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Math Help - Test for Convergence

  1. #1
    Member VitaX's Avatar
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    Test for Convergence

    Test for convergence:

    \int_4^{\infty} \frac{1}{\sqrt{x} - 1} dx

    0\leq \frac{1}{\sqrt{x}} \leq \frac{1}{\sqrt{x} - 1}

    \lim_{b \rightarrow \infty} \int_4^b \frac{1}{\sqrt{x}} dx

    \lim_{b \rightarrow \infty} \int_4^b x^{-\frac{1}{2}}

    \lim_{b \rightarrow \infty} 2x^{\frac{1}{2}}|_4^b

    \lim_{b \rightarrow \infty} \left[2(b)^{\frac{1}{2}} - 2(4)^{\frac{1}{2}}\right] = -4

    Since the limit is finite, the function converges. Did I do this one right?

    \int_{\pi}^{\infty} \frac{1 + sinx}{x^2} dx

    How would I go about solving this one? Which function would I choose to integrate to see if a finite value is found?
    Last edited by VitaX; March 23rd 2010 at 09:38 PM. Reason: Posted too soon
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  2. #2
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    That integral can be split up, one of which converges to zero. The other should be easy to evaluate: the sine function alternates between -1 and 1, however the quadratic increases without bound. However, you have -1 to 1 divided by a large number, it eventually converges to 0.
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  3. #3
    Member VitaX's Avatar
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    Quote Originally Posted by ANDS! View Post
    That integral can be split up, one of which converges to zero. The other should be easy to evaluate: the sine function alternates between -1 and 1, however the quadratic increases without bound. However, you have -1 to 1 divided by a large number, it eventually converges to 0.
    I'm a little confused by what you're saying, I was told to use either Direct Comparison Test or Limit Comparison Test in which both tell me to come up with a new function thats less than or greater than the original function (I'm not too sure if it matters) and integrate said function. But from what you're saying, you just integrate the function directly? Either way did I atleast do the first one correctly?
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  4. #4
    Super Member General's Avatar
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    Quote Originally Posted by VitaX View Post
    Test for convergence:

    \int_4^{\infty} \frac{1}{\sqrt{x} - 1} dx

    0\leq \frac{1}{\sqrt{x}} \leq \frac{1}{\sqrt{x} - 1}

    \lim_{b \rightarrow \infty} \int_4^b \frac{1}{\sqrt{x}} dx

    \lim_{b \rightarrow \infty} \int_4^b x^{-\frac{1}{2}}

    \lim_{b \rightarrow \infty} 2x^{\frac{1}{2}}|_4^b

    \lim_{b \rightarrow \infty} \left[2(b)^{\frac{1}{2}} - 2(4)^{\frac{1}{2}}\right] = -4
    The final answer for the limit is wrong.

    Since the limit is finite, the function converges. Did I do this one right?
    No

    \int_{\pi}^{\infty} \frac{1 + sinx}{x^2} dx

    How would I go about solving this one? Which function would I choose to integrate to see if a finite value is found?
    Since \color{red} sin(x) \leq 1 \, \forall \, x \in R \,\ then \,\ \color{red} \frac{1+sin(x)}{x^2} \leq \frac{2}{x^2}
    So .. ?
    .
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  5. #5
    Member VitaX's Avatar
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    Ok I understand what mistake I made on the first one, was dumb of me. Infinity minus a constant is still Infinity.

    For the second one I know that sinx oscillates between -1 and 1 but how were you able to come up with the new function that is greater than the original? Is it that since sinx \leq 1 so therefore it means you can subsitute 1 in for sinx? That part to me is a little confusing. Either way, I integrated it and \frac{2}{x^2} converges the limit is \frac{2}{\pi}.
    Last edited by VitaX; March 24th 2010 at 11:47 PM.
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  6. #6
    Super Member General's Avatar
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    Quote Originally Posted by VitaX View Post
    1 but how were you able to come up with the new function that is greater than the original? Is it that since sinx \leq 1 so therefore it means you can subsitute 1 in for sinx?
    .
    Yes.
    And by the comparison theorem, the original improper integral is .. ?
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  7. #7
    Member VitaX's Avatar
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    Quote Originally Posted by General View Post
    Yes.
    And by the comparison theorem, the original improper integral is .. ?
    Not sure I quite understand what you mean here
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