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Math Help - Series convergent or divergent by Comparison Test.

  1. #1
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    Series convergent or divergent by Comparison Test.

    Please deternine whether the series is convergent or divergent by Comparison Test.
    Series sign (2^(n+1))/(3^(n+1))
    My resoning: bn=2^n/3^n= (2/3)^n, r=2/3 less than 1, geometric series converges. It is true that an is less than bn, it means that the initial series converges.

    If this is wrong, pleaseexplain how to do it by Comparison Test.
    Thank you very much
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  2. #2
    MHF Contributor matheagle's Avatar
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    There's no need to compare this to anything.
    This is a geometric with r=2/3. This is just \biggl({2\over 3}\biggr)^{n+1}.
    You can do the ratio test and get 2/3, but that's basically a limit comparison test to the geo's.
    You can find what the sum converges to, but you need to tell me where n starts.
    Last edited by matheagle; March 13th 2009 at 10:19 PM.
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  3. #3
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    n starts at 1

    I do not have a choice in the method. The assignment says I must do it using Comparison Test. n=1,
    Is the answer that the series converges correct?
    Thank you
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  4. #4
    MHF Contributor matheagle's Avatar
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    Sure compare it to itself...

    \biggl({2\over 3}\biggr)^{n+1}= \biggl({2\over 3}\biggr)\biggl({2\over 3}\biggr)^n.

    So

    \sum_{n=1}^{\infty}\biggl({2\over 3}\biggr)^{n+1}= \biggl({2\over 3}\biggr)\sum_{n=1}^{\infty}\biggl({2\over 3}\biggr)^n<\sum_{n=1}^{\infty}\biggl({2\over 3}\biggr)^n<\infty.

    But the answer is...

    \biggl({2\over 3}\biggr)\biggl({2/3\over 1-2/3}\biggr)={4/9\over 1/3}={4\over 3}.
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