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Math Help - covergense questuin mmn11 2b

  1. #1
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    covergense questuin mmn11 2b

    \sum_{k=2}^{infinity}(-1)^{k}\frac{(2k+3)^{k+1}}{(3k-4)^{k-1}}
    is it converges absolutly ,or with condition(dont know the english term),
    or is it diverges?


    from nth root method of cauchy
    the limit of (2k+3)(3k-4) is infinity

    ??
    Last edited by transgalactic; March 2nd 2011 at 01:10 AM.
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  2. #2
    Ted
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    \displaystyle \lim_{k\to \infty} \frac{2k+3}{3k-4} = \frac{2}{3}

    So?
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  3. #3
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    Quote Originally Posted by transgalactic View Post
    \sum_{k=2}^{infinity}(-1)^{k}\frac{(2k+3)^{k+1}}{(3k-4)^{k-1}}
    is it converges absolutly ,or with condition(dont know the english term),
    The term is, in fact, "conditionally convergent".

    or is it diverges?


    from nth root method of cauchy
    the limit of (2k+3)(3k-4) is infinity

    ??
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  4. #4
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    no they have different powers
    numerator will be 1
    denominator is -1
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  5. #5
    Ted
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    I did not say anything about your series.
    I've just answered the question which you edited it.

    In order to test the series for absolutely/condintionally convergent, you should check the absolute series.

    Did try anything for \displaystyle \sum_{k=2}^{\infty} \frac{(2k+3)^{k+1}}{(3k-4)^{k-1}} ?
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  6. #6
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    first we need to test the absolute value series
    the n'th root doesnt work because i get a limit of infinity

    what you did what takng the expression witout the powers on numerator and denominator
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