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Homework Help: Absolute converge of a series

  1. May 28, 2006 #1
    Hello

    I have this problem here:

    Given the series

    [tex]\sum_{n=1} ^{\infty} \frac{x^{n+1}}{n(n+1)}[/tex]

    show that this converges for every [tex]x \in \{ w \in \mathbb{C} \| \|w \| \leq 1 \}[/tex]

    Solution:

    Since

    [tex] \sum_{n=1} ^{\infty} \left| \frac{x^{n+1}}{n(n+1)} \right| = \sum_{n=1} ^{\infty} \frac{1}{n(n+1)} \right| [/tex]

    Since [tex]\sum_{n=1} ^{\infty} \frac{1}{n(n+1)} \right| [/tex] convergence then according the definition then

    [tex]\sum_{n=1} ^{\infty} \frac{x^{n+1}}{n(n+1)}[/tex] is absolute convergent.

    Thus

    [tex]x \in \{ w \in \mathbb{C} \| \|w \| \leq 1 \}[/tex]. Right?

    Best Regards
    Fred
     
    Last edited: May 28, 2006
  2. jcsd
  3. May 28, 2006 #2

    shmoe

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    This:

    [tex] \sum_{n=1} ^{\infty} \left| \frac{x^{n+1}}{n(n+1)} \right| = \sum_{n=1} ^{\infty} \frac{1}{n(n+1)} \right| [/tex]

    shouldn't be an equality. You hopefully meant an inequality here which is valid for x on the unit disc. Otherwise it looks good.
     
  4. May 28, 2006 #3
    Then it should have say ?

    [tex] \sum_{n=1} ^{\infty} \left| \frac{x^{n+1}}{n(n+1)} \right| \leq \sum_{n=1} ^{\infty} \frac{1}{n(n+1)} \right| [/tex]

    Regards Humminbird

    p.s. Do I use uniform continuety to show that

    [tex]g(x) = \sum_{n=1} ^{\infty} \frac{x^{n+1}}{n(n+1)}[/tex]

    is continious on [tex]x \in \{ w \in \mathbb{C} \| \|w \| \leq 1\}[/tex] ??


     
    Last edited: May 28, 2006
  5. May 28, 2006 #4

    HallsofIvy

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    Science Advisor

    No, if
    [tex]\sum_{n=1} ^{\infty} \frac{x^{n+1}}{n(n+1)}[/tex]
    then all you can say is "if x= 1 this is
    [tex]\sum_{n=1} ^{\infty} \frac{1}{n(n+1)}[/tex]
    which converges absolutely" (since all terms are positive anyway).

    Of course, I'm not sure why you chose to make x= 1. Have you determined what the radius of convergence is? And if the radius of convergence is 1, what happens when x= -1?
     
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