Results 1 to 5 of 5

Math Help - Showing a series is less than another

  1. #1
    Newbie
    Joined
    Dec 2012
    From
    Davis
    Posts
    22

    Showing a series is less than another

    I have some issues with this problem.

    Let (the sum of an from n=1 to infinity) and (the sum of bn from n=1 to infinity) be two convergent series. If an (less than or equal to) bn for all n in Natural Number, and am < bm for some m in Natural Number, show that (the sum of an from n=1 to infinity) < (the sum of bn from n=1 to infinity).

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,548
    Thanks
    1418

    Re: Showing a series is less than another

    Sorry but isn't this statement obvious? If all terms in the a series are less than the terms in the b series, then surely the a series must be smaller than the b series...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2012
    From
    Davis
    Posts
    22

    Re: Showing a series is less than another

    Well, ish, but a lot of the elementary analysis I'm dealing with seems to be proving somewhat obvious things with a more roundabout method. I mean, if a problem like this came up on the test, I couldn't really just state the statement is self-explanatory. But for me, I have more of an issue of figuring out how to start my proofs.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,548
    Thanks
    1418

    Re: Showing a series is less than another

    Well in this case, since \displaystyle \begin{align*} a_m < b_m \end{align*}, that means \displaystyle \begin{align*} b_m = a_m + k_m \end{align*} where \displaystyle \begin{align*} k_m > 0 \end{align*}.

    That means \displaystyle \begin{align*} \sum b_m = \sum \left( a_m + k_m \right) = \sum a_m + \sum k_m > \sum a_m \end{align*}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2012
    From
    Planet Earth
    Posts
    194
    Thanks
    49

    Re: Showing a series is less than another

    Quote Originally Posted by Prove It View Post
    Sorry but isn't this statement obvious? If all terms in the a series are less than the terms in the b series, then surely the a series must be smaller than the b series...
    Depends on who it's obvious to. Half of the things that are "proven" these days as exercises for relevant classes are obvious. The demand for rigor never seems to end.

    Well I guess you could argue that a_n will in at least once instance be less than b_n, and then will never possibly exceed or equal b_n, therefore the sum is smaller... depends what kind of arguments you are allowed to use.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Showing being even
    Posted in the Geometry Forum
    Replies: 2
    Last Post: September 7th 2011, 10:06 AM
  2. showing
    Posted in the Algebra Forum
    Replies: 4
    Last Post: June 18th 2011, 04:54 AM
  3. Replies: 0
    Last Post: February 17th 2010, 12:27 PM
  4. Showing that Z<x,y> is not a UFD
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: January 27th 2010, 09:37 AM
  5. showing f'(-x)=f'(x)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 26th 2007, 08:14 AM

Search Tags


/mathhelpforum @mathhelpforum