Results 1 to 7 of 7

Math Help - convergence from comparison test

  1. #1
    Newbie
    Joined
    Nov 2009
    Posts
    18

    convergence from comparison test

    Here is the problem:

    If  \sum a_{n} with  a_{n}>0 is convergent, then is  \sum \sqrt{a_{n}a_{n+1}} always convergent? if not provide a counter example.


    Now I don't think that it is also convergent. for example,  2 \equiv \sqrt{2*2} < \sqrt {2*3} . If I am right, what would be a counter example in terms of a? or anything else for that matter?

    But if it is truly convergent, I know you have to approach this with the fact that a square root is non negative and you have to use a comparison test. but i just don't see how or why
     \sum \sqrt{a_{n}a_{n+1}}   <    \sum a_{n}
    Last edited by osudude; December 1st 2009 at 01:25 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by osudude View Post
    Here is the problem:

    If  \sum a_{n} with  a_{n}>0 is convergent, then is  \sum \sqrt{a_{n}a_{n+1}} always convergent? if not provide a counter example.


    Now I don't think that it is also convergent. for example,  2 \equiv \sqrt{2*2} < \sqrt {2*3} . If I am right, what would be a counter example in terms of a? or anything else for that matter?

    But if it is truly convergent, I know you have to approach this with the fact that a square root is non negative and you have to use a comparison test. but i just don't see how or why
     \sum \sqrt{a_{n}a_{n+1}} < \sum a_{n}
    AMGM!The AM-GM inequality
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2009
    Posts
    18
    so in application of the amgm, would i just get that  a_{n} < \sqrt{a_{n}a_{n+1}} and thus it is divergent from the comparison test??
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Quote Originally Posted by osudude View Post
    so in application of the amgm, would i just get that  a_{n} < \sqrt{a_{n}a_{n+1}} and thus it is divergent from the comparison test??
    I do not think that you understand the inequality.
    a^2+b^2\ge2ab.
    So in this case 2\sqrt{a_na_{n+1}}\le a_n+a_{n+1}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by osudude View Post
    so in application of the amgm, would i just get that  a_{n} < \sqrt{a_{n}a_{n+1}} and thus it is divergent from the comparison test??
    Quote Originally Posted by Plato View Post
    I do not think that you understand the inequality.
    a^2+b^2\ge2ab.
    So in this case 2\sqrt{a_na_{n+1}}\le a_n+a_{n+1}
    As Plato pointed out the specific case of the AMGM here is that \left(\sqrt{x}-\sqrt{y}\right)^2\ge 0 ( then of course we may expand this to get \left(\sqrt{x}\right)^2-2\sqrt{xy}+\left(\sqrt{y}\right)^2\implies x+y\ge 2\sqrt{xy}. I pointed out the general case so that you may note that there is nothing special about there being two terms...except that it is particularly easy to prove the lemma needed.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Nov 2009
    Posts
    18
    ok so I am still confused. So I understand that

     2\sqrt{a_na_{n+1}}\le a_n+a_{n+1}

    which equals

    \sqrt{a_na_{n+1}}\le \frac {a_n+a_{n+1}}{2}

    now we know that

     \frac{a_n+a_{n+1}}{2} > \frac{a_{n}+a_{n}}{2} = a_{n}

    . So if

     \frac{a_n+a_{n+1}}{2} > \sqrt{a_na_{n+1}}

    and

     \frac{a_n+a_{n+1}}{2} > \frac{a_{n}+a_{n}}{2} = a_{n}


    Then what does that tell us about the relation between

    \sqrt{a_na_{n+1}} and {a_n} ? I mean I want to show, somehow, that \sqrt{a_na_{n+1}} \leq a_n and thus \sqrt{a_na_{n+1}} is convergent by the comparison test
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by osudude View Post
    ok so I am still confused. So I understand that

     2\sqrt{a_na_{n+1}}\le a_n+a_{n+1}

    which equals

    \sqrt{a_na_{n+1}}\le \frac {a_n+a_{n+1}}{2}

    now we know that

     \frac{a_n+a_{n+1}}{2} > \frac{a_{n}+a_{n}}{2} = a_{n}

    . So if

     \frac{a_n+a_{n+1}}{2} > \sqrt{a_na_{n+1}}

    and

     \frac{a_n+a_{n+1}}{2} > \frac{a_{n}+a_{n}}{2} = a_{n}


    Then what does that tell us about the relation between

    \sqrt{a_na_{n+1}} and {a_n} ? I mean I want to show, somehow, that \sqrt{a_na_{n+1}} \leq a_n and thus \sqrt{a_na_{n+1}} is convergent by the comparison test
    \sum\sqrt{a_n a_{n+1}}\le\frac{1}{2}\sum\left\{a_n+a_{n-1}\right\}=a_0+2\sum a_n
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Comparison test for convergence/divergence
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: January 9th 2012, 07:55 AM
  2. [SOLVED] Comparison test and Limit Comparison test for series
    Posted in the Calculus Forum
    Replies: 5
    Last Post: November 25th 2010, 12:54 AM
  3. Comparison or Limit Comparison Test Problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 12th 2010, 07:46 AM
  4. Comparison & Limit Comparison test for series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 25th 2009, 04:00 PM
  5. convergence and comparison test help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 15th 2009, 01:32 AM

Search Tags


/mathhelpforum @mathhelpforum