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  1. #1
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    Post Series

    I have only done simpler series before so this is new to me. Grateful for help, Thanks!

    Determine the value of M so that the given series converges.
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  2. #2
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    Quote Originally Posted by Duffman View Post
    I have only done simpler series before so this is new to me. Grateful for help, Thanks!

    Determine the value of M so that the given series converges.
    If you simplify

    \frac{n+1}{n-6} + \frac{n+1}{n+7} - 2 - \frac{M}{n} = \frac{42M-(M-85)n-(M-1)n^2}{n(n-6)(n+7)}

    LCT with \sum_{n=7}^{\infty} \frac{1}{n} will show it diverges for all values of M except M = 1 and if so, then

    \frac{n+1}{n-6} + \frac{n+1}{n+7} - 2 - \frac{M}{n} = \frac{42+84n}{n(n-6)(n+7)}

    LCT with \sum_{n=7}^{\infty} \frac{1}{n^2} will show it converges.
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  3. #3
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    By the divergence theorem the limit of the inside term needs to be zero in order to have convergence.
    Therefore,
    \lim \frac{42M-(M-85)n-(M-1)n^2}{n(n-6)(n+7)} = 0 \implies M = 1.
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