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For which values of x does this series converge?

  1. Oct 30, 2011 #1
    1. The problem statement, all variables and given/known data
    For what real values of x does this series converge?

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


    3. The attempt at a solution

    I've rewritten the series as [tex]\sum_{n=1}^{∞}{\frac{x^n}{1 + x^{2n}}}[/tex] and I know that I can make each term larger, say
    [tex]{\frac{x^n}{1 + x^{2n}}} < {\frac{x^n}{x^{2n}}} = \left({\frac{x}{x^{2}}}\right)^n = \left(\frac{1}{x}\right)^n[/tex]

    Then this will converge, by comparison to the geometric series, when |x| > 1. But surely this can't be a method for finding ALL values of x for which the original series converges, can it? Because I changed the series , so all I've found is the values of x for which [tex]\sum_{n=1}^{∞}{\frac{x^n}{x^{2n}}}[/tex] converges. How can I go about this in a different way?

    Then there's the question of uniform convergence. Can I just use the M-test for both parts, and say that

    [tex]\left|\frac{x^n}{1 + x^{2n}}\right| < \left|\frac{x^n}{x^{2n}}\right| = \left|\frac{1}{x}\right|^n = M_n[/tex]

    so, [tex]\sum_{n=1}^{∞}{M_n} = \sum_{n=1}^{∞} \left|\frac{1}{x}\right|^n[/tex] converges uniformly when |x| > 1?

    I just fear that I'm simplifying things too much and not getting ALL of the possible values of x such that the series converges. Any suggestions?
     
    Last edited: Oct 30, 2011
  2. jcsd
  3. Oct 30, 2011 #2

    lanedance

    User Avatar
    Homework Helper

    this problem might be well suited to a comparison

    say 0<x<1, then for all n
    [tex]
    \frac{1}{x^n} + x^n > \frac{1}{x^n}
    [/tex]

    Hence
    [tex]
    \frac{1}{\frac{1}{x^n} + x^n}< \frac{1}{\frac{1}{x^n}}
    [/tex]

    note you could do similar thing for x>1 and negative values
     
  4. Oct 30, 2011 #3
    Okay, that makes sense. But since I'm only comparing to other series, can I be sure that I'm catching all of the values of x where the original series converges?

    EDIT: Nevermind...since I'm comparing the original series based on specific intervals of x, I'll be fine.

    So I'll get the same interval for convergence and uniform convergence by the M-test?
     
  5. Oct 31, 2011 #4

    lanedance

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    Homework Helper

    not quite

    consider x=1/2, using the fact below do you think the series will converge?
    [tex]
    \frac{1}{\frac{1}{(\frac{1}{2})^n} + (\frac{1}{2})^n}< \frac{1}{\frac{1}{(\frac{1}{2})^n}}= \frac{1}{2^n}
    [/tex]


    now consider x=2, using the fact below do you think the series will converge?
    [tex]
    \frac{1}{\frac{1}{(2)^n} + (2)^n}< \frac{1}{2^n}
    [/tex]

    note the symmetry in arguments, for any x>1, you can use the same argument above for y=1/x, with 0<y<1
     
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