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Math Help - Sequence

  1. #1
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    Sequence

    How do I do this problem?

    Is the sequence An = sqrt(n^2+n+1) -sqrt(n^2-n+1) convergent? Find it's limit.
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  2. #2
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    Hello, Nichelle14!

    Is the sequence: A_n \:= \:\sqrt{n^2+n+1} - \sqrt{n^2-n+1} convergent? .Find its limit.

    Multiply top and bottom by the conjugate:

    . . \frac{\sqrt{n^2+n+1} - \sqrt{n^2-n+1}}{1}\cdot\frac{\sqrt{n^2+n+1} + \sqrt{n^2 - n+1}}{\sqrt{n^2+n+1} + \sqrt{n^2-n+1}}

    . . = \;\frac{(n^2+n+1) - (n^2-n+1)}{\sqrt{n^2+n+1} + \sqrt{n^2-n+1}} \;= \;\frac{2n}{\sqrt{n^2+n+1} + \sqrt{n^2-n+1}}


    Divide the top by n; divide the bottom by \sqrt{n^2} = n

    . . \frac{\frac{2n}{n}}{\frac{\sqrt{n^2+n+1}}{\sqrt{n^  2}} + \frac{\sqrt{n^2-n+1}}{\sqrt{n^2}}} \;=\;\frac{2}{\sqrt{\frac{n^2}{n^2} + \frac{n}{n^2} + \frac{1}{n^2}} + \sqrt{\frac{n^2}{n^2} - \frac{n}{n^2} + \frac{1}{n^2}}}

    . . =\;\frac{2}{\sqrt{1 + \frac{1}{n} + \frac{1}{n^2}} + \sqrt{1 - \frac{1}{n} + \frac{1}{n^2}}}


    Then: . \lim_{n\to\infty}\left[\frac{2}{\sqrt{1 + \frac{1}{n} + \frac{1}{n^2}} + \sqrt{1 - \frac{1}{n} + \frac{1}{n^2}}}\right]\;= \frac{2}{\sqrt{1+0+0} + \sqrt{1 + 0 + 0}}\;=\;\frac{2}{1+1}


    Therefore: . \boxed{\lim_{n\to\infty}\left(\sqrt{n^2+n+1} - \sqrt{n^2-n+1}\right)\;=\;1\,}

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