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Math Help - alternating series test

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

    \sum^{\infty}_{n=3} \frac{cos(n\pi)}{n^{\frac{3}{5}}}

    What would I compare this to? \frac{cos(n)}{n} ?

    Also, when n=3 how does this change things?

    Thanks!
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  2. #2
    o_O
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    Notice that: \cos (n\pi) = (-1)^{n}

    So really, your series is just: \sum_{n=3}^{\infty} \frac{(-1)^n}{n^{\frac{3}{5}}}

    And there really isn't anything special about starting from n = 3.
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  3. #3
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by o_O View Post
    Notice that: \cos (n\pi) = (-1)^{n}

    So really, your series is just: \sum_{n=3}^{\infty} \frac{(-1)^n}{n^{\frac{3}{5}}}

    And there really isn't anything special about starting from n = 3.

    Ok, so I would say that:

    \sum a_n > 0 so it converges and since 0 < \sum a_{n+1} < \sum a_n, then \sum a_{n+1} also converges?
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