Results 1 to 5 of 5

Thread: Comparison Test

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    65

    Comparison Test

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

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

    What I have so far is

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

    Now I need an easy, smaller function that I can prove that if that smaller function diverges, then $\displaystyle f(\theta)$ diverges, am missing something here?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Feb 2009
    From
    stgo
    Posts
    84
    Quote Originally Posted by Latszer View Post
    $\displaystyle \int_1^\infty \frac{1}{\sqrt{\theta+1}}d\theta$

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

    What I have so far is

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

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

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

    I think now its correct
    Last edited by felper; Feb 16th 2010 at 07:11 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2009
    Posts
    65
    $\displaystyle \frac{1}{x}$ is not less than $\displaystyle \frac{1}{x+1}$ as x approaches infinity, so while your logic makes sense it technically does not "prove" the integral.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Sep 2009
    Posts
    65
    Is it a sufficient proof if I just sub the integral like....

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

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

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

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


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

    -Tyler
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by Latszer View Post
    Is it a sufficient proof if I just sub the integral like....

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

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

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

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


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

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


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

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

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


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

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

    This is clearly divergent, as the square root function is always increasing.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Comparison test and Limit Comparison test for series
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Nov 25th 2010, 12:54 AM
  2. Comparison or Limit Comparison Test Problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Mar 12th 2010, 07:46 AM
  3. Limit comparison/comparison test series
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Mar 25th 2009, 08:27 PM
  4. Comparison & Limit Comparison test for series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Mar 25th 2009, 04:00 PM
  5. Integral Test/Comparison Test
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jan 27th 2009, 10:58 AM

Search Tags


/mathhelpforum @mathhelpforum