Results 1 to 6 of 6

Math Help - to prove that the sequence is unbounded.

  1. #1
    Senior Member abhishekkgp's Avatar
    Joined
    Jan 2011
    From
    India
    Posts
    495
    Thanks
    1

    to prove that the sequence is unbounded.

    QUESTION: show that the sequence defined recursively as

    MY SOLUTION:
    Let the sequence be bounded above. It then has a least upper bound say .

    now,


    hence:


    It can be computed that

    so we have:


    This is true for all so we have:

    Contradiction.

    Is this correct?

    also if you have a good method to do it then please post it.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    44
    Quote Originally Posted by abhishekkgp View Post
    also if you have a good method to do it then please post it.

    Prove by induction (very easy):

    x_n>0\;\forall n=1,2,\ldots

    The sequence is increasing:

    x_{n+1}-x_n=\dfrac{1}{x_n}>0

    If the sequence is upper bounded, denote l > 0 its limit. Then,

    x_{n+1}=x_n+\dfrac{1}{x_n}\Rightarrow l=l+\dfrac{1}{l}\Rightarrow 0=\dfrac{1}{l}\neq 0

    Contradiction.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member abhishekkgp's Avatar
    Joined
    Jan 2011
    From
    India
    Posts
    495
    Thanks
    1
    Quote Originally Posted by FernandoRevilla View Post
    Prove by induction (very easy):

    x_n>0\;\forall n=1,2,\ldots

    The sequence is increasing:

    x_{n+1}-x_n=\dfrac{1}{x_n}>0

    If the sequence is upper bounded, denote l > 0 its limit. Then,

    x_{n+1}=x_n+\dfrac{1}{x_n}\Rightarrow l=l+\dfrac{1}{l}\Rightarrow 0=\dfrac{1}{l}\neq 0

    Contradiction.
    in my book the concept of limit has not so far been introduced. so i am not sure how you got . i believe the way i did it was correct too. thanks anyway.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    44
    Quote Originally Posted by abhishekkgp View Post
    now,


    hence:

    You need to prove that the terms of the sequence are positive in order to obtain an equivalent inequality.


    so we have:

    It should be:

    2x_n \leq L+\sqrt{L^2-4}


    This is true for all so we have:

    Contradiction.

    That is true if you choose L as the least upper bound of the sequence.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member abhishekkgp's Avatar
    Joined
    Jan 2011
    From
    India
    Posts
    495
    Thanks
    1
    Quote Originally Posted by FernandoRevilla View Post
    You need to prove that the terms of the sequence are positive in order to obtain an equivalent inequality.





    It should be:

    2x_n \leq L+\sqrt{L^2-4}

    oh yes right!



    That is true if you choose L as the least upper bound of the sequence.
    yes i had mentioned that L is the least upper bound of the sequence.
    thanks for that.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    44
    Quote Originally Posted by abhishekkgp View Post
    yes i had mentioned that L is the least upper bound of the sequence.

    Right.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. unbounded sequence
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: December 9th 2009, 07:56 AM
  2. Cauchy sequence w/ unbounded subsequence
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: November 13th 2009, 03:58 PM
  3. Unbounded Sequence
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: May 4th 2009, 12:30 AM
  4. Unbounded sequence?
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: October 4th 2008, 04:35 AM
  5. Unbounded sequence
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 6th 2008, 07:59 AM

/mathhelpforum @mathhelpforum