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Math Help - alternating series #3

  1. #1
    Senior Member mollymcf2009's Avatar
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    alternating series #3

    One more that I am hung up on...

    <br />
\sum^{\infty}_{n=2}  (-\frac{n}{8})^n
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  2. #2
    MHF Contributor matheagle's Avatar
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    The root test is best here
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  3. #3
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by matheagle View Post
    The root test is best here
    So I would have

    - \sqrt[n]{(\frac{n}{8})^n}

    = -\frac{n}{8} = - \infty ????
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  4. #4
    MHF Contributor matheagle's Avatar
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    you need abs values here, just like the ratio test
    BUT inside the root. You don't want the root of a negative number.
    Whenever the entire term is to the n ^{th} power it's easy to use the root test.
    Last edited by matheagle; March 29th 2009 at 06:47 PM.
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  5. #5
    MHF Contributor matheagle's Avatar
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    you place the abs values here....

    Quote Originally Posted by mollymcf2009 View Post
    So I would have

    \biggl|\biggl({-n\over 8}\biggr)^n\biggr|^{1/n}=\frac{n}{8}\to\infty>1
    Basically it shows you, like in the ratio test, that the terms (in absolute value) are getting bigger.
    Hence they cannot be approaching zero.
    Last edited by matheagle; March 29th 2009 at 07:36 PM.
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