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Convergence of a series

  1. May 23, 2006 #1
    Hello

    I have this question here which has puzzled me.

    Given a series

    [tex]\sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2}[/tex]

    Show that the series converge for every [tex]y \in \mathbb{R}[/tex]

    By the test of comparison

    [tex]\left| \sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2} \right | \leq \frac{1}{n^2}[/tex]

    Since its know that

    [tex]\sum \limit_{n=0} ^{\infty} \frac{1}{n^2}[/tex] converge, then the series [tex]\sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2}[/tex] converge for every [tex]y \in \mathbb{R}[/tex]

    Second show that the series converge Uniformt on [tex]\mathbb{R}[/tex]

    Again since

    [tex]\left| \sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2} \right | \leq \frac{1}{n^2}[/tex]

    and since [tex]\sum \limit_{n=0} ^{\infty} \frac{1}{n^2}[/tex] converge.

    Then by Weinstrass M-Test, then series Converge Uniformt on [tex]\mathrm{R}[/tex]

    Third show that the sum-function

    [tex]f: \mathbb{R} \rightarrow \mathbb{R}[/tex]

    [tex]f(y) = \sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2}[/tex]

    is continuous on [tex]\mathbb{R}[/tex]

    Can I conclude here that since the series converge Uniformly on R, then its sum-functions is continuous on [tex]\mathbb{R}[/tex] ???

    /Frank
     
    Last edited: May 23, 2006
  2. jcsd
  3. May 23, 2006 #2
    So what is your question? What is puzzling you?
     
  4. May 23, 2006 #3

    LeonhardEuler

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    y can be 0, so this should be
    [tex]\left| \sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2} \right | \leq \frac{1}{n^2}[/tex]
     
  5. May 23, 2006 #4

    Curious3141

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    There's a problem - the sequence you're comparing with is zeta(2), which converges (the sum is pi^2/6). However, that sum goes from n = 1 to infinity. When you start with n = 0, the first term is infinite, so obviously the sequence diverges.

    I would suggest amending it like so :

    [tex]\sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2} < \frac{1}{y^2} + \sum \limit_{n=1} ^{\infty} \frac{1}{n^2} = \frac{1}{y^2} + \frac{\pi^2}{6} [/tex]

    BTW, y must be nonzero for convergence.
     
    Last edited: May 23, 2006
  6. May 23, 2006 #5
    So What can I conclude that the series

    doesn't converge for every [tex]y \in \mathbb{R}[/tex] ??

    /Frank

     
  7. May 23, 2006 #6

    Curious3141

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    It converges for all [tex]y \in \mathbb{R}[/tex]\{0}
     
  8. May 23, 2006 #7
    Okay,

    what about my other conclusions regarding the series ?
    Do they look okay?

    /Frank
     
    Last edited: May 23, 2006
  9. May 23, 2006 #8

    Curious3141

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    I honestly do not know enough analysis to comment on those, I'm sure someone more knowledgeable will be along to handle those. :smile:
     
  10. May 23, 2006 #9

    benorin

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    For all real [tex]y\neq 0,[/tex] we have

    [tex]f(y) = \sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2}=\frac{\pi}{2y} \mbox{coth}( \pi y) [/tex]​

    EDIT: my bad, I forgot to start with n=0, but hey: who's counting.
     
    Last edited: May 23, 2006
  11. May 23, 2006 #10
    Hello benorin,

    Could You please check if my other statements about Uniform Convergence and continuety for the series, are correct?

    /Frank


     
    Last edited: May 23, 2006
  12. May 23, 2006 #11

    benorin

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    Convergence is uniform for all non-zero real y, since

    [tex]\left| \frac{1}{y^2 + n^2}\right| \leq \frac{1}{n^2}[/tex] for all [tex]n\geq 1[/tex]

    and [tex]\sum_{n=1}^{\infty}\frac{1}{n^2}[/tex] converges, thus

    EDIT: forgot a + sign in this one:

    [tex]f(y) = \sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2}=\frac{1}{y^2}+ \sum \limit_{n=1} ^{\infty} \frac{1}{y^2 + n^2} [/tex]

    converges uniformly for all [tex]y\neq 0[/tex] EDIT: by the Weierstrass M-test. Furthermore, as each term in the series is a continuous function of y, the sum function is also continuous by uniform convergence.

    Note that you can derive the formula

    [tex]f(y) = \sum \limit_{n=0} ^{\infty} \frac{1}{y^2 + n^2}
    =\frac{\pi}{2y} \mbox{coth}( \pi y) [/tex]​

    from the partial fraction expansion of coth(x), which is

    [tex]\mbox{coth}( x)= 2x\sum \limit_{n=1} ^{\infty} \frac{1}{x^2 + \pi ^2n^2}+\frac{1}{x} [/tex]​
     
    Last edited: May 23, 2006
  13. May 23, 2006 #12
    Okay Thank You for correcting me :)

    One final question though

    I need to show that

    [tex]lim _{y \rightarrow \infty} f(y) = 0 [/tex]

    Can I Do this by Wierstrass M-Test?

    /Frank

     
  14. May 23, 2006 #13

    benorin

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    Use uniform convergence, the main idea of uniform convergence is to allow the interchange of limiting processes, such as

    [tex]\lim_{x\rightarrow \infty}\sum_{n=0}^{\infty}f_n(x) = \sum_{n=0}^{\infty}\lim_{x\rightarrow \infty}f_n(x)[/tex]

    if the series is uniformly convergent for all x. Try it.
     
  15. May 23, 2006 #14
    Then by Uniform Convergence

    [tex]{\lim \sup_{n \rightarrow \infty}} \left| f_n(y) - f(y) \right| = 0[/tex]

    Can I then conclude that

    [tex]{\lim_{n \rightarrow \infty}}f(y) = 0[/tex] ??


     
    Last edited: May 23, 2006
  16. May 23, 2006 #15

    benorin

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    What is meant is, do this:

    [tex]\lim_{y\rightarrow \infty}f(y)=\lim_{y\rightarrow \infty}\sum_{n=0}^{\infty} \frac{1}{y^2 + n^2}= \sum_{n=0}^{\infty}\lim_{y\rightarrow \infty} \frac{1}{y^2 + n^2} = \sum_{n=0}^{\infty}0=0[/tex]​
     
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