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Math Help - Finding the nth partial sum and taking the limit?

  1. #1
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    Post Finding the nth partial sum and taking the limit?

    I need to find the sum of each series by computing their n-th partial sum and then taking the limit as n -> infinity.

    1) Σ (infinity being on top, n=0 on bottom) 5/2^n - 4/3^n

    I already took the limit by separating the two, and ending up with:

    5/(1-(1/2)) - 4(1-(1/3)) = 4

    I just need the n-th partial sum.

    2) Σ (infinity being on top, n=1 on bottom) 1/(n^2-(1/4))

    I really don't know how to do this one.

    Thanks!
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    First hint : \sum_{j=0}^nx^j = \frac{1-x^{n+1}}{1-x}

    Second hint : \frac{1}{n^2-1/4}=\frac{1}{n-1/2}-\frac{1}{n+1/2}
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by lindz View Post
    I need to find the sum of each series by computing their n-th partial sum and then taking the limit as n -> infinity.

    1) Σ (infinity being on top, n=0 on bottom) 5/2^n - 4/3^n

    I already took the limit by separating the two, and ending up with:

    5/(1-(1/2)) - 4(1-(1/3)) = 4

    I just need the n-th partial sum.

    2) Σ (infinity being on top, n=1 on bottom) 1/(n^2-(1/4))

    I really don't know how to do this one.

    Thanks!
    \sum_{\ell=0}^n\omega^\ell=\frac{1-\omega^{n+1}}{1-\omega}. Does that help?
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