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Math problem (analysis)

  1. Apr 21, 2005 #1
    Hi,

    I have to show that the function

    [tex]f(x) = \sum_{n=1}^{\infty}\frac1{x^2+n^2}[/tex] ​

    tends to 0 as [tex]x \rightarrow \infty[/tex], i.e. [tex]\lim_{x\rightarrow\infty}f(x) = 0[/tex]. How can I do this?

    There is a hint that says I should use the inequality [tex] f(x) \leq \sum_{n=1}^N\tfrac1{x^2+n^2} + \sum_{n=N+1}^\infty\tfrac1{n^2} [/tex]. It is obvious that the first term approaches 0 as [tex]x \rightarrow \infty[/tex], but what about the second term?
     
  2. jcsd
  3. Apr 21, 2005 #2

    dextercioby

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    How do you know that?

    How do you know that

    [tex] \lim_{x\rightarrow +\infty}\sum_{n=1}^{N}\frac{1}{x^{2}+n^{2}}=? 0 [/tex]

    My Maple says it's undefined,as well as the initial limit...


    Daniel.
     
  4. Apr 21, 2005 #3

    dextercioby

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    Denote the initial limit by [itex] F =:\lim_{x\rightarrow +\infty} f(x) [/itex]

    U can show that [itex] f(0)=\zeta\left(2\right)=\frac{\pi^{2}}{6} [/itex] and [itex] \forall x\in\mathbb{R}, f(x)\leq f(0) [/itex].

    Daniel.
     
  5. Apr 21, 2005 #4

    dextercioby

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    I've looked at the graph and the function quicky goes to 0 as "x" increases.

    It's continuous everywhere on [itex] \mathbb{R} [/itex] and has the derivative

    [tex] f'(x)=\sum_{n=1}^{\infty}\left(-\frac{2x}{\left(x^{2}+n^{2}\right)^{2}}\right) [/tex]

    which is negative on the positive semiaxis.Therefore the function decreases continuously on the positive semiaxis.Since it's continuous,even and strictly positive,the asymptotic limits have to be zero.

    Looking forward for critique.

    Daniel.
     
  6. Apr 21, 2005 #5

    Hurkyl

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    It's a function of N.

    (Notice that f(x) is not a function of N...)
     
  7. Apr 21, 2005 #6

    Hurkyl

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    Why wouldn't the asymptotic limit be 1/10, or π4/792?
     
  8. Apr 21, 2005 #7

    saltydog

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    You know what, I would give a whole dollar if someone could explain to me why:

    [tex]\sum_{n=1}^{\infty}\frac{1}{a^2+n^2}=\frac{a\pi Coth[a\pi]-1}{2a^2}[/tex]

    This is what Mathematica returns and I don't wish to suggest using Mathemtaica to prove things but I don't know how to prove the original question, and in my attempts to learn how, I've reduced it to the above. Now, you know what happens to the RHS as a goes to infinity right?
     
  9. Apr 21, 2005 #8

    dextercioby

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    Interesting.My ancient version of Maple returned a difference of psi functions of complex arguments multiplied with 1/(ix).


    Daniel.
     
  10. Apr 21, 2005 #9
    I can let N -> infinity and thereby make the second term vanish but then what about the first term which also depends on N?
     
  11. Apr 21, 2005 #10

    Hurkyl

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    It's a finite sum.
     
  12. Apr 21, 2005 #11
    Hey in response to your question, I have observed that it's very easy to tell that the sum tends to 0 as n goes to infinity. As n^2 gets large, the x^2 term gets very large too at the same rate. Any # with an inreasing denominator decreases in value, therefore the total sum goes to zero. :rofl:
     
    Last edited: Apr 21, 2005
  13. Apr 21, 2005 #12

    cepheid

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    "n" is not what is tending to infinity here. "x" is, as far as I can see. But even it it were n doing the increasing, the following still makes no sense:

    Why? How does the value of n affect x in any way?
     
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