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Math Help - Convergence/Divergence of a series

  1. #1
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    Convergence/Divergence of a series

    How would I determine and demonstrate whether or not the following series converges or diverges for a selected (arbitrary) real value for x:
    \sum_{n=0}^{\infty}\cos(2\pi\cdot(2^{n}x))?
    I've found clearly that it diverges for any x of the form x=\frac{m}{2^k} with m and k being integers. I've also found that it diverges for x=\frac{m}{3\cdot2^k}. However, I don't know if it diverge for all x (as I suspect), or if there are real values of x for which the series converges. How do I find out conclusively?

    --Kevin C.
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    Quote Originally Posted by TwistedOne151 View Post
    How would I determine and demonstrate whether or not the following series converges or diverges for a selected (arbitrary) real value for x:
    \sum_{n=0}^{\infty}\cos(2\pi\cdot(2^{n}x))?
    I've found clearly that it diverges for any x of the form x=\frac{m}{2^k} with m and k being integers. I've also found that it diverges for x=\frac{m}{3\cdot2^k}. However, I don't know if it diverge for all x (as I suspect), or if there are real values of x for which the series converges. How do I find out conclusively?

    --Kevin C.
    Off the top of my head, perhaps a proof along the following lines:

    Using the double angle formula:

    \cos(2\pi\cdot(2^{n}x) = \cos(2[\pi\cdot(2^{n}x)]) = 1 - 2 \sin^2(\pi\cdot(2^{n}x)).

    The sum of 1's certainly diverges. If the sum of sin^2 's converges then overall you've got divergence ..... On the other hand, if twice the sum of sin^2 's diverges, it'll diverge much more slowly than the sum of 1's and so you've got divergence again ....

    Sloppy I know (and the 'other hand' claim might prove to be bogus) - over to you (or others) to dot the i's and cross the t's (and put in the missing words and supply the punctuation)
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    Quote Originally Posted by TwistedOne151 View Post
    How would I determine and demonstrate whether or not the following series converges or diverges for a selected (arbitrary) real value for x:
    \sum_{n=0}^{\infty}\cos(2\pi\cdot(2^{n}x))?
    I've found clearly that it diverges for any x of the form x=\frac{m}{2^k} with m and k being integers. I've also found that it diverges for x=\frac{m}{3\cdot2^k}. However, I don't know if it diverge for all x (as I suspect), or if there are real values of x for which the series converges. How do I find out conclusively?

    --Kevin C.
    Or, of course, an even simpler proof that applies the limit test (also known as the nth term test) .......

    Let a_n = cos(2\pi\cdot(2^{n}x)). Since the sequence \{a_n\} doesn't converge to zero, the series is divergent.
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