Showing a sequence is decreasing and bounded

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I need to show that the sequence given by
$$a_1=2$$
$$a_n=frac12Bigg(a_n-1+frac2a_n-1Bigg)$$
is monotonically decreasing and bounded. But every time I try, I can't make the bounds tight enough to prove it. How would I solve this?






real-analysis sequences-and-series recursion







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  • iirc it alternates between being above the limit and below the limit
    – mathworker21
    1 hour ago










  • The lower bound is easy to find by AM-GM inequality. The monotonicity is determined by directly compute $a_n+1-a_n$.
    – xbh
    1 hour ago














up vote
3
down vote

favorite












I need to show that the sequence given by
$$a_1=2$$
$$a_n=frac12Bigg(a_n-1+frac2a_n-1Bigg)$$
is monotonically decreasing and bounded. But every time I try, I can't make the bounds tight enough to prove it. How would I solve this?






real-analysis sequences-and-series recursion







share|cite|improve this question









New contributor




arow257 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.



















  • iirc it alternates between being above the limit and below the limit
    – mathworker21
    1 hour ago










  • The lower bound is easy to find by AM-GM inequality. The monotonicity is determined by directly compute $a_n+1-a_n$.
    – xbh
    1 hour ago












up vote
3
down vote

favorite









up vote
3
down vote

favorite











I need to show that the sequence given by
$$a_1=2$$
$$a_n=frac12Bigg(a_n-1+frac2a_n-1Bigg)$$
is monotonically decreasing and bounded. But every time I try, I can't make the bounds tight enough to prove it. How would I solve this?






real-analysis sequences-and-series recursion







share|cite|improve this question









New contributor




arow257 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I need to show that the sequence given by
$$a_1=2$$
$$a_n=frac12Bigg(a_n-1+frac2a_n-1Bigg)$$
is monotonically decreasing and bounded. But every time I try, I can't make the bounds tight enough to prove it. How would I solve this?






real-analysis sequences-and-series recursion




real-analysis sequences-and-series recursion






share|cite|improve this question









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arow257 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









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Check out our Code of Conduct.









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edited 1 hour ago









Chinnapparaj R

2,325520




2,325520






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arow257

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arow257 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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Check out our Code of Conduct.











  • iirc it alternates between being above the limit and below the limit
    – mathworker21
    1 hour ago










  • The lower bound is easy to find by AM-GM inequality. The monotonicity is determined by directly compute $a_n+1-a_n$.
    – xbh
    1 hour ago
















  • iirc it alternates between being above the limit and below the limit
    – mathworker21
    1 hour ago










  • The lower bound is easy to find by AM-GM inequality. The monotonicity is determined by directly compute $a_n+1-a_n$.
    – xbh
    1 hour ago















iirc it alternates between being above the limit and below the limit
– mathworker21
1 hour ago




iirc it alternates between being above the limit and below the limit
– mathworker21
1 hour ago












The lower bound is easy to find by AM-GM inequality. The monotonicity is determined by directly compute $a_n+1-a_n$.
– xbh
1 hour ago




The lower bound is easy to find by AM-GM inequality. The monotonicity is determined by directly compute $a_n+1-a_n$.
– xbh
1 hour ago










3 Answers
3






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up vote
2
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The limit is $sqrt2$. In fact, the convergence is quite fast, about doubling the number of digits each iteration:
$$a_n-sqrt2=frac12a_n-1left(a_n-1-sqrt2right)^2$$






share|cite|improve this answer



























    up vote
    2
    down vote













    Here $a_n$ satisfies the quadratic equation $$a_n^2-(2a_n+1)a_n+2=0$$ This equation has a real root and hence the discriminant $$4(a_n+1^2-2) geq 0$$ That is, $$a_n+1^2 geq 2$$ for $n geq 1$.



    Now,
    $a_n-a_n+1=a_n-frac12Bigg( a_n+frac2a_n Bigg)=frac12Bigg(fraca_n^2-2a_n Bigg)geq 0$



    for all $n geq 1$.



    Hence $$a_n+1 leq a_n, forall n geq 2$$



    and so by MCT it converges to its glb!






    share|cite|improve this answer





























      up vote
      1
      down vote













      $a_n+1-a_n=frac12left(a_n+frac2a_nright)-a_n=-frac12left (a_n-frac2a_nright) =-frac12left(fraca_n^2-2a_nright)le0$, as $a_1=2$.
      Hence $(a_n)$ is monotonically decreasing.



      The limit $a$, if it exists, satisfies $$a=frac12left(a+frac2aright)implies 2a^2=a^2+2implies a^2=2$$.



      Now we can easily see that $a_nge0$ (it is the average of two positive numbers).



      Hence $a_ntosqrt2$.






      share|cite|improve this answer






















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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        2
        down vote













        The limit is $sqrt2$. In fact, the convergence is quite fast, about doubling the number of digits each iteration:
        $$a_n-sqrt2=frac12a_n-1left(a_n-1-sqrt2right)^2$$






        share|cite|improve this answer
























          up vote
          2
          down vote













          The limit is $sqrt2$. In fact, the convergence is quite fast, about doubling the number of digits each iteration:
          $$a_n-sqrt2=frac12a_n-1left(a_n-1-sqrt2right)^2$$






          share|cite|improve this answer






















            up vote
            2
            down vote










            up vote
            2
            down vote









            The limit is $sqrt2$. In fact, the convergence is quite fast, about doubling the number of digits each iteration:
            $$a_n-sqrt2=frac12a_n-1left(a_n-1-sqrt2right)^2$$






            share|cite|improve this answer












            The limit is $sqrt2$. In fact, the convergence is quite fast, about doubling the number of digits each iteration:
            $$a_n-sqrt2=frac12a_n-1left(a_n-1-sqrt2right)^2$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            robjohn♦

            259k26299616




            259k26299616




















                up vote
                2
                down vote













                Here $a_n$ satisfies the quadratic equation $$a_n^2-(2a_n+1)a_n+2=0$$ This equation has a real root and hence the discriminant $$4(a_n+1^2-2) geq 0$$ That is, $$a_n+1^2 geq 2$$ for $n geq 1$.



                Now,
                $a_n-a_n+1=a_n-frac12Bigg( a_n+frac2a_n Bigg)=frac12Bigg(fraca_n^2-2a_n Bigg)geq 0$



                for all $n geq 1$.



                Hence $$a_n+1 leq a_n, forall n geq 2$$



                and so by MCT it converges to its glb!






                share|cite|improve this answer


























                  up vote
                  2
                  down vote













                  Here $a_n$ satisfies the quadratic equation $$a_n^2-(2a_n+1)a_n+2=0$$ This equation has a real root and hence the discriminant $$4(a_n+1^2-2) geq 0$$ That is, $$a_n+1^2 geq 2$$ for $n geq 1$.



                  Now,
                  $a_n-a_n+1=a_n-frac12Bigg( a_n+frac2a_n Bigg)=frac12Bigg(fraca_n^2-2a_n Bigg)geq 0$



                  for all $n geq 1$.



                  Hence $$a_n+1 leq a_n, forall n geq 2$$



                  and so by MCT it converges to its glb!






                  share|cite|improve this answer
























                    up vote
                    2
                    down vote










                    up vote
                    2
                    down vote









                    Here $a_n$ satisfies the quadratic equation $$a_n^2-(2a_n+1)a_n+2=0$$ This equation has a real root and hence the discriminant $$4(a_n+1^2-2) geq 0$$ That is, $$a_n+1^2 geq 2$$ for $n geq 1$.



                    Now,
                    $a_n-a_n+1=a_n-frac12Bigg( a_n+frac2a_n Bigg)=frac12Bigg(fraca_n^2-2a_n Bigg)geq 0$



                    for all $n geq 1$.



                    Hence $$a_n+1 leq a_n, forall n geq 2$$



                    and so by MCT it converges to its glb!






                    share|cite|improve this answer














                    Here $a_n$ satisfies the quadratic equation $$a_n^2-(2a_n+1)a_n+2=0$$ This equation has a real root and hence the discriminant $$4(a_n+1^2-2) geq 0$$ That is, $$a_n+1^2 geq 2$$ for $n geq 1$.



                    Now,
                    $a_n-a_n+1=a_n-frac12Bigg( a_n+frac2a_n Bigg)=frac12Bigg(fraca_n^2-2a_n Bigg)geq 0$



                    for all $n geq 1$.



                    Hence $$a_n+1 leq a_n, forall n geq 2$$



                    and so by MCT it converges to its glb!







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 1 hour ago

























                    answered 1 hour ago









                    Chinnapparaj R

                    2,325520




                    2,325520




















                        up vote
                        1
                        down vote













                        $a_n+1-a_n=frac12left(a_n+frac2a_nright)-a_n=-frac12left (a_n-frac2a_nright) =-frac12left(fraca_n^2-2a_nright)le0$, as $a_1=2$.
                        Hence $(a_n)$ is monotonically decreasing.



                        The limit $a$, if it exists, satisfies $$a=frac12left(a+frac2aright)implies 2a^2=a^2+2implies a^2=2$$.



                        Now we can easily see that $a_nge0$ (it is the average of two positive numbers).



                        Hence $a_ntosqrt2$.






                        share|cite|improve this answer


























                          up vote
                          1
                          down vote













                          $a_n+1-a_n=frac12left(a_n+frac2a_nright)-a_n=-frac12left (a_n-frac2a_nright) =-frac12left(fraca_n^2-2a_nright)le0$, as $a_1=2$.
                          Hence $(a_n)$ is monotonically decreasing.



                          The limit $a$, if it exists, satisfies $$a=frac12left(a+frac2aright)implies 2a^2=a^2+2implies a^2=2$$.



                          Now we can easily see that $a_nge0$ (it is the average of two positive numbers).



                          Hence $a_ntosqrt2$.






                          share|cite|improve this answer
























                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            $a_n+1-a_n=frac12left(a_n+frac2a_nright)-a_n=-frac12left (a_n-frac2a_nright) =-frac12left(fraca_n^2-2a_nright)le0$, as $a_1=2$.
                            Hence $(a_n)$ is monotonically decreasing.



                            The limit $a$, if it exists, satisfies $$a=frac12left(a+frac2aright)implies 2a^2=a^2+2implies a^2=2$$.



                            Now we can easily see that $a_nge0$ (it is the average of two positive numbers).



                            Hence $a_ntosqrt2$.






                            share|cite|improve this answer














                            $a_n+1-a_n=frac12left(a_n+frac2a_nright)-a_n=-frac12left (a_n-frac2a_nright) =-frac12left(fraca_n^2-2a_nright)le0$, as $a_1=2$.
                            Hence $(a_n)$ is monotonically decreasing.



                            The limit $a$, if it exists, satisfies $$a=frac12left(a+frac2aright)implies 2a^2=a^2+2implies a^2=2$$.



                            Now we can easily see that $a_nge0$ (it is the average of two positive numbers).



                            Hence $a_ntosqrt2$.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 8 mins ago

























                            answered 1 hour ago









                            Chris Custer

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