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Math Help - Comparison theorem convergence help.

  1. #1
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    Comparison theorem convergence help.

    Use comparison theorem to see if the following converges.

    integral from 1 to infinity:

    5[((1+(x)^1/2)^1/2)/(x)^1/2]

    or in words

    5 [root(1+root x)]/root x

    This is what I did.
    comparison theorem states that for f(x) > g(x) for all x > a, if f(x) converges, so does g(x).

    I set f(x) as 5 root (1+x)

    is that right?
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  2. #2
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    Quote Originally Posted by Kuma View Post
    Use comparison theorem to see if the following converges.

    integral from 1 to infinity:

    5[((1+(x)^1/2)^1/2)/(x)^1/2]

    or in words

    5 [root(1+root x)]/root x

    This is what I did.
    comparison theorem states that for f(x) > g(x) for all x > a, if f(x) converges, so does g(x).

    I set f(x) as 5 root (1+x)

    is that right?
    the constant 5 is inconsequential ... pull it out of the integrand.

    note for x \ge 1 ...

    \dfrac{\sqrt{1+\sqrt{x}}}{\sqrt{x}} \le \dfrac{\sqrt{\sqrt{x} + \sqrt{x}}}{\sqrt{x}} = \dfrac{\sqrt{2}}{\sqrt[4]{x}}
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by skeeter View Post
    the constant 5 is inconsequential ... pull it out of the integrand.

    note for x \ge 1 ...

    \dfrac{\sqrt{1+\sqrt{x}}}{\sqrt{x}} \le \dfrac{\sqrt{\sqrt{x} + \sqrt{x}}}{\sqrt{x}} = \dfrac{\sqrt{2}}{\sqrt[4]{x}}
    You have demonstrated that the function to be integrated is less than a function the integral of which diverges... so that Your 'demonstration' demonstrates nothing... may be is more useful this inequality, valid for x>1...

    \displaystyle \frac{\sqrt{1+\sqrt{x}}}{\sqrt{x}} > \frac{1}{\sqrt{x}}

    Kind regards

    \chi \sigma
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  4. #4
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    chisigma is correct ... sorry for the error.
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