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Math Help - Comparison Test

  1. #1
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    Comparison Test

    \int_1^\infty \frac{1}{\sqrt{\theta+1}}d\theta

    My professor tells us to create an intuition and then create a proof.

    What I have so far is

    Let f(\theta) = \frac{1}{\sqrt{\theta+1}}d\theta
    When \theta \rightarrow \infty; f(\theta) acts like \frac{1}{\sqrt{\theta}}
    So I guess that f(\theta) diverges because p = \frac{1}{2}

    Now I need an easy, smaller function that I can prove that if that smaller function diverges, then f(\theta) diverges, am missing something here?
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  2. #2
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    Quote Originally Posted by Latszer View Post
    \int_1^\infty \frac{1}{\sqrt{\theta+1}}d\theta

    My professor tells us to create an intuition and then create a proof.

    What I have so far is

    Let f(\theta) = \frac{1}{\sqrt{\theta+1}}d\theta
    When \theta \rightarrow \infty; f(\theta) acts like \frac{1}{\sqrt{\theta}}
    So I guess that f(\theta) diverges because p = \frac{1}{2}

    Now I need an easy, smaller function that I can prove that if that smaller function diverges, then f(\theta) diverges, am missing something here?

    \int \dfrac{dx}{\sqrt{x+x}}= \dfrac{1}{\sqrt{2}}\int \dfrac{dx}{\sqrt{x}} < \int \dfrac{dx}{\sqrt{x+1}}

    I think now its correct
    Last edited by felper; February 16th 2010 at 07:11 PM.
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  3. #3
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     \frac{1}{x} is not less than \frac{1}{x+1} as x approaches infinity, so while your logic makes sense it technically does not "prove" the integral.
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  4. #4
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    Is it a sufficient proof if I just sub the integral like....

    \int_1^\infty\frac{d\theta}{\sqrt{\theta+1}} = \lim_{c\to\infty}\int_1^c \frac{d\theta}{\sqrt{\theta+1}}

    let u=\theta + 1 ; du = d\theta
    when  \theta = c \rightarrow u = c + 1 ; when  \theta = 1 \rightarrow u = 2
    Let d = c+1
    When  c \rightarrow\infty ,  d \rightarrow\infty

    \lim_{d\to\infty}\int_2^d\frac{du}{\sqrt{u}} = \lim_{d\to\infty}\int_1^\infty\frac{du}{\sqrt{u}}d  u - \int_1^2\frac{du}{\sqrt{u}}du

    And then since \lim_{d\to\infty}\int_1^d\frac{du}{\sqrt{u}} diverges, then \int_1^\infty\frac{d\theta}{\sqrt{\theta+1}} diverges.


    Did I do anything "illegal"? Is that a sufficient proof?

    -Tyler
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  5. #5
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    Quote Originally Posted by Latszer View Post
    Is it a sufficient proof if I just sub the integral like....

    \int_1^\infty\frac{d\theta}{\sqrt{\theta+1}} = \lim_{c\to\infty}\int_1^c \frac{d\theta}{\sqrt{\theta+1}}

    let u=\theta + 1 ; du = d\theta
    when  \theta = c \rightarrow u = c + 1 ; when  \theta = 1 \rightarrow u = 2
    Let d = c+1
    When  c \rightarrow\infty ,  d \rightarrow\infty

    \lim_{d\to\infty}\int_2^d\frac{du}{\sqrt{u}} = \lim_{d\to\infty}\int_1^\infty\frac{du}{\sqrt{u}}d  u - \int_1^2\frac{du}{\sqrt{u}}du

    And then since \lim_{d\to\infty}\int_1^d\frac{du}{\sqrt{u}} diverges, then \int_1^\infty\frac{d\theta}{\sqrt{\theta+1}} diverges.


    Did I do anything "illegal"? Is that a sufficient proof?

    -Tyler
    Why not just realise that \frac{1}{\sqrt{\theta + 1}} = (\theta + 1)^{-\frac{1}{2}}?


    So \int{\frac{1}{\sqrt{\theta + 1}}\,d\theta} = \int{(\theta + 1)^{-\frac{1}{2}}\,d\theta}

     = 2(\theta + 1)^{\frac{1}{2}} + C

     = 2\sqrt{\theta + 1} + C.


    So \int_1^{\infty}{\frac{1}{\sqrt{\theta + 1}}\,d\theta} = \lim_{\epsilon \to \infty}\left[2\sqrt{\theta + 1}\right]_1^{\epsilon}

     = \lim_{\epsilon \to \infty}2\sqrt{\epsilon + 1} - 2\sqrt{1 + 1}

    This is clearly divergent, as the square root function is always increasing.
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