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Convergent series

  1. Mar 12, 2015 #1
    I was goofing around with Mathematica and found that Sum_(k>=1)(sin(k)/k)=Sum_(k>=1)(sin(k)/k)^2. In other words a convergent series such that if you square each of its terms the sum is the same. Question is: is this a unique property or are there other convergent series with the property? Cheers.
     
  2. jcsd
  3. Mar 12, 2015 #2

    mfb

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    Are you sure they are exactly the same?

    It is easy to construct series with this property, if you ask for the partial sum to be the same every two steps for example. This boils down to solutions for a^2+b^2 = a+b with the additional constraint that the sequence made out of those pairs should converge to zero. On this circle, your (a,b) points have to converge to (0,0).
    That is just a very small subset of all series with this property.
     
  4. Mar 12, 2015 #3
    Yes thank you that is so. And as for those sums, yes, they are equal. I'm sort of amazed at it, but wondering if I should be.
     
  5. Mar 13, 2015 #4

    jbunniii

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    They are exactly the same. One way to see this is to observe that ##\sin(x)/x## is the Fourier transform of a rectangle function with height equal to ##\pi## and support equal to ##[-1/(2\pi),1/(2\pi)]##, and ##\sin^2(x)/x^2## is the Fourier transform of a triangle function (the convolution of the above rectangle with itself). The Poisson summation formula, assuming we can justify its application, gives us
    $$\sum_{k=-\infty}^{\infty} \frac{\sin(k)}{k} = \sum_{k=-\infty}^{\infty}r(k)$$
    and
    $$\sum_{k=-\infty}^{\infty} \frac{\sin^2(k)}{k^2} = \sum_{k=-\infty}^{\infty}t(k)$$
    where ##r## and ##t## are the rectangle and triangle functions described above. The right hand side in both cases is equal to ##\pi##, so
    $$\sum_{k=-\infty}^{\infty} \frac{\sin(k)}{k} = \sum_{k=-\infty}^{\infty} \frac{\sin^2(k)}{k^2} = \pi$$
    Then, since both summands are symmetric around ##k=0## and both are equal to ##1## at ##k=0##, we conclude that
    $$\sum_{k=1}^{\infty} \frac{\sin(k)}{k} = \sum_{k=1}^{\infty} \frac{\sin^2(k)}{k^2} = \frac{\pi - 1}{2}$$
     
    Last edited: Mar 13, 2015
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