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Math Help - Complex series

  1. #1
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    Complex series

    I have this question, it asks to find the set of C numbers (z) where the series below converges.



    What I've done so far is found the series by trial and error then taken the absolute value and found that it converges for (x^2 +y^2)<2 so the set is a circle of radius 2, but I'm not sure if that is correct.

    I've tried doing the ratio test on the series but it got ugly, so I split it into two series and found (z/2)<1 and 2z<1 which is close to the mod of what I got before.

    Is this the right way to do it?

    Thanks, Daniel
    Last edited by Daniiel; August 31st 2011 at 08:42 PM.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Complex series

    Use the Hadamard's formula: the radius of convergence is \rho=(\lim\sup \sqrt [n]{|a_n|})^{-1} . In this case, is easy to find because we have two evident convergent subsequences.
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  3. #3
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    Re: Complex series

    Thanks Fernando,

    I get pretty much what i got before but inverted,

    Is this correct, the series can be written as 3 + sum (z/2)^2n - sum 2z^(2n-1)

    Then using ratio test on each seperately the one of the left gives |z|<2 and the right |z|<1/2 for the series to converge

    the left one is the same as what I got doing the ratio test on the orignial series but not the right

    is that right, a circle of radius 2?
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: Complex series

    Quote Originally Posted by Daniiel View Post
    is that right, a circle of radius 2?
    No, the sequence \sqrt [n]{|a_n|} (avoiding \sqrt [n]{|a_0|}) is (2,1/2,2,1/2,\ldots) . The set of all subsequence limits is A=\{2,1/2\} so, \lim\sup\sqrt [n]{|a_n|}=\sup A=2 and \rho=1/2 .
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  5. #5
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    Re: Complex series

    Sweet, Thanks a lot Fernando

    Hey just out of curiousity, because all the tests deal with |an|, if the question asked find z such that it is absolutely convergent, would the answer be exactly the same? Can you use the tests in the same way to solve for a z which makes the series absolutely convergent?
    Last edited by Daniiel; September 1st 2011 at 12:13 AM.
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Re: Complex series

    Quote Originally Posted by Daniiel View Post
    Hey just out of curiousity, because all the tests deal with |an|, if the question asked find z such that it is absolutely convergent, would the answer be exactly the same?
    Now we are dealing with power series, this means that only in the boundary of the convergence disk we can find a convergent series which is not absolutely convergent.
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  7. #7
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    Re: Complex series

    If I wanted to determine the sum of this series when it converges,

    Would i first write the series as a proper series just by trial and error

    so
    3 +  \sum_{n=1}^{00} (-1)^n (z)^n(2^n)^{{(-1)}^{(n+1)}}


    Or the series as two seperate sums Then add the two sums so 3 +s1 + s2?

    Would that be the right way to go?

    I get 3 + \frac{4}{4-z^2} + \frac{2z}{1-(2z)^2} -1

    -1 because i made both sums start from n=0 and when n=0 one of the sums =1
    Last edited by Daniiel; September 5th 2011 at 06:39 PM.
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