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I need help badly in yet another uniform convergence problem

  1. Mar 4, 2008 #1
    I hate, HATE uniform convergence for series... hate this, really do. I'm trying hard, but our prof gives us very difficult borderline problems.

    Now, I am almost 100% sure this is not uniformly convergence at x=1, but since it's a stinking series, it's hard to figure out it's summation

    1. The problem statement, all variables and given/known data
    data[/b]
    1) Test the following series for Uniform Convergence on [0,1]
    [tex]
    (1-x)\sum\limits_{n = 1}^{\inf } {\frac{x^n}{1+x^n}}
    [/tex]

    3. The attempt at a solution

    I honestly have not a clue where to start. I want to try the following, and I don't know if it's true.

    [tex]
    \frac{1}{2} < \frac{1}{1+x^n}
    [/tex]
    for x in [0,1)

    So [tex]
    (1-x)\sum\limits_{n = 1}^{\inf } {\frac{x^n}{2}} <
    (1-x)\sum\limits_{n = 1}^{\inf } {\frac{x^n}{1+x^n}}
    [/tex]

    and for x in [0,1) [tex]
    (1-x)\sum\limits_{n = 1}^{\inf } {\frac{x^n}{2}} = \frac{1}{2}
    [/tex]

    and we know that and for x = 0 [tex]
    (1-x)\sum\limits_{n = 1}^{\inf } {\frac{x^n}{1+x^n}} = 0
    [/tex]

    Thus the limit function is not continuous, so it can't be uniformly convergent.

    Sound proof or bad proof?
     
  2. jcsd
  3. Mar 4, 2008 #2

    quasar987

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    The idea was good, but you assumed that

    [tex]\sum_{n=1}^{\infty}x^n = \frac{1}{1-x}[/tex]

    while actually this is true if the index starts at n=0 instead of n=1.
     
  4. Mar 4, 2008 #3
    The logic should still work, however, correct? Or is there no fix?
     
  5. Mar 4, 2008 #4
    Isn't that simply going to come out to x/2 then? So it should still work correct?

    (The limit function is not continuous, since it's greater than or equal to 1/2 at 1, and not 0)
     
    Last edited: Mar 4, 2008
  6. Mar 5, 2008 #5

    quasar987

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    Now I can't find a fault. :)
     
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