Results 1 to 4 of 4

Math Help - Convergence proof...

  1. #1
    Junior Member
    Joined
    Jan 2011
    Posts
    58

    Convergence proof...

    I am a little stuck on the following:
    suppose that:
    1) sum(n..infinity) (a_n)^2 and sum(n..infinity)(b_n)^2 converges, then prove sum(n..infinity) (a_n)*(b_n) converges.

    I am guessing it has something to do with the comparison test but i can't quite put my finger on it, something with 0<=(abs(a_n)-abs(b_n))^2

    In addition am i on the right lines with thinking that, as the sum of (b_n)^2 converges, there must exist a number N such that sum(b_n)^2<1 for all n>=N
    Then look at the sum that goes from N to infinity of a_n*b_n ? this is the part im stuck at,
    any help would much appreciated, many thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member girdav's Avatar
    Joined
    Jul 2009
    From
    Rouen, France
    Posts
    678
    Thanks
    32

    Re: Convergence proof...

    0\leq |a_nb_n|\leq \frac{a_n^2+b_n^2}2.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5

    Re: Convergence proof...

    Quote Originally Posted by breitling View Post
    I am a little stuck on the following:
    suppose that:
    1) sum(n..infinity) (a_n)^2 and sum(n..infinity)(b_n)^2 converges, then prove sum(n..infinity) (a_n)*(b_n) converges.

    I am guessing it has something to do with the comparison test but i can't quite put my finger on it, something with 0<=(abs(a_n)-abs(b_n))^2

    In addition am i on the right lines with thinking that, as the sum of (b_n)^2 converges, there must exist a number N such that sum(b_n)^2<1 for all n>=N
    Then look at the sum that goes from N to infinity of a_n*b_n ? this is the part im stuck at,
    any help would much appreciated, many thanks.
    Setting c_{n}=\text{max}[|a_{n}|,|b_{n}|] it is clear that \sum_{n} c^{2}_{n} converges. But because is |a_{n}\ b_{n}|\le c^{2}_{n} also \sum_{n} a_{n}\ b_{n} converges...



    Marry Christmas from Serbia

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2011
    Posts
    58

    Re: Convergence proof...

    Many thanks to both of you
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Convergence proof...
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 1st 2010, 05:46 AM
  2. Convergence proof 2...
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 27th 2010, 12:58 PM
  3. Proof of Convergence
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: August 17th 2009, 03:09 PM
  4. Convergence Proof
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 20th 2008, 01:16 PM
  5. Convergence proof
    Posted in the Advanced Math Topics Forum
    Replies: 1
    Last Post: October 31st 2007, 02:17 PM

Search Tags


/mathhelpforum @mathhelpforum