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Math Help - The Integral Test

  1. #1
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    Exclamation The Integral Test

    The question:

    \sum _{n=2}^{\infty }{\frac {1}{n \left( \ln  \left( n \right) <br />
 \right) ^{2}}}

    I need help in finding if f \left( x \right) ={\frac {1}{x \left( \ln  \left( x \right) <br />
 \right) ^{2}}}<br />
is decreasing.

    I found that {\frac {d}{dx}}f \left( x \right) =-{\frac { \left( \ln  \left( x<br />
 \right)  \right) ^{2}+2\,\ln  \left( x \right) }{{x}^{2} \left( \ln <br />
 \left( x \right)  \right) ^{4}}}. Then I made -{\frac { \left( \ln  \left( x \right)  \right) ^{2}+2\,\ln  \left( x<br />
 \right) }{{x}^{2} \left( \ln  \left( x \right)  \right) ^{4}}}<0<br />
    which turns out to be {e}^{-2}<x

    but when I graph it, f(x) is still increasing at {e}^{-2}. What am I doing wrong?

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  2. #2
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    Quote Originally Posted by s3n4te View Post
    The question:

    \sum _{n=2}^{\infty }{\frac {1}{n \left( \ln \left( n \right) <br />
\right) ^{2}}}

    I need help in finding if f \left( x \right) ={\frac {1}{x \left( \ln \left( x \right) <br />
\right) ^{2}}}<br />
is decreasing.

    I found that {\frac {d}{dx}}f \left( x \right) =-{\frac { \left( \ln \left( x<br />
\right) \right) ^{2}+2\,\ln \left( x \right) }{{x}^{2} \left( \ln <br />
\left( x \right) \right) ^{4}}}. Then I made -{\frac { \left( \ln \left( x \right) \right) ^{2}+2\,\ln \left( x<br />
\right) }{{x}^{2} \left( \ln \left( x \right) \right) ^{4}}}<0<br />
    which turns out to be {e}^{-2}<x

    but when I graph it, f(x) is still increasing at {e}^{-2}. What am I doing wrong?

    But your sum starts at n = 2 so you only need to consider values of x \ge 2
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  3. #3
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    Quote Originally Posted by danny arrigo View Post
    But your sum starts at n = 2 so you only need to consider values of x \ge 2
    Thanks for the reply.

    I know, but I'm just puzzled why it's not decreasing at {e}^{-2}
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  4. #4
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    Quote Originally Posted by s3n4te View Post
    Thanks for the reply.

    I know, but I'm just puzzled why it's not decreasing at {e}^{-2}

    y = \frac{1}{x(\ln{x})^2}

    y' = \frac{-[2\ln{x} + (\ln{x})^2]}{x^2(\ln{x})^4}

    y' = 0 ...

    \ln{x}(2 + \ln{x}) = 0

    \ln{x} = 0 at x = 1 ... y is undefined there

    \ln{x} = -2

    x = e^{-2} ... y is not decreasing at x = e^{-2} since y' = 0 there.
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  5. #5
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    Quote Originally Posted by s3n4te View Post
    Thanks for the reply.

    I know, but I'm just puzzled why it's not decreasing at {e}^{-2}
    Your derivative (for x \ne 1)

    -{\frac { \left( \ln \left( x \right) \right) ^{2}+2\,\ln \left( x<br />
\right) }{{x}^{2} \left( \ln \left( x \right) \right) ^{4}}} = -{\frac { \ln x +2}{{x}^{2} \left( \ln \left( x \right) \right) ^{3}}}

    and for x < e^{-2} your denominator is negative make the entire expression positive.
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  6. #6
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    Quote Originally Posted by skeeter View Post
    y = \frac{1}{x(\ln{x})^2}

    y' = \frac{-[2\ln{x} + (\ln{x})^2]}{x^2(\ln{x})^4}

    y' = 0 ...

    \ln{x}(2 + \ln{x}) = 0

    \ln{x} = 0 at x = 1 ... y is undefined there

    \ln{x} = -2

    x = e^{-2} ... y is not decreasing at x = e^{-2} since y' = 0 there.
    but if you make y' < 0 ...
    {e}^{-2}<x

    doesn't that mean right after {e}^{-2} it should decrease?
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  7. #7
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    Quote Originally Posted by s3n4te View Post
    but if you make y' < 0 ...
    {e}^{-2}<x

    doesn't that mean right after {e}^{-2} it should decrease?
    between the two critical values, x = e^{-2} and x = 1 is x = e^{-1}

    ... y'(e^{-1}) > 0 , which indicates y is increasing on that interval.
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  8. #8
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    Quote Originally Posted by skeeter View Post
    between the two critical values, x = e^{-2} and x = 1 is x = e^{-1}

    ... y'(e^{-1}) > 0 , which indicates y is increasing on that interval.
    Cool, but how am I supposed to figure that out with the equality test -{\frac { \left( \ln  \left( x \right)  \right) ^{2}+2\,\ln  \left( x<br />
 \right) }{{x}^{2} \left( \ln  \left( x \right)  \right) ^{4}}}<0<br />
?
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  9. #9
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    Quote Originally Posted by s3n4te View Post
    Cool, but how am I supposed to figure that out with the equality test -{\frac { \left( \ln  \left( x \right)  \right) ^{2}+2\,\ln  \left( x<br />
 \right) }{{x}^{2} \left( \ln  \left( x \right)  \right) ^{4}}}<0<br />
?
    Never heard of the "equality" test. How do you do an "equality" test using an inequality?

    Do you know the first derivative and second derivative tests for extrema?
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  10. #10
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    Quote Originally Posted by skeeter View Post
    Never heard of the "equality" test. How do you do an "equality" test using an inequality?

    Do you know the first derivative and second derivative tests for extrema?
    oops sorry I meant inequality :P

    Yea but that was last term.... OH I get it now lol

    Thanks for your help skeeter!
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