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Thread: Alternating Series test

  1. October 23rd 2012, 06:42 PM #1
    higgins19
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    Alternating Series test

    sin[(2n-1)pi]/2
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  2. October 23rd 2012, 07:01 PM #2
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    Re: Alternating Series test

    Do you really need to use the alternating series test here? You should know that \displaystyle \begin{align*} \sin{\left[ \frac{\left( 2n - 1 \right) \pi}{2} \right]} = (-1)^{n + 1} \end{align*}, and so adding them, the series can't possibly converge because it can only ever take the values of 1 and 0.
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