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Math Help - test for convergence of series

  1. #1
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    test for convergence of series

    sigma from 1 to infinite of 1/[(squareroot n+1) -nsquareootn]
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    Quote Originally Posted by twilightstr View Post
    sigma from 1 to infinite of 1/[(squareroot n+1) -nsquareootn]
    Do you mean \sum_{n=1}^\infty \frac{1}{\sqrt{n+1}- n\sqrt{n}}?

    An obvious first step would be to multiply numerator and denominator by \sqrt{n+1}+ n\sqrt{n} so that each term is \frac{\sqrt{n+1}+ n\sqrt{n}}{n+1- n^3}. Now a comparison test should do it.

    Or, now that I think about it, just a comparison to \frac{1}{n^{3/2}}!
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    no. 1/[(squareroot (n+1)) - (squareroot (n-1))]
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    Quote Originally Posted by twilightstr View Post
    no. 1/[(squareroot (n+1)) - (squareroot (n-1))]
    If you mean

     <br />
\sum_{n=1}^{\infty} \frac{1}{\sqrt{n+1}-\sqrt{n-1}}<br />

    then HallsofIvy's suggestion still works - multiply btop and bottom by  \sqrt{n+1}+\sqrt{n-1}

    giving

     <br />
\sum_{n=1}^{\infty} \frac{\sqrt{n+1}+\sqrt{n-1}}{2}<br />

    which diverges (terms don't go to zero)

    You could also do a limit comparison with \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}
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  5. #5
    Senior Member Pinkk's Avatar
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    \frac{1}{\sqrt{n+1}-\sqrt{n-1}} \ge \frac{1}{\sqrt{n+1}}

    Since the second is divergent (You should be able to figure that \sum_{n=1}^{\infty} \frac{1}{\sqrt{n+1}} is divergent). By the comparison test, \sum_{n=1}^{\infty}\frac{1}{\sqrt{n+1}-\sqrt{n-1}} must also diverge.
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