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Convergence of Series Sinx/x

  1. Jul 31, 2013 #1
    1. The problem statement, all variables and given/known data
    Determine whether the series Ʃ(1 to infinity) sinx / x converges or diverges.


    2. Relevant equations
    This question appears in the integral test section, but as far as i know the integral test can only be used for decreasing functions, right?


    3. The attempt at a solution
    Using the ratio test, limx->infinity sin(x+1)/(x+1)*(sinx/x)=limx->infinity x*sin(x+1)/((x+1)(sinx))
    This is where i got stuck-this limit oscillates between positive infinity and negative infinity.
    Using the root test, i need to find the limit of (sinx)^(1/x) as x approaches infinity, which also gets me nowhere.

    we have not done taylor series yet so i'm sure there is a relatively simple approach to this question...please help?
     
  2. jcsd
  3. Jul 31, 2013 #2

    Dick

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    The answer to this is actually not very simple. You need something like http://en.wikipedia.org/wiki/Dirichlet's_test Have you covered anything like that? It's a generalization of the alternating series test. None of the elementary tests will work. As you've figured out.
     
  4. Jul 31, 2013 #3
    Thank you for your reply. The text we are using only covers the elementary tests, and the textbook solved this question via the integral test, which leads me to believe that the book has made a mistake as this is not a monotonic decreasing function.
     
  5. Aug 1, 2013 #4

    Dick

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    I'd agree with that. You can't really even integrate it in any elementary sense anyway, the integral is Si(x), the sine integral, which doesn't mean much beyond being a formal symbol for the integral. What actually does the textbook say? BTW the Dirichlet test will tell it does converge. Needs a bit of work to show that though.
     
    Last edited: Aug 1, 2013
  6. Aug 1, 2013 #5

    marcusl

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    FYI, it's possible to show that the envelope of sinx/x is monotonically decreasing--it decays as 1/x. I will grant, however, that this is not very helpful in solving your problem.
     
  7. Aug 1, 2013 #6
    Ok, so I have read about the dirichlet's test. so I can prove that sinx/x converges because 1/x is decreasing and approaching zero, while Ʃ (1 to N) sinx ≤ N for any positive integer N since sinx is bounded above by 1, right?
     
  8. Aug 1, 2013 #7

    Dick

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    Not quite. You need to prove that the sine sum has a bound that's independent of N. You can actually derive a formula for the sum and show it's bounded.
     
  9. Aug 1, 2013 #8
    I surfed the internet and the only result that does not involve complex exponents is this:

    By the sum and difference identities for Cosine, we obtain 2*sin(x)*sin(0.5)=cos(x-0.5)-cos(x+0.5)

    When we sum x from 1 to infinity, we get 2*sin(x)*sin(0.5)=cos(0.5)-cos(x+0.5) because there
    are mass cancellations on the right side of the equation.

    Thus, Ʃ 1 to infinity sin(x) = lim x→infinity (cos(0.5)-cos(x+0.5))/2*sin(0.5), which is bounded above and below because the cosine function is bounded above by 1 and below by -1.

    Would this be considered a proof that sinx/x converges?
     
  10. Aug 1, 2013 #9

    lurflurf

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    first we have
    $$\int_0^\infty \! \sin(x)/x \, \mathrm{dx}=\pi/2$$
    but the integral test does not require that we know the integral, only that it converges to a finite value.
    The monotone issue can be handled by transforming the function

    Here is what we said last time someone asked this
    I suggested
    $$\sum_{n=1}^N \frac{\sin(n \, a)}{n}=-\frac{\cos(N \, a+a/2)}{2(N+1) \sin(a/2)}+\frac{\cot(a/2)}{2}-\sum_{n=1}^N \frac{\cos(n \, a+a/2)}{2n(n+1) \sin(a/2)} $$
    The second sum being clearly convergent
    or
    use
    http://en.wikipedia.org/wiki/Abel's_summation_formula
     
  11. Aug 1, 2013 #10

    jbunniii

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    I'll mention as a side note that if you know a few facts about Fourier transforms, not only can we easily show that ##\sum_{n=0}^{\infty} \sin(n)/n## converges, but we can actually find the value of the sum.

    First, recognize that
    $$f(x) = \begin{cases}\pi & \text{if }-\frac{1}{2\pi} < x < \frac{1}{2\pi} \\
    0 & \text{otherwise}\end{cases}$$
    and
    $$\hat{f}(y) = \frac{\sin(y )}{y}$$
    are a Fourier transform pair, then we can use the Poisson summation formula:
    $$\sum_{n = -\infty}^{\infty} f(n) = \sum_{k=-\infty}^{\infty}\hat{f}(k)$$
    I'll omit the details because this is probably beyond the scope of the OP's assignment, but once you learn a bit of Fourier analysis, this clearly becomes the "right" way to solve the problem.
     
    Last edited: Aug 1, 2013
  12. Aug 1, 2013 #11

    mfb

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    The integral test needs a monotone decreasing function. f(x)=sin(x)/x does not satisfy that.
     
  13. Aug 1, 2013 #12

    Zondrina

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    Indeed I believe the second sum should be from ##-∞## to ##∞##. I'd prefer this over Dirichlet's test.
     
  14. Aug 1, 2013 #13

    jbunniii

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    Yes, thanks. I'll fix it now.
     
  15. Aug 1, 2013 #14
    No way to "force-fit" an "integral-like" test on this series right?

    Suppose that was the exercise: force-fit an integral-like test to the series [itex]\sum_{n=1}^{\infty}\frac{\sin(n)}{n}[/itex].

    How about I try?

    Consider:

    [tex] \frac{u(x)}{x}[/tex]

    where [itex]u(x)[/itex] is a uniformly-periodic function with [itex]lim_{x\to 0} \frac{u(x)}{x}=k[/itex] and [itex]u(x)=u(x+a)[/itex] with [itex]u(b)=-u(b+(ka+a/2))[/itex] (like the sine function)

    then if:

    [tex]\int_1^{\infty}\frac{u(x)}{x}dx[/tex]

    converges then the sum:

    [tex]\sum_1^{\infty} \frac{u(n)}{n}[/tex]

    converges as well.

    You guys telling me I can't do that? Granted, I need to spruce it up, prove it for one, but it looks reasonable to try.
     
    Last edited: Aug 1, 2013
  16. Aug 1, 2013 #15

    mfb

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    No, you need stronger constraints on u(x). Consider u(x)=cos(2 pi x)...

    Personal guess: Assume u is continuous and periodic with an irrational period. I would expect that this is sufficient to let ##\sum \frac{u(n)}{n} converge. If you drop one of those two conditions, I can find a counterexample.
     
  17. Aug 1, 2013 #16
    But [itex]\lim_{x\to 0} \frac{\cos(2\pi x)}{x}=\infty[/itex] so is not sine-like.


    I have doubts about that: suppose it's periodic but not uniformly so. That is, suppose in one period, the function is primarily positive and only negative for a short interval. I'd rather stipulate the function is uniformly periodic: just as much positive area as negative area in one period.

    Can anyone show me a counter example where the sum using such a function would diverge?

    Also, I like the irrational period: this introduces essentially a random distribution of positive and negative values in the limit as n goes to infinity (I think).
     
  18. Aug 1, 2013 #17

    jbunniii

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    I assume you want a continuous counterexample. Otherwise, I think this works:
    $$u(x) = \begin{cases}
    \sin(\pi x/2) & \text{ if } 0 \leq x \leq 1 \\
    -\sin(\pi x/2) & \text{ if } 1 < x \leq 2 \\
    \end{cases}$$
    Now extend ##u## periodically outside the interval ##[0,2]##.

    Then we have ##u(n) = 0## if ##n## is even, and ##u(n) = 1## if ##n## is odd, and ##u(x)/x \to 1## as ##x \to 0##. So ##\sum_{n=0}^{\infty} u(n)/n = 1 + 1 + \frac{1}{3} + \frac{1}{5} + \cdots = \infty##, but (I think) ##\int_{0}^{\infty} u(x)/x## converges.

    [edit] Fixed some typos
     
  19. Aug 1, 2013 #18
    Hi hbunnii,

    But that example has an integer period. I like the irrational period requirement. I think maybe this may turn out to be essential for the proof.
     
  20. Aug 1, 2013 #19

    jbunniii

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    I think I also have a continuous counterexample, but I'm too lazy to write an explicit piecewise formula.

    Suppose we start with the positive part of a single period of sine:
    $$s(x) = \begin{cases}
    \sin(x) & \text{ if }0 \leq x \leq \pi \\
    0 & \text{ otherwise }\\
    \end{cases}$$
    Now, first scale ##s## so it has width 1 and height 1, and place it so it is centered at ##x = 1##. So the support of this part is ##[\frac{1}{2}, \frac{3}{2}]##.

    Next scale ##s## so it has width ##\frac{1}{2}## and height ##-h##, and place one copy of this at ##[0, \frac{1}{2}]## and another at ##[\frac{3}{2}, 2]##. Thus we have covered all of ##[0,2]##. We choose ##h## so that ##\int_{0}^{2} u(x) dx = 0##. (I'll attach a sketch in a minute if I can work out how to do it.)

    Then ##u## is continuous, and it has the property that ##u(n) = 0## for even ##n##, ##u(n) = 1## for odd ##n##, and ##u(x)/x \to 1## as ##x \to 0##, just as in the previous example. So again we have ##\sum_{n=0}^{\infty}u(n)/n = 1 + 1 + \frac{1}{3} + \frac{1}{5} + \cdots = \infty##, but ##\int_{0}^{2}u(x)/x## converges.

    [edit] attached picture, if I did it correctly
     

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    Last edited: Aug 1, 2013
  21. Aug 1, 2013 #20

    jbunniii

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    Yes, I suspect it is essential. Proving that it is sufficient is another story...
     
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