How to prove $a_{n+1}=(1+a_n+a^2_{n-1})/3$ is a non-decreasing sequence?












0












$begingroup$


$$a_1=a_2=0.5$$



It isn't hard to show that $0.5le a_nlt 1$, and that if the sequence converge, the limit is 1.



But how to prove it's monotone?



I've tried:
$$a_{n+1}-a_n=frac{1-2a_n+a^2_{n-1}}{3}ge 0$$
$$frac{1+a^2_{n-1}}{2}ge a_n$$
$$1-a_n ge a_n - a^2_{n-1}$$



But couldn't prove any of the inequalities.






real-analysis sequences-and-series convergence induction monotone-functions







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  • 1




    $begingroup$
    @Somos not yet, will try now
    $endgroup$
    – User
    Jan 6 at 3:08


















0












$begingroup$


$$a_1=a_2=0.5$$



It isn't hard to show that $0.5le a_nlt 1$, and that if the sequence converge, the limit is 1.



But how to prove it's monotone?



I've tried:
$$a_{n+1}-a_n=frac{1-2a_n+a^2_{n-1}}{3}ge 0$$
$$frac{1+a^2_{n-1}}{2}ge a_n$$
$$1-a_n ge a_n - a^2_{n-1}$$



But couldn't prove any of the inequalities.






real-analysis sequences-and-series convergence induction monotone-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    @Somos not yet, will try now
    $endgroup$
    – User
    Jan 6 at 3:08
















0












0








0





$begingroup$


$$a_1=a_2=0.5$$



It isn't hard to show that $0.5le a_nlt 1$, and that if the sequence converge, the limit is 1.



But how to prove it's monotone?



I've tried:
$$a_{n+1}-a_n=frac{1-2a_n+a^2_{n-1}}{3}ge 0$$
$$frac{1+a^2_{n-1}}{2}ge a_n$$
$$1-a_n ge a_n - a^2_{n-1}$$



But couldn't prove any of the inequalities.






real-analysis sequences-and-series convergence induction monotone-functions







share|cite|improve this question











$endgroup$




$$a_1=a_2=0.5$$



It isn't hard to show that $0.5le a_nlt 1$, and that if the sequence converge, the limit is 1.



But how to prove it's monotone?



I've tried:
$$a_{n+1}-a_n=frac{1-2a_n+a^2_{n-1}}{3}ge 0$$
$$frac{1+a^2_{n-1}}{2}ge a_n$$
$$1-a_n ge a_n - a^2_{n-1}$$



But couldn't prove any of the inequalities.






real-analysis sequences-and-series convergence induction monotone-functions




real-analysis sequences-and-series convergence induction monotone-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 4:17









Michael Rozenberg

98.9k1590189




98.9k1590189










asked Jan 6 at 2:41









UserUser

452312




452312








  • 1




    $begingroup$
    @Somos not yet, will try now
    $endgroup$
    – User
    Jan 6 at 3:08
















  • 1




    $begingroup$
    @Somos not yet, will try now
    $endgroup$
    – User
    Jan 6 at 3:08










1




1




$begingroup$
@Somos not yet, will try now
$endgroup$
– User
Jan 6 at 3:08






$begingroup$
@Somos not yet, will try now
$endgroup$
– User
Jan 6 at 3:08












2 Answers
2






active

oldest

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5












$begingroup$

The hint.



By induction:
$$
begin{align}
a_{n+1}-a_n&=frac{1-2a_n+a_{n-1}^2}{3}
\&=frac{frac{3}{3}-2frac{1+a_{n-1}+a^2_{n-2}}{3}+frac{3a_{n-1}^2}{3}}{3}
\&=frac{1-2a_{n-1}+a_{n-1}^2+2(a_{n-1}^2-a_{n-2}^2)}{9}
\&geq0
end{align}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    In my shame, I could not follow your arithmetic. But it gave me the right direction. Thanks!
    $endgroup$
    – User
    Jan 6 at 3:28










  • $begingroup$
    @User You are welcome!
    $endgroup$
    – Michael Rozenberg
    Jan 6 at 3:31



















1












$begingroup$

Following the implicit advice from previous answer to use the definition twice together with induction,



$$a_{n+1}=(1+a_n+a^2_{n-1})/3$$
$$a_{n}=(1+a_{n-1}+a^2_{n-2})/3$$
$$a_{n+1}-a_{n}=(a_n-a_{n-1}+a^2_{n-1}-a^2_{n-2})/3$$



If we assume $a_n ge a_{n-1} ge a_{n-2}$ we get $a_{n+1} ge a_n$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    The hint.



    By induction:
    $$
    begin{align}
    a_{n+1}-a_n&=frac{1-2a_n+a_{n-1}^2}{3}
    \&=frac{frac{3}{3}-2frac{1+a_{n-1}+a^2_{n-2}}{3}+frac{3a_{n-1}^2}{3}}{3}
    \&=frac{1-2a_{n-1}+a_{n-1}^2+2(a_{n-1}^2-a_{n-2}^2)}{9}
    \&geq0
    end{align}
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      In my shame, I could not follow your arithmetic. But it gave me the right direction. Thanks!
      $endgroup$
      – User
      Jan 6 at 3:28










    • $begingroup$
      @User You are welcome!
      $endgroup$
      – Michael Rozenberg
      Jan 6 at 3:31
















    5












    $begingroup$

    The hint.



    By induction:
    $$
    begin{align}
    a_{n+1}-a_n&=frac{1-2a_n+a_{n-1}^2}{3}
    \&=frac{frac{3}{3}-2frac{1+a_{n-1}+a^2_{n-2}}{3}+frac{3a_{n-1}^2}{3}}{3}
    \&=frac{1-2a_{n-1}+a_{n-1}^2+2(a_{n-1}^2-a_{n-2}^2)}{9}
    \&geq0
    end{align}
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      In my shame, I could not follow your arithmetic. But it gave me the right direction. Thanks!
      $endgroup$
      – User
      Jan 6 at 3:28










    • $begingroup$
      @User You are welcome!
      $endgroup$
      – Michael Rozenberg
      Jan 6 at 3:31














    5












    5








    5





    $begingroup$

    The hint.



    By induction:
    $$
    begin{align}
    a_{n+1}-a_n&=frac{1-2a_n+a_{n-1}^2}{3}
    \&=frac{frac{3}{3}-2frac{1+a_{n-1}+a^2_{n-2}}{3}+frac{3a_{n-1}^2}{3}}{3}
    \&=frac{1-2a_{n-1}+a_{n-1}^2+2(a_{n-1}^2-a_{n-2}^2)}{9}
    \&geq0
    end{align}
    $$






    share|cite|improve this answer











    $endgroup$



    The hint.



    By induction:
    $$
    begin{align}
    a_{n+1}-a_n&=frac{1-2a_n+a_{n-1}^2}{3}
    \&=frac{frac{3}{3}-2frac{1+a_{n-1}+a^2_{n-2}}{3}+frac{3a_{n-1}^2}{3}}{3}
    \&=frac{1-2a_{n-1}+a_{n-1}^2+2(a_{n-1}^2-a_{n-2}^2)}{9}
    \&geq0
    end{align}
    $$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 7 at 12:34









    dvb

    1337




    1337










    answered Jan 6 at 3:13









    Michael RozenbergMichael Rozenberg

    98.9k1590189




    98.9k1590189












    • $begingroup$
      In my shame, I could not follow your arithmetic. But it gave me the right direction. Thanks!
      $endgroup$
      – User
      Jan 6 at 3:28










    • $begingroup$
      @User You are welcome!
      $endgroup$
      – Michael Rozenberg
      Jan 6 at 3:31


















    • $begingroup$
      In my shame, I could not follow your arithmetic. But it gave me the right direction. Thanks!
      $endgroup$
      – User
      Jan 6 at 3:28










    • $begingroup$
      @User You are welcome!
      $endgroup$
      – Michael Rozenberg
      Jan 6 at 3:31
















    $begingroup$
    In my shame, I could not follow your arithmetic. But it gave me the right direction. Thanks!
    $endgroup$
    – User
    Jan 6 at 3:28




    $begingroup$
    In my shame, I could not follow your arithmetic. But it gave me the right direction. Thanks!
    $endgroup$
    – User
    Jan 6 at 3:28












    $begingroup$
    @User You are welcome!
    $endgroup$
    – Michael Rozenberg
    Jan 6 at 3:31




    $begingroup$
    @User You are welcome!
    $endgroup$
    – Michael Rozenberg
    Jan 6 at 3:31











    1












    $begingroup$

    Following the implicit advice from previous answer to use the definition twice together with induction,



    $$a_{n+1}=(1+a_n+a^2_{n-1})/3$$
    $$a_{n}=(1+a_{n-1}+a^2_{n-2})/3$$
    $$a_{n+1}-a_{n}=(a_n-a_{n-1}+a^2_{n-1}-a^2_{n-2})/3$$



    If we assume $a_n ge a_{n-1} ge a_{n-2}$ we get $a_{n+1} ge a_n$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Following the implicit advice from previous answer to use the definition twice together with induction,



      $$a_{n+1}=(1+a_n+a^2_{n-1})/3$$
      $$a_{n}=(1+a_{n-1}+a^2_{n-2})/3$$
      $$a_{n+1}-a_{n}=(a_n-a_{n-1}+a^2_{n-1}-a^2_{n-2})/3$$



      If we assume $a_n ge a_{n-1} ge a_{n-2}$ we get $a_{n+1} ge a_n$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Following the implicit advice from previous answer to use the definition twice together with induction,



        $$a_{n+1}=(1+a_n+a^2_{n-1})/3$$
        $$a_{n}=(1+a_{n-1}+a^2_{n-2})/3$$
        $$a_{n+1}-a_{n}=(a_n-a_{n-1}+a^2_{n-1}-a^2_{n-2})/3$$



        If we assume $a_n ge a_{n-1} ge a_{n-2}$ we get $a_{n+1} ge a_n$






        share|cite|improve this answer









        $endgroup$



        Following the implicit advice from previous answer to use the definition twice together with induction,



        $$a_{n+1}=(1+a_n+a^2_{n-1})/3$$
        $$a_{n}=(1+a_{n-1}+a^2_{n-2})/3$$
        $$a_{n+1}-a_{n}=(a_n-a_{n-1}+a^2_{n-1}-a^2_{n-2})/3$$



        If we assume $a_n ge a_{n-1} ge a_{n-2}$ we get $a_{n+1} ge a_n$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 6 at 3:27









        UserUser

        452312




        452312






























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