Proving that $lim_{n to infty}nx_n=frac{1}{q}$












1












$begingroup$


Suppose $0<x_1<1$, $x_{n+1}=x_n(1-x_n)$, prove $lim_{n to infty} nx_n=1$. Suppose now $0<x_1<frac{1}{q}$ and $0<qleq 1$ and $x_{n+1}=x_n(1-qx_n)$, prove $lim_{n to infty}nx_n=frac{1}{q}$.



The first one can be done by showing the sequence is decreasing and bounded and therefore has a limit. What about the second one?






real-analysis







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  • 4




    $begingroup$
    Substitute$$y_n = qx_n$$
    $endgroup$
    – Jakobian
    Dec 10 '18 at 18:02






  • 1




    $begingroup$
    How is showing that the sequence $x_n$ is bounded and decreasing sufficient to show $lim_{ntoinfty} nx_n=1$?? You need a bit more work to verify the coveted limit.
    $endgroup$
    – Mark Viola
    Dec 10 '18 at 21:16


















1












$begingroup$


Suppose $0<x_1<1$, $x_{n+1}=x_n(1-x_n)$, prove $lim_{n to infty} nx_n=1$. Suppose now $0<x_1<frac{1}{q}$ and $0<qleq 1$ and $x_{n+1}=x_n(1-qx_n)$, prove $lim_{n to infty}nx_n=frac{1}{q}$.



The first one can be done by showing the sequence is decreasing and bounded and therefore has a limit. What about the second one?






real-analysis







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Substitute$$y_n = qx_n$$
    $endgroup$
    – Jakobian
    Dec 10 '18 at 18:02






  • 1




    $begingroup$
    How is showing that the sequence $x_n$ is bounded and decreasing sufficient to show $lim_{ntoinfty} nx_n=1$?? You need a bit more work to verify the coveted limit.
    $endgroup$
    – Mark Viola
    Dec 10 '18 at 21:16
















1












1








1





$begingroup$


Suppose $0<x_1<1$, $x_{n+1}=x_n(1-x_n)$, prove $lim_{n to infty} nx_n=1$. Suppose now $0<x_1<frac{1}{q}$ and $0<qleq 1$ and $x_{n+1}=x_n(1-qx_n)$, prove $lim_{n to infty}nx_n=frac{1}{q}$.



The first one can be done by showing the sequence is decreasing and bounded and therefore has a limit. What about the second one?






real-analysis







share|cite|improve this question











$endgroup$




Suppose $0<x_1<1$, $x_{n+1}=x_n(1-x_n)$, prove $lim_{n to infty} nx_n=1$. Suppose now $0<x_1<frac{1}{q}$ and $0<qleq 1$ and $x_{n+1}=x_n(1-qx_n)$, prove $lim_{n to infty}nx_n=frac{1}{q}$.



The first one can be done by showing the sequence is decreasing and bounded and therefore has a limit. What about the second one?






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 18:00









Rebellos

14.8k31248




14.8k31248










asked Dec 10 '18 at 17:55









user42493user42493

1877




1877








  • 4




    $begingroup$
    Substitute$$y_n = qx_n$$
    $endgroup$
    – Jakobian
    Dec 10 '18 at 18:02






  • 1




    $begingroup$
    How is showing that the sequence $x_n$ is bounded and decreasing sufficient to show $lim_{ntoinfty} nx_n=1$?? You need a bit more work to verify the coveted limit.
    $endgroup$
    – Mark Viola
    Dec 10 '18 at 21:16
















  • 4




    $begingroup$
    Substitute$$y_n = qx_n$$
    $endgroup$
    – Jakobian
    Dec 10 '18 at 18:02






  • 1




    $begingroup$
    How is showing that the sequence $x_n$ is bounded and decreasing sufficient to show $lim_{ntoinfty} nx_n=1$?? You need a bit more work to verify the coveted limit.
    $endgroup$
    – Mark Viola
    Dec 10 '18 at 21:16










4




4




$begingroup$
Substitute$$y_n = qx_n$$
$endgroup$
– Jakobian
Dec 10 '18 at 18:02




$begingroup$
Substitute$$y_n = qx_n$$
$endgroup$
– Jakobian
Dec 10 '18 at 18:02




1




1




$begingroup$
How is showing that the sequence $x_n$ is bounded and decreasing sufficient to show $lim_{ntoinfty} nx_n=1$?? You need a bit more work to verify the coveted limit.
$endgroup$
– Mark Viola
Dec 10 '18 at 21:16






$begingroup$
How is showing that the sequence $x_n$ is bounded and decreasing sufficient to show $lim_{ntoinfty} nx_n=1$?? You need a bit more work to verify the coveted limit.
$endgroup$
– Mark Viola
Dec 10 '18 at 21:16












2 Answers
2






active

oldest

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1












$begingroup$


We need first show that $lim_{ntoinfty}n x_n=1$.





To begin, note that for $xin (0,1)$, $0<x(1-x)<frac14$. Hence, provided $x_1in(0,1)$, $x_n>0$. Moreover, $x_{n+1}-x_n=-x_n^2<0$.




Inasmuch as $x_n$ is bounded below by $0$ and monotonically decreasing, the sequence converges. Suppose it's limit is $L$. Clearly $L=L(1-L)$. Hence $L=0$ and we find $lim_{nto infty}x_n=0$.




Let $a_n=1/x_n$. Since $x_nto 0$, then $a_ntoinfty$.



Hence, the Stolz-Cesaro Theorem guarantees that



$$ begin{align}
lim_{ntoinfty}frac1{nx_n} &=lim_{ntoinfty}frac{a_n}n\\
&=lim_{ntoinfty}(a_{n+1}-a_n)\\
&=lim_{ntoinfty}frac1{1-x_n}\\
&=1
end{align}$$



as was to be shown.




For the second part, let $y_n=qx_n$ as was already mentioned in a comment and another solution.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Hint :



    It's essentialy the same as the first part, as :



    $$lim_{n to infty} nx_n = frac{1}{q} Leftrightarrow q lim_{n to infty} nx_n = 1 Leftrightarrow lim_{nto infty} nqx_n = 1 $$



    But isn't that $lim_{nto infty} nz_n$ = 1 where $z_n equiv qx_n$ ?






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$


      We need first show that $lim_{ntoinfty}n x_n=1$.





      To begin, note that for $xin (0,1)$, $0<x(1-x)<frac14$. Hence, provided $x_1in(0,1)$, $x_n>0$. Moreover, $x_{n+1}-x_n=-x_n^2<0$.




      Inasmuch as $x_n$ is bounded below by $0$ and monotonically decreasing, the sequence converges. Suppose it's limit is $L$. Clearly $L=L(1-L)$. Hence $L=0$ and we find $lim_{nto infty}x_n=0$.




      Let $a_n=1/x_n$. Since $x_nto 0$, then $a_ntoinfty$.



      Hence, the Stolz-Cesaro Theorem guarantees that



      $$ begin{align}
      lim_{ntoinfty}frac1{nx_n} &=lim_{ntoinfty}frac{a_n}n\\
      &=lim_{ntoinfty}(a_{n+1}-a_n)\\
      &=lim_{ntoinfty}frac1{1-x_n}\\
      &=1
      end{align}$$



      as was to be shown.




      For the second part, let $y_n=qx_n$ as was already mentioned in a comment and another solution.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$


        We need first show that $lim_{ntoinfty}n x_n=1$.





        To begin, note that for $xin (0,1)$, $0<x(1-x)<frac14$. Hence, provided $x_1in(0,1)$, $x_n>0$. Moreover, $x_{n+1}-x_n=-x_n^2<0$.




        Inasmuch as $x_n$ is bounded below by $0$ and monotonically decreasing, the sequence converges. Suppose it's limit is $L$. Clearly $L=L(1-L)$. Hence $L=0$ and we find $lim_{nto infty}x_n=0$.




        Let $a_n=1/x_n$. Since $x_nto 0$, then $a_ntoinfty$.



        Hence, the Stolz-Cesaro Theorem guarantees that



        $$ begin{align}
        lim_{ntoinfty}frac1{nx_n} &=lim_{ntoinfty}frac{a_n}n\\
        &=lim_{ntoinfty}(a_{n+1}-a_n)\\
        &=lim_{ntoinfty}frac1{1-x_n}\\
        &=1
        end{align}$$



        as was to be shown.




        For the second part, let $y_n=qx_n$ as was already mentioned in a comment and another solution.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$


          We need first show that $lim_{ntoinfty}n x_n=1$.





          To begin, note that for $xin (0,1)$, $0<x(1-x)<frac14$. Hence, provided $x_1in(0,1)$, $x_n>0$. Moreover, $x_{n+1}-x_n=-x_n^2<0$.




          Inasmuch as $x_n$ is bounded below by $0$ and monotonically decreasing, the sequence converges. Suppose it's limit is $L$. Clearly $L=L(1-L)$. Hence $L=0$ and we find $lim_{nto infty}x_n=0$.




          Let $a_n=1/x_n$. Since $x_nto 0$, then $a_ntoinfty$.



          Hence, the Stolz-Cesaro Theorem guarantees that



          $$ begin{align}
          lim_{ntoinfty}frac1{nx_n} &=lim_{ntoinfty}frac{a_n}n\\
          &=lim_{ntoinfty}(a_{n+1}-a_n)\\
          &=lim_{ntoinfty}frac1{1-x_n}\\
          &=1
          end{align}$$



          as was to be shown.




          For the second part, let $y_n=qx_n$ as was already mentioned in a comment and another solution.






          share|cite|improve this answer









          $endgroup$




          We need first show that $lim_{ntoinfty}n x_n=1$.





          To begin, note that for $xin (0,1)$, $0<x(1-x)<frac14$. Hence, provided $x_1in(0,1)$, $x_n>0$. Moreover, $x_{n+1}-x_n=-x_n^2<0$.




          Inasmuch as $x_n$ is bounded below by $0$ and monotonically decreasing, the sequence converges. Suppose it's limit is $L$. Clearly $L=L(1-L)$. Hence $L=0$ and we find $lim_{nto infty}x_n=0$.




          Let $a_n=1/x_n$. Since $x_nto 0$, then $a_ntoinfty$.



          Hence, the Stolz-Cesaro Theorem guarantees that



          $$ begin{align}
          lim_{ntoinfty}frac1{nx_n} &=lim_{ntoinfty}frac{a_n}n\\
          &=lim_{ntoinfty}(a_{n+1}-a_n)\\
          &=lim_{ntoinfty}frac1{1-x_n}\\
          &=1
          end{align}$$



          as was to be shown.




          For the second part, let $y_n=qx_n$ as was already mentioned in a comment and another solution.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 10 '18 at 21:34









          Mark ViolaMark Viola

          132k1276174




          132k1276174























              1












              $begingroup$

              Hint :



              It's essentialy the same as the first part, as :



              $$lim_{n to infty} nx_n = frac{1}{q} Leftrightarrow q lim_{n to infty} nx_n = 1 Leftrightarrow lim_{nto infty} nqx_n = 1 $$



              But isn't that $lim_{nto infty} nz_n$ = 1 where $z_n equiv qx_n$ ?






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Hint :



                It's essentialy the same as the first part, as :



                $$lim_{n to infty} nx_n = frac{1}{q} Leftrightarrow q lim_{n to infty} nx_n = 1 Leftrightarrow lim_{nto infty} nqx_n = 1 $$



                But isn't that $lim_{nto infty} nz_n$ = 1 where $z_n equiv qx_n$ ?






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Hint :



                  It's essentialy the same as the first part, as :



                  $$lim_{n to infty} nx_n = frac{1}{q} Leftrightarrow q lim_{n to infty} nx_n = 1 Leftrightarrow lim_{nto infty} nqx_n = 1 $$



                  But isn't that $lim_{nto infty} nz_n$ = 1 where $z_n equiv qx_n$ ?






                  share|cite|improve this answer









                  $endgroup$



                  Hint :



                  It's essentialy the same as the first part, as :



                  $$lim_{n to infty} nx_n = frac{1}{q} Leftrightarrow q lim_{n to infty} nx_n = 1 Leftrightarrow lim_{nto infty} nqx_n = 1 $$



                  But isn't that $lim_{nto infty} nz_n$ = 1 where $z_n equiv qx_n$ ?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 10 '18 at 18:09









                  RebellosRebellos

                  14.8k31248




                  14.8k31248






























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