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Homework Help: Convergent Series

  1. Jul 13, 2010 #1
    1. The problem statement, all variables and given/known data

    Is this series convergent for all real x:

    [tex]\sum[/tex][tex]^{\infty}_{k=2}[/tex][tex]\frac{sin(kx)}{ln(k)}[/tex]

    2. Relevant equations
    3. The attempt at a solution

    This series is less than

    [tex]\frac{1}{ln(2)}[/tex][tex]\sum[/tex][tex]^{\infty}_{k=2}[/tex]sin(kx)

    which is less than [tex]\frac{\pi}{x ln2}[/tex]. So, the series is bounded for all x. I'm thinking that the Dirichlet Test would show that this series converges.
     
    Last edited: Jul 13, 2010
  2. jcsd
  3. Jul 13, 2010 #2

    Hurkyl

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    Er, but sin(kx) doesn't converge to zero as k goes to infinity.
     
  4. Jul 13, 2010 #3
    But the sum of the sin(kx) would still be bounded, right?

    If you mean the sum of sin(kx), I found that it was less than 2pi/x in this way:

    Consider the interval [0,2pi] with x in this interval. Then there are at most 2pi/x terms, with some of them being < 0. Since sin(y) <= 1 for all y, we have that the sum of the |sin(kx)| < 2pi/x.

    Since the partial sums of sin(kx) is bounded and 1/ln(k) is a nonincreasing nullsequence, would this mean that the the series converges by Dirichlet's Test?
     
    Last edited: Jul 13, 2010
  5. Jul 13, 2010 #4

    Hurkyl

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    The sum you wrote is over all k, not just those k for which [itex]0 \leq kx \leq 2\pi[/itex].



    If x is not a rational multiple of 2pi, I'm not sure if I expect the sequence of partial sums of sin(kx) to be bounded or not. But if that sequence is bounded, then your argument is valid.
     
  6. Jul 14, 2010 #5
    Take a look at Example 1, under corollary 6, in this PDF: http://people.oregonstate.edu/~peterseb/mth311/docs/311abel.pdf [Broken]

    And yes you're right, [tex] \sum sin(kx) [/tex] is bounded.
     
    Last edited by a moderator: May 4, 2017
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