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Math Help - Area - Log Inequality

  1. #1
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    Area - Log Inequality

    Using f(x) = lnx, and f(n-1) < Sn n-1 f(x) dx < f(n)

    Show that for n>1

    (n-1)! < n^n (n-1)! / (e(n-1)^(n-1)) < n!

    Have tried to approach this using graphs, but could not get middle thing. Done this by adding the rectangle below f(x) and rectangles above f(x) with lengths of 1 unit square, starting with x =1
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  2. #2
    A Plied Mathematician
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    Clarification

    To restate the problem:

    Using f(x)=\ln(x), and

    \displaystyle{f(n-1)<\int_{n-1}^{n}f(x)\,dx<f(n)},

    show that for n>1, we have

    \displaystyle{(n-1)!<\frac{n^{n}(n-1)!}{e(n-1)^{n-1}}<n!.}

    Is that correct?

    If so, I would compute the integral

    \displaystyle{\int_{n-1}^{n}f(x)\,dx=\int_{n-1}^{n}\ln(x)\,dx},

    and combine all the terms into one logarithm.

    That should give you some direction.
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Lukybear View Post
    Using f(x) = lnx, and f(n-1) < Sn n-1 f(x) dx < f(n)

    Show that for n>1

    (n-1)! < n^n (n-1)! / (e(n-1)^(n-1)) < n!

    Have tried to approach this using graphs, but could not get middle thing. Done this by adding the rectangle below f(x) and rectangles above f(x) with lengths of 1 unit square, starting with x =1
    To start, I would take the logs of your three terms in the desired inequalities. Now what's  \ln[(n-1)!] ?
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  4. #4
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    To both above posters thxs for replying. I have tried previously adding areas above the f(x) and below the f(x) but however, i do not get the middle term. Instead i get n^n/e^(n-1) when simplifying.
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  5. #5
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    Are you sure you're simplifying correctly? For the integral I mentioned, I get

    \displaystyle{\ln\left(\frac{n^{n}}{e\,(n-1)^{n-1}}\right).}

    Can you show your line-by-line simplification of the integral?
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  6. #6
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    Yea sorry my bad. Reading it wrongly, through the limits were n - 1
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  7. #7
    A Plied Mathematician
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    So have you gotten the needed result?
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