Results 1 to 4 of 4

Math Help - comparison test with 0 as upperbound

  1. #1
    Junior Member
    Joined
    Feb 2009
    Posts
    31

    comparison test with 0 as upperbound

    We all know the rules for comparison test with 0 <= asubk <= bsubk. Then

    I. If my series of a's diverges then so does my series of b's.
    II. if my series of b's converges then so does my series of a's.

    However, I want to know what these two rules would look like if

    asubk <= bsubk <= 0. I know the test still works.

    I'm using this for a proof in analysis. I thought of the hard part to modify traditional comparison test, but now can't complete it because my brain is exploiding.

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by grandunification View Post
    asubk <= bsubk <= 0. I know the test still works.

    I'm using this for a proof in analysis. I thought of the hard part to modify traditional comparison test, but now can't complete it because my brain is exploiding.

    Thanks
    if the a's, then the b's will. if the b's diverge, then the a's will.

    try to modify the original comparison test by multiplying the system through by a negative power. if 0 is a lower bound, that is.

    it would do you well to actually post the problem you are having trouble with
    Follow Math Help Forum on Facebook and Google+

  3. #3
    tah
    tah is offline
    Junior Member
    Joined
    Feb 2009
    Posts
    51
    Instead of the series of a's, try to use the series of (-a)'s, and so is for b
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by grandunification View Post
    We all know the rules for comparison test with 0 <= asubk <= bsubk. Then

    I. If my series of a's diverges then so does my series of b's.
    II. if my series of b's converges then so does my series of a's.

    However, I want to know what these two rules would look like if

    asubk <= bsubk <= 0. I know the test still works.

    I'm using this for a proof in analysis. I thought of the hard part to modify traditional comparison test, but now can't complete it because my brain is exploiding.

    Thanks
    A series of non-negative terms converges if and only the sequence of partial sums are bounded. Therefore, if \sum_{n\geq 1}b_n converges then \sum_{k=1}^n b_k is bounded, but so is \sum_{k=1}^n a_k, and so \sum_{n\geq 1}a_n converges.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Comparison test and Limit Comparison test for series
    Posted in the Calculus Forum
    Replies: 5
    Last Post: November 25th 2010, 12:54 AM
  2. Comparison or Limit Comparison Test Problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 12th 2010, 07:46 AM
  3. Limit comparison/comparison test series
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 25th 2009, 08:27 PM
  4. Comparison & Limit Comparison test for series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 25th 2009, 04:00 PM
  5. Integral Test/Comparison Test
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 27th 2009, 10:58 AM

Search Tags


/mathhelpforum @mathhelpforum