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Math Help - Test the series for convergence.

  1. #1
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    Test the series for convergence.

    Here is it:
    \sum\limits_{n=1}^\infty \frac{ \sqrt{n+1} - \sqrt{n} }{n}.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by TWiX View Post
    Here is it:
    \sum\limits_{n=1}^\infty \frac{ \sqrt{n+1} - \sqrt{n} }{n}.
    Use the LCT with the convergent \frac{1}{n^{3/2}}
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  3. #3
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    Your series will coverge if \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_{n}}\right|<1
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  4. #4
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    Quote Originally Posted by VonNemo19 View Post
    Use the LCT with the convergent \frac{1}{n^{3/2}}
    Failed.
    The limit =\infty
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  5. #5
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    Quote Originally Posted by pickslides View Post
    Your series will coverge if \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_{n}}\right|<1
    I didnt try it.
    But its algebric function.
    I think it will be 1 ---> Failed.
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by pickslides View Post
    Your series will coverge if \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_{n}}\right|<1
    That's hard to show, no?
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  7. #7
    Math Engineering Student
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    \frac{\sqrt{n+1}-\sqrt{n}}{n}=\frac{1}{n\left( \sqrt{n+1}+\sqrt{n} \right)}<\frac{1}{2n^{\frac{3}{2}}}.
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  8. #8
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    In general not that bad, but this question is a bit sticky yep
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