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Math Help - Check my Limit

  1. #1
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    Check my Limit

    Please tell me if this is correct:

    Find the limit of (E k=1 to infinity) of 1/ (4n^2 -1)

    I believe this is 0.
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  2. #2
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    Re: Check my Limit

    Quote Originally Posted by veronicak5678 View Post
    Please tell me if this is correct:

    Find the limit of (E k=1 to infinity) of 1/ (4n^2 -1)

    I believe this is 0.
    Hi veronicak5678,

    Do you mean, \sum_{k=1}^{\infty}\frac{1}{4n^2-1} ?
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  3. #3
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    Re: Check my Limit

    Yes, that's what I mean. Sorry it's not clear.
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  4. #4
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    Re: Check my Limit

    \displaystyle \frac{1}{4n^2 - 1} = \frac{1}{(2n - 1)(2n+1)} = \frac{1}{2}\left(\frac{1}{2n - 1} - \frac{1}{2n + 1}\right), so

    \displaystyle \begin{align*} \sum_{n = 1}^{\infty}\frac{1}{4n^2 - 1} &= \frac{1}{2}\lim_{k \to \infty}\sum_{n = 1}^{k}\left(\frac{1}{2n-1} - \frac{1}{2n + 1}\right) \\ &= \frac{1}{2}\lim_{k \to \infty}\left[\left(1 - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{7}\right) + \dots + \left(\frac{1}{2k - 1} - \frac{1}{2k + 1}\right)\right] \\ &= \frac{1}{2}\lim_{k \to \infty}\left(1 - \frac{1}{2k + 1}\right) \\ &= \frac{1}{2}\left(1 - 0\right) \\ &= \frac{1}{2} \end{align*}
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  5. #5
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    Re: Check my Limit

    Got it. Thanks very much.
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