +1
                $endgroup$
                – Timbuc
                Apr 5 '15 at 12:31


















              • $begingroup$
                This is, as asked by the OP, the simplest, most basic answer here imo, with only basic stuff about sequences. +1
                $endgroup$
                – Timbuc
                Apr 5 '15 at 12:31
















              $begingroup$
              This is, as asked by the OP, the simplest, most basic answer here imo, with only basic stuff about sequences. +1
              $endgroup$
              – Timbuc
              Apr 5 '15 at 12:31




              $begingroup$
              This is, as asked by the OP, the simplest, most basic answer here imo, with only basic stuff about sequences. +1
              $endgroup$
              – Timbuc
              Apr 5 '15 at 12:31











              2












              $begingroup$

              If $|a| < 1,$ we claim that the series $sum_{1}^{infty}na^{n}$ converges, so that
              $na^{n} to 0$. It suffices to show that the series is absolutely convergent. But, since
              $$frac{(n+1)|a|^{n+1}}{n|a|^{n}} = (1+frac{1}{n})|a| to |a| < 1,$$ by the ratio test the series converges. Thus we have
              $$lim_{n to infty}na^{n} = 0.$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                If $|a| < 1,$ we claim that the series $sum_{1}^{infty}na^{n}$ converges, so that
                $na^{n} to 0$. It suffices to show that the series is absolutely convergent. But, since
                $$frac{(n+1)|a|^{n+1}}{n|a|^{n}} = (1+frac{1}{n})|a| to |a| < 1,$$ by the ratio test the series converges. Thus we have
                $$lim_{n to infty}na^{n} = 0.$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  If $|a| < 1,$ we claim that the series $sum_{1}^{infty}na^{n}$ converges, so that
                  $na^{n} to 0$. It suffices to show that the series is absolutely convergent. But, since
                  $$frac{(n+1)|a|^{n+1}}{n|a|^{n}} = (1+frac{1}{n})|a| to |a| < 1,$$ by the ratio test the series converges. Thus we have
                  $$lim_{n to infty}na^{n} = 0.$$






                  share|cite|improve this answer









                  $endgroup$



                  If $|a| < 1,$ we claim that the series $sum_{1}^{infty}na^{n}$ converges, so that
                  $na^{n} to 0$. It suffices to show that the series is absolutely convergent. But, since
                  $$frac{(n+1)|a|^{n+1}}{n|a|^{n}} = (1+frac{1}{n})|a| to |a| < 1,$$ by the ratio test the series converges. Thus we have
                  $$lim_{n to infty}na^{n} = 0.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Apr 5 '15 at 12:28









                  Gary MooreGary Moore

                  17.3k21546




                  17.3k21546























                      1












                      $begingroup$

                      You can use the relationship of sequences limits and function limits, just let $0<a<1$ for convenient, and study $xa^x,xrightarrowinfty$ you can use L.Hospital theorem, $lim frac{x}{a^{-x}}=lim frac{1}{-a^{-x}}frac{1}{lna}=0 $






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        just sequences, thanks
                        $endgroup$
                        – Stabilo
                        Apr 5 '15 at 12:19










                      • $begingroup$
                        Oh, that's ok... in other conditions where you can use L.Hospital, don't forget it and it's so powerful.
                        $endgroup$
                        – Hans
                        Apr 5 '15 at 12:23
















                      1












                      $begingroup$

                      You can use the relationship of sequences limits and function limits, just let $0<a<1$ for convenient, and study $xa^x,xrightarrowinfty$ you can use L.Hospital theorem, $lim frac{x}{a^{-x}}=lim frac{1}{-a^{-x}}frac{1}{lna}=0 $






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        just sequences, thanks
                        $endgroup$
                        – Stabilo
                        Apr 5 '15 at 12:19










                      • $begingroup$
                        Oh, that's ok... in other conditions where you can use L.Hospital, don't forget it and it's so powerful.
                        $endgroup$
                        – Hans
                        Apr 5 '15 at 12:23














                      1












                      1








                      1





                      $begingroup$

                      You can use the relationship of sequences limits and function limits, just let $0<a<1$ for convenient, and study $xa^x,xrightarrowinfty$ you can use L.Hospital theorem, $lim frac{x}{a^{-x}}=lim frac{1}{-a^{-x}}frac{1}{lna}=0 $






                      share|cite|improve this answer









                      $endgroup$



                      You can use the relationship of sequences limits and function limits, just let $0<a<1$ for convenient, and study $xa^x,xrightarrowinfty$ you can use L.Hospital theorem, $lim frac{x}{a^{-x}}=lim frac{1}{-a^{-x}}frac{1}{lna}=0 $







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Apr 5 '15 at 12:17









                      HansHans

                      12210




                      12210












                      • $begingroup$
                        just sequences, thanks
                        $endgroup$
                        – Stabilo
                        Apr 5 '15 at 12:19










                      • $begingroup$
                        Oh, that's ok... in other conditions where you can use L.Hospital, don't forget it and it's so powerful.
                        $endgroup$
                        – Hans
                        Apr 5 '15 at 12:23


















                      • $begingroup$
                        just sequences, thanks
                        $endgroup$
                        – Stabilo
                        Apr 5 '15 at 12:19










                      • $begingroup$
                        Oh, that's ok... in other conditions where you can use L.Hospital, don't forget it and it's so powerful.
                        $endgroup$
                        – Hans
                        Apr 5 '15 at 12:23
















                      $begingroup$
                      just sequences, thanks
                      $endgroup$
                      – Stabilo
                      Apr 5 '15 at 12:19




                      $begingroup$
                      just sequences, thanks
                      $endgroup$
                      – Stabilo
                      Apr 5 '15 at 12:19












                      $begingroup$
                      Oh, that's ok... in other conditions where you can use L.Hospital, don't forget it and it's so powerful.
                      $endgroup$
                      – Hans
                      Apr 5 '15 at 12:23




                      $begingroup$
                      Oh, that's ok... in other conditions where you can use L.Hospital, don't forget it and it's so powerful.
                      $endgroup$
                      – Hans
                      Apr 5 '15 at 12:23











                      0












                      $begingroup$

                      Here a very elementary way:





                      • $|a| < 1 stackrel{a neq 0}{Rightarrow} |a| = frac{1}{1+p}$ for a $p > 0$.

                      • $(1+p)^n = 1 + np + color{blue}{frac{n(n-1)}{2}p^2} + cdots + p^n color{blue}{>frac{n(n-1)}{2}p^2}$


                      So, it follows
                      $$0 leq |na^n| = n|a|^n = frac{n}{(1+p)^n} < frac{n}{color{blue}{frac{n(n-1)}{2}p^2}} = frac{2}{(n-1)p^2} stackrel{n to infty}{longrightarrow} 0$$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Here a very elementary way:





                        • $|a| < 1 stackrel{a neq 0}{Rightarrow} |a| = frac{1}{1+p}$ for a $p > 0$.

                        • $(1+p)^n = 1 + np + color{blue}{frac{n(n-1)}{2}p^2} + cdots + p^n color{blue}{>frac{n(n-1)}{2}p^2}$


                        So, it follows
                        $$0 leq |na^n| = n|a|^n = frac{n}{(1+p)^n} < frac{n}{color{blue}{frac{n(n-1)}{2}p^2}} = frac{2}{(n-1)p^2} stackrel{n to infty}{longrightarrow} 0$$






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Here a very elementary way:





                          • $|a| < 1 stackrel{a neq 0}{Rightarrow} |a| = frac{1}{1+p}$ for a $p > 0$.

                          • $(1+p)^n = 1 + np + color{blue}{frac{n(n-1)}{2}p^2} + cdots + p^n color{blue}{>frac{n(n-1)}{2}p^2}$


                          So, it follows
                          $$0 leq |na^n| = n|a|^n = frac{n}{(1+p)^n} < frac{n}{color{blue}{frac{n(n-1)}{2}p^2}} = frac{2}{(n-1)p^2} stackrel{n to infty}{longrightarrow} 0$$






                          share|cite|improve this answer









                          $endgroup$



                          Here a very elementary way:





                          • $|a| < 1 stackrel{a neq 0}{Rightarrow} |a| = frac{1}{1+p}$ for a $p > 0$.

                          • $(1+p)^n = 1 + np + color{blue}{frac{n(n-1)}{2}p^2} + cdots + p^n color{blue}{>frac{n(n-1)}{2}p^2}$


                          So, it follows
                          $$0 leq |na^n| = n|a|^n = frac{n}{(1+p)^n} < frac{n}{color{blue}{frac{n(n-1)}{2}p^2}} = frac{2}{(n-1)p^2} stackrel{n to infty}{longrightarrow} 0$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 9 '18 at 17:35









                          trancelocationtrancelocation

                          12.9k1827




                          12.9k1827






























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