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Math Help - Divergence by comparison

  1. #1
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    Divergence by comparison

    Can someone help me prove that the following integral diverges by comparison I'm kind of stuck

    <br />
\int_{0}^{\infty }{\frac{dx}{\sqrt{1+x^{2}}}}<br />


    Thanks
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  2. #2
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    Quote Originally Posted by jarny View Post
    Can someone help me prove that the following integral diverges by comparison I'm kind of stuck

    <br />
\int_{0}^{\infty }{\frac{dx}{\sqrt{1+x^{2}}}}<br />


    Thanks
    Hi

    \frac{1}{\sqrt{1+x^2}} is equivalent to \frac{1}{x} in +\infty
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  3. #3
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    That wont work. I know that but that integral is mathematically greater than my integral and a comparison won't work
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  4. #4
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by running-gag View Post
    Hi

    \frac{1}{\sqrt{1+x^2}} is equivalent to \frac{1}{x} in +\infty

    Let  a_{n}= \frac{1}{\sqrt{1+n^2}} and let  b_{n}=\frac{1}{\sqrt{n^2}} = \frac{1}{n}


    The Limit Comparison test says that if the ratio  \frac{a_{n}}{b_{n}} tends to a finite positive number (such as 1?), then the two series behave the same way. The Harmonic diverges, so your series must diverge as well.

    I hope this helps.
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  5. #5
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    Not really I need something in the form

    1+x^2 < _____

    so

    1/sqrt(1+x^2) > __________


    and S __________ diverges so by comparison S1/sqrt(1+x^2) must diverge
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  6. #6
    Senior Member apcalculus's Avatar
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    Hmm... I understand what you are asking, but I can't think of a candidate for the Comparison Test to be applied because the Harmonic Series is 'above' the given series, so comparing term by term does not help conclude that what you have diverges. The Limit Comparison Test, however, helps answer this question as I have done above.
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  7. #7
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    We know this x \geqslant 1 \Rightarrow \quad \frac{1}<br />
{{\sqrt {1 + x^2 } }} \geqslant \frac{1}<br />
{{\sqrt 2 x}} \Rightarrow \quad \int_1^\infty  {\frac{{dx}}<br />
{{\sqrt {1 + x^2 } }} \geqslant \int_1^\infty  {\frac{{dx}}<br />
{{\sqrt 2 x}}} }

    We also know that \int_0^1 {\frac{{dx}}{{\sqrt {1 + x^2 } }}} is finite.
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  8. #8
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    Plato 1/ sqrt(2x) is greater than 1/sqrt( 1+ x^2)

    I think you screwed up
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  9. #9
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    Actually I think I got it
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  10. #10
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    Quote Originally Posted by jarny View Post
    Plato 1/ sqrt(2x) is greater than 1/sqrt( 1+ x^2)
    I think you screwed up
    It is \frac{1}{{\left( {\sqrt 2 } \right)x}}.
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