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    sequence...

    determine whether sequence is convergent or divergent

    an = (square root n) - square root(n^2-1)
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    Quote Originally Posted by twilightstr View Post
    determine whether sequence is convergent or divergent

    an = (square root n) - square root(n^2-1)
    Note that \sqrt{n} - \sqrt{n^2 - 1} = \frac{(\sqrt{n} - \sqrt{n^2 - 1}) \cdot (\sqrt{n} + \sqrt{n^2 - 1})}{\sqrt{n} + \sqrt{n^2 - 1}} = \frac{n - n^2 + 1}{\sqrt{n} + \sqrt{n^2 - 1}} = \frac{1 - n + \frac{1}{n}}{\frac{1}{\sqrt{n}} + \sqrt{1 - \frac{1}{n^2}}}
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