1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Absolute converge of a series
Loading...