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Series again

  1. Aug 22, 2005 #1
    Okay, I have this function defined as an infinite series:

    [tex]f(x) = \sum_{n=1}^{\infty}\frac{\sin(nx)}{x+n^4}[/tex]​

    which is converges uniformly and absolutely for x > 0. I have shown that f is continous and has a derivative for x > 0. Now I have to show that [tex]f(x) \rightarrow 0[/tex] as [tex]x \rightarrow \infty.[/tex] It's obvious that it is the case, but how do I prove it. I've tried putting 1/x outside the sum, but then I don't know about the remaining part. Any ideas?
    Last edited: Aug 22, 2005
  2. jcsd
  3. Aug 22, 2005 #2
    set [tex]f_n(x) = \frac{sin(nx)}{x+n^4}[/tex], then [tex]|f_n(x)| \leq \frac{1}{n^4}[/tex] for all x>0
  4. Aug 23, 2005 #3
    Hmm.. How does that show that f(x) -> 0 as x -> infinity?
  5. Aug 23, 2005 #4


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    by the comparison test
  6. Aug 23, 2005 #5
    Maybe I'm a little slow here, but [tex]\sum_{n=1}^{\infty}\frac1{n^4}[/tex] does not equal 0, so how come?
  7. Aug 23, 2005 #6


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    My mistake. Have you any theorems concerning the interchange of limits for uniformly convergent series?
  8. Aug 23, 2005 #7
    No, I can't find any relevant theorems to apply. I'm lost at sea.
  9. Aug 23, 2005 #8
    This series converges absolutely by the p-series test. I dont see why you have to look for any limits or to check it is decreases. If it comverges absolutely, then it converges.


  10. Aug 27, 2005 #9


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    [tex]f(x) = \frac{sin(x)}{x+1} + \sum_{n=2}^{\infty}\frac{\sin(nx)}{x+n^4}[/tex]

    This is clearly bounded by:

    [tex]\frac{1}{x+1} + \int_{n=1}^\infty \frac{1}{x+n^4}\,\,dn[/tex]

    which can be shown to go to zero.

  11. Aug 27, 2005 #10
    isn't that a much harder way to do it though? & judging by brogger's other posts i would say he's in the standard advanced calculus course where you'd learn the usual tests for convergence. (like comparison)
  12. Aug 28, 2005 #11


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    It's been a few decades, but I suppose that the integral test is still taught in freshman calculus. The advantage of using the integral test is that in addition to convergence, since the bounding integral goes to zero as x to infinity, you also get that the series goes to zero.

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