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Math Help - Could someone help me with this partial limit of a series

  1. #1
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    Could someone help me with this partial limit of a series

    What are the partial limits of that series ?


    tried for hours, couldn't figure this out. Thanks in advance.
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  2. #2
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    Partial sums, you mean?

    \sum \Big(6(-1)^{n(n+1)/2} + 8 \cos ({n\pi\over 2})\Big) = 6\sum(-1)^{n(n+1)/2} + 8 \sum \cos ({n\pi\over 2}). Suppose n is 0 or 3 mod 4. Then n(n+1)/2 is even. Otherwise, n(n+1)/2 is odd. This should tell you how to handle the first sum. Now suppose n is 0 mod 4. Then \cos ({n\pi\over 2}) is 1. If n is 2 mod 4, then it is -1. Otherwise, it is 0. This will let you handle the second sum.
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  3. #3
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    Quote Originally Posted by azarue View Post
    What are the partial limits of that series ?


    tried for hours, couldn't figure this out. Thanks in advance.

    Note that \cos\left(\frac{n\pi}{2}\right)=\left\{\begin{arra  y}{cl}0&,\,if\,\, n=1,3\!\!\!\pmod 4\\\!\!\!-1&,\,if\,\,n=2\!\!\!\pmod 4\\1&,\,if\,\,n=0\!\!\!\pmod 4\end{array}\right. , so :

    n=0\!\!\!\pmod 4\Longrightarrow a_n=14\,,\,\,n=1\!\!\!\pmod 4\Longrightarrow a_n=-6 , n=2\!\!\!\pmod 4\Longrightarrow a_n=-14\,,\,\,n=3\!\!\!\pmod 4\Longrightarrow a_n=6 .

    Check the above closely and proceed.

    Tonio
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