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Math Help - infinite Limit

  1. #1
    Junior Member Tclack's Avatar
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    infinite Limit

    So I'm supposed to prove that
     lim_n_\rightharpoonup_\infty \frac{\sqrt[n]{n!}}{n}

    And I'm doing so by using the squeezing theorem (with guided help from the book)

    I know for certain that

     \frac{1}{e^(1-\frac{1}{n})}  <    \frac{\sqrt[n]{n!}}{n}  <    \frac{(1+\frac{1}{n})(n+1)^\frac{1}{n}}{e}

    (btw the left is 1 over e to the power of all that stuff, Latex is making it look weird)

    This is correct. My book confirmed so. And I can see how the left side of the equation approaches 1/e. Somehow the right side does and I can't see how....anybody?
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  2. #2
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    Quote Originally Posted by Tclack View Post
    So I'm supposed to prove that
     lim_n_\rightharpoonup_\infty \frac{\sqrt[n]{n!}}{n}

    And I'm doing so by using the squeezing theorem (with guided help from the book)

    I know for certain that

     \frac{1}{e^(1-\frac{1}{n})}  <    \frac{\sqrt[n]{n!}}{n}  <    \frac{(1+\frac{1}{n})(n+1)^\frac{1}{n}}{e}

    (btw the left is 1 over e to the power of all that stuff, Latex is making it look weird)

    This is correct. My book confirmed so. And I can see how the left side of the equation approaches 1/e. Somehow the right side does and I can't see how....anybody?
    If you let x = \frac{1}{n} then \displaystyle \lim_{x \to 0} \frac{(1+x)(1+\frac{1}{x})^x}{e} = \frac{1}{e} \lim_{x \to 0} (1+x) \cdot \lim_{x \to 0} (1+\frac{1}{x})^x.
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  3. #3
    MHF Contributor Unknown008's Avatar
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    For the LaTeX code, you should wrap the whole 'power' in brackets like this:

    e^{(1 - \frac{1}{n})}

    As for the right side, 1/infinity tends to 0, and anything to the power of zero becomes 1.

    So, you are left with:

    \frac{1+\frac{1}{n}}{e}

    Once again, 1/infinity tends to zero, and 1 + 0 = 1. So, you get 1/e.

    I hope it helps!
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  4. #4
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    \frac{\left(1 + \frac{1}{n}\right)(1 + n)^{\frac{1}{n}}}{e} = \frac{1}{e}\cdot e^{\ln{\left[\left(1 + \frac{1}{n}\right)(1 + n)^{\frac{1}{n}}\right]}}

     = \frac{1}{e}\cdot e^{\ln{\left(1 + \frac{1}{n}\right)} + \frac{1}{n}\cdot\ln{(1 + n)}}

     = \frac{1}{e}\cdot e^{\ln{\left(1 + \frac{1}{n}\right)}}\cdot e^{ \frac{\ln{(1 + n)}}{n}}.


    Therefore

    \lim_{n \to \infty}\frac{\left(1 + \frac{1}{n}\right)(1 + n)^{\frac{1}{n}}}{e} = \lim_{n \to \infty}\frac{1}{e}\cdot e^{\ln{\left(1 + \frac{1}{n}\right)}}\cdot e^{\frac{\ln{(1 + n)}}{n}}

     = \frac{1}{e}\lim_{n \to \infty}e^{\ln{\left(1 + \frac{1}{n}\right)}}\lim_{n \to \infty}e^{\frac{\ln{(1 + n)}}{n}}

     = \frac{1}{e}\cdot e^{\ln{1}} \cdot e^{\lim_{n \to \infty}\frac{\ln{(1+n)}}{n}}

     = \frac{1}{e} \cdot e^{\lim_{n \to \infty}\frac{\frac{1}{1+n}}{1}} by L'Hospital's Rule

     = \frac{1}{e} \cdot e^{\lim_{n \to \infty}\frac{1}{1+n}}

     = \frac{1}{e} \cdot e^0

     = \frac{1}{e}.
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  5. #5
    Junior Member Tclack's Avatar
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    Man, you guys replied fast, I was playing around with this for about 4 hours.

    I'm trying to get better. Does anyone know of a book or source of really hard problems? I've found this:

    http://www.maths.cam.ac.uk/undergrad...tep/advpcm.pdf

    Any other sources?
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  6. #6
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    When you have indeterminate limits (like in this case, \infty^{0}) you need to apply a transformation to get \frac{0}{0} or \frac{\infty}{\infty} so that you can then apply L'Hospital's Rule.

    Indeterminate form - Wikipedia, the free encyclopedia
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  7. #7
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    there's a way by using Riemann sums, can you find it?
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