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Math Help - Uniform convergence again

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Uniform convergence again

    Hello everyone, how does one go about proving uniform convergence of a series which is conditionally convergent by the Weierstrass M-test?

    For example we know that





    So we proceed with the M-test to show uniform convergence on this interval. We know it is because it is inthe interval of convergence and a series is uniformally convergent forall x an element of its interval of convergence..but we know this retrospectively...we must prove it by the M-test

    So we start with



    But here is where I get a little stuck...since using the M-test we must have that



    We would have that


    Or could you say virtue of the less than or EQUAL to that



    It must be the former otherwise you cannot prove convergence of the comparison series. So could someone confirm that I may use this last inequality for my comparison series opposed to the one without the (-1)^n.

    EDIT: This cannot be right, because if x=1 and n=3 then we have that a positive value is less than a negative value...ok so what now?


    Secondly I do not know how to prove that the following series converges uniformly on [-1,0].



    We once again see that this interval is a subset of the interval of convergence thus uniformly convergent but my problem is that



    Is there an easy one that I am missing...or what should I do here? Remember although there are other tests like Dirchlects(must have messed up that spelling) as well as others, the directions clearly state Weierstrass M-test.


    Any help appreciated.

    Alex
    Last edited by Mathstud28; July 27th 2008 at 11:11 PM.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    For the first one I got it using Abel's test which says


    So for the first one seeing that we may define



    That

    converges

    and

    this should be [-1,1]

    And finally
    THis should be [-1,1]

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