Results 1 to 9 of 9
Like Tree1Thanks
  • 1 Post By Plato

Math Help - Absolute Convergence

  1. #1
    Newbie
    Joined
    Oct 2013
    From
    Washington, DC
    Posts
    2

    Absolute Convergence

    Let {\sum\limits_{n = 1}^{\infty}{a_n}} and {\sum\limits_{n = 1}^{\infty}{b_n}} be an absolutely convergent series of real numbers. Prove that the series {\sum\limits_{n = 1}^{\infty}{\sqrt{|(a_n)(b_n)|}}} converges.

    Now by the definition of absolute convergence we know that:
    {\sum\limits_{n = 1}^{\infty}{|(a_n)|} = g} for some real number g and {\sum\limits_{n = 1}^{\infty}{|(b_n)|} = h} for some real number h.

    So we have {\sum\limits_{n = 1}^{\infty}{|(a_n)||(b_n)|}}={\sum\limits_{n = 1}^{\infty}{|(a_n)(b_n)|} = gh}.

    How do I tie in square roots?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Oct 2012
    From
    Ireland
    Posts
    503
    Thanks
    136

    Re: Absolute Convergence

    Quote Originally Posted by hughmistler View Post
    [TEX]So we have {\sum\limits_{n = 1}^{\infty}{|(a_n)||(b_n)|}}={\sum\limits_{n = 1}^{\infty}{|(a_n)(b_n)|} = gh}.
    You cannot do this step because \sum a_n \times \sum b_n \neq \sum a_nb_n

    \sum a_n \times \sum b_n= (a_1+a_2+a_3...)(b_1+b_2+b_3+...)

    \sum a_nb_n= a_1b_1+a_2b_2+a_3b_3+...

    Sorry I don't know how to go about doing the required proof.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,383
    Thanks
    1474
    Awards
    1

    Re: Absolute Convergence

    Quote Originally Posted by hughmistler View Post
    Let {\sum\limits_{n = 1}^{\infty}{a_n}} and {\sum\limits_{n = 1}^{\infty}{b_n}} be an absolutely convergent series of real numbers. Prove that the series {\sum\limits_{n = 1}^{\infty}{\sqrt{|(a_n)(b_n)|}}} converges.
    For any real numbers a~\&~b can you prove that 2\sqrt{|ab|}\le |a|+|b|~?
    Hint: look at (\sqrt{|a|}-\sqrt{|b|})^2\ge 0.

    What can you do with that?
    Thanks from johng
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2013
    From
    DC
    Posts
    6

    Re: Absolute Convergence

    Quote Originally Posted by Plato View Post
    For any real numbers a~\&~b can you prove that 2\sqrt{|ab|}\le |a|+|b|~?
    Hint: look at (\sqrt{|a|}-\sqrt{|b|})^2\ge 0.

    What can you do with that?
    I mean, I understand how to derive that, i.e.,

    (\sqrt{|a|}-\sqrt{|b|})^2\ge 0.
    (\sqrt{|a|}-\sqrt{|b|})(\sqrt{|a|}-\sqrt{|b|})\ge 0.
    (|a| + |b| - 2\sqrt{|a||b|}\ge 0.
    2\sqrt{|ab|}\le |a| + |b|.

    But I am not really sure how to use that. Do I need to use the definition of lim of sequence? (a number L is called the limit of a sequence if for any \varepsilon > 0, \exists N \in \mathbb{N} such that for all n > N, one has |a_n - L| < \varepsilon.)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member Aryth's Avatar
    Joined
    Feb 2007
    From
    USA
    Posts
    651
    Thanks
    1
    Awards
    1

    Re: Absolute Convergence

    Quote Originally Posted by bobcantor1983 View Post
    I mean, I understand how to derive that, i.e.,

    (\sqrt{|a|}-\sqrt{|b|})^2\ge 0.
    (\sqrt{|a|}-\sqrt{|b|})(\sqrt{|a|}-\sqrt{|b|})\ge 0.
    (|a| + |b| - 2\sqrt{|a||b|}\ge 0.
    2\sqrt{|ab|}\le |a| + |b|.

    But I am not really sure how to use that. Do I need to use the definition of lim of sequence? (a number L is called the limit of a sequence if for any \varepsilon > 0, \exists N \in \mathbb{N} such that for all n > N, one has |a_n - L| < \varepsilon.)
    Note: \frac{h + g}{2} = \frac{\sum_{n=1}^{\infty}|a_n| + \sum_{n=1}^{\infty}|b_n|}{2} = \sum_{n=1}^{\infty}\left(\frac{|a_n|}{2} + \frac{|b_n|}{2}\right)

    Note that the series on the right side is still convergent. What does this tell us about \sum_{n=1}^{\infty}\sqrt{|a_n||b_n|}?
    Last edited by Aryth; October 1st 2013 at 08:22 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Oct 2013
    From
    DC
    Posts
    6

    Re: Absolute Convergence

    I understand how you can do that with addition and that the sum of two convergent series converges, but how can that apply to multiplying two convergent series?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member Aryth's Avatar
    Joined
    Feb 2007
    From
    USA
    Posts
    651
    Thanks
    1
    Awards
    1

    Re: Absolute Convergence

    Quote Originally Posted by bobcantor1983 View Post
    I understand how you can do that with addition and that the sum of two convergent series converges, but how can that apply to multiplying two convergent series?
    By the absolute value property (and the how the square root is defined), we have that \left(\frac{|a_n| + |b_n|}{2}\right)*\sqrt{|a_n||b_n|}\geq 0, \forall n. Following from my previous post, we have that:
    \sqrt{|a_n||b_n|} \leq \frac{|a_n| + |b_n|}{2}, \forall n.
    By the comparison test, \sum_{n=1}^{\infty}\sqrt{|a_n||b_n|} is convergent.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Oct 2013
    From
    DC
    Posts
    6

    Re: Absolute Convergence

    So let me put this together, does this all make sense, am I missing anything?

    Suppose {\sum\limits_{n = 1}^{\infty}{a_n}} and {\sum\limits_{n = 1}^{\infty}{b_n}} are absolutely convergent series of real numbers. We need to show that {\sum\limits_{n = 1}^{\infty}{\sqrt{|(a_n)(b_n)|}}} converges. We will show this using the comparison test.

    The comparison test states: Suppose that we have two series, {\sum\limits_{n = 1}^{\infty}{a_n}} and {\sum\limits_{n = 1}^{\infty}{b_n}}, with a_n, b_n \ge 0 for all n and a_n \le b_n for all n; then if {\sum\limits_{n = 1}^{\infty}{b_n}} is convergent so is {\sum\limits_{n = 1}^{\infty}{a_n}}.

    Now by the definition of absolute convergence we know that:
    {\sum\limits_{n = 1}^{\infty}{|(a_n)|} = g} for some real number g and {\sum\limits_{n = 1}^{\infty}{|(b_n)|} = h} for some real number h.

    Let us consider that \sqrt{|a_n b_n|} \le \frac{|a_n| + |b_n|}{2}
    We show this is true:

    (\sqrt{|a_n|}-\sqrt{|b_n|})^2\ge 0.

    (\sqrt{|a_n|}-\sqrt{|b_n|})(\sqrt{|a_n|}-\sqrt{|b|})\ge 0.

    (|a_n| + |b| - 2\sqrt{|a_n||b_n|}\ge 0.

    2\sqrt{|a_n b_n|}\le |a_n| + |b_n|.

    \sqrt{|a_n b_n|}\le \frac{|a_n| + |b_n|}{2}.

    Now consider that,
    \frac{g + h}{2} = \frac{\sum_{n=1}^{\infty}|a_n| + \sum_{n=1}^{\infty}|b_n|}{2} = \sum_{n=1}^{\infty}\left(\frac{|a_n|}{2} + \frac{|b_n|}{2}\right)

    Since we have shown that every element \sqrt{|a_n b_n|} is less than or equal to this new series, and it is clear that this new series converges since it is the sum of two convergent series multiplied by a constant. Hence by the comparison test, \sum_{n=1}^{\infty}\sqrt{|a_n||b_n|} is convergent.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,383
    Thanks
    1474
    Awards
    1

    Re: Absolute Convergence

    Quote Originally Posted by bobcantor1983 View Post
    So let me put this together, does this all make sense, am I missing anything?
    Suppose {\sum\limits_{n = 1}^{\infty}{a_n}} and {\sum\limits_{n = 1}^{\infty}{b_n}} are absolutely convergent series of real numbers. We need to show that {\sum\limits_{n = 1}^{\infty}{\sqrt{|(a_n)(b_n)|}}} converges. We will show this using the comparison test.
    The comparison test states: Suppose that we have two series, {\sum\limits_{n = 1}^{\infty}{a_n}} and {\sum\limits_{n = 1}^{\infty}{b_n}}, with a_n, b_n \ge 0 for all n and a_n \le b_n for all n; then if {\sum\limits_{n = 1}^{\infty}{b_n}} is convergent so is {\sum\limits_{n = 1}^{\infty}{a_n}}.
    Now by the definition of absolute convergence we know that:
    {\sum\limits_{n = 1}^{\infty}{|(a_n)|} = g} for some real number g and {\sum\limits_{n = 1}^{\infty}{|(b_n)|} = h} for some real number h.
    Let us consider that \sqrt{|a_n b_n|} \le \frac{|a_n| + |b_n|}{2}
    We show this is true:
    (\sqrt{|a_n|}-\sqrt{|b_n|})^2\ge 0.
    (\sqrt{|a_n|}-\sqrt{|b_n|})(\sqrt{|a_n|}-\sqrt{|b|})\ge 0.
    (|a_n| + |b| - 2\sqrt{|a_n||b_n|}\ge 0.
    2\sqrt{|a_n b_n|}\le |a_n| + |b_n|.
    \sqrt{|a_n b_n|}\le \frac{|a_n| + |b_n|}{2}.
    Now consider that,
    \frac{g + h}{2} = \frac{\sum_{n=1}^{\infty}|a_n| + \sum_{n=1}^{\infty}|b_n|}{2} = \sum_{n=1}^{\infty}\left(\frac{|a_n|}{2} + \frac{|b_n|}{2}\right)
    Since we have shown that every element \sqrt{|a_n b_n|} is less than or equal to this new series, and it is clear that this new series converges since it is the sum of two convergent series multiplied by a constant. Hence by the comparison test, \sum_{n=1}^{\infty}\sqrt{|a_n||b_n|} is convergent.
    There is nothing wrong with what you have done. It is just overly complicated.
    From the given you know that both \sum\limits_n {\left| {{a_n}} \right|} \;\& \;\sum\limits_n {\left| {{b_n}} \right|} converge.

    From the lemma you know that
    \sum\limits_n {\left| {\sqrt {\left| {{a_n}{b_n}} \right|} } \right|}  \le \sum\limits_n {\left| {2\sqrt {\left| {{a_n}{b_n}} \right|} } \right|}  \le \sum\limits_n {\left| {{a_n}} \right|} \; + \;\sum\limits_n {\left| {{b_n}} \right|} .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. absolute convergence
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 29th 2010, 02:58 AM
  2. Absolute convergence
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 17th 2008, 11:41 PM
  3. Absolute convergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 9th 2008, 07:04 PM
  4. absolute convergence
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 26th 2007, 06:04 PM
  5. help with sum for absolute convergence
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 17th 2007, 10:32 PM

Search Tags


/mathhelpforum @mathhelpforum