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Convergence test question

  1. Nov 17, 2012 #1
    Hi,
    How may I know whether the series ((-1)^n)[cos (3^n)x]^3/(3^n) converges/diverges?Should I use the Leibniz Criterion?
    It is stated that (cos a)^3 = (1/4)(3cos a + cos 3a)
     
  2. jcsd
  3. Nov 17, 2012 #2

    Mark44

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    The series is not alternating, despite the factor of (-1)n, so I don't see that the Leibniz Criterion applies. I also don't see that the hint is helpful.

    For any n, the terms in your series are bounded below by -1/(3n) and bounded above by 1/(3n), both of which are terms in geometric series.
     
  4. Nov 17, 2012 #3

    SammyS

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    Is that really an x in the series?
    [itex]\displaystyle (-1)^n\frac{\cos^3( (3^n)x)}{3^n}[/itex]​

    If so, the divergence/convergence certainly may depend upon x.
     
  5. Nov 17, 2012 #4
    That is indeed x in the formula.
    Mark44, I am still not sure how to go about it. May you please assist and/or elaborate further?
     
  6. Nov 17, 2012 #5

    haruspex

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    The hint is very useful. You have a cos3() in the expression for the series, and the hint gives you a substitution for that. Try it.
     
  7. Nov 17, 2012 #6

    SammyS

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    As Mark said, the series is not an alternating series.

    In fact it is alternating for a few specific values of x, namely when x is an integer multiple of π .

    It can also be a constant series with 0 as the constant if x is an odd half-integer multiple of π .

    ... But to help you carry on with the problem:

    What is the range of cos(x), and cos3(x) ?

    Use the squeeze theorem.
     
  8. Nov 21, 2012 #7
    Why would the sum of this series be equal to 0.75cos(x)? I managed to express this series as a telescopic summation and thus merely (1/4)[-cos3x - ((1/3)^n)cos(3^(n+1)x)] remain in the sum. Won't ((1/3)^n)cos(3^(n+1)x) be equal to zero as n->infinity? Hence, why won't the sum be -0.25cos3x?
     
  9. Nov 21, 2012 #8

    haruspex

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    Does the sum start at n=0 or n=1? That would account for the different results.
     
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