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Math Help - [SOLVED] A limit with a factorial

  1. #1
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    [SOLVED] A limit with a factorial

    I know that  \lim \limits_{n \to \infty } \frac{n!}{{n^n }} = 0
    since  \lim \limits_{n \to \infty } \frac{n!}{{n^n }} =<br />
\lim_{n \to \infty} \left( \frac{1}{n} \cdot \frac{2}{n} \cdot \ \cdots \ \cdot \frac{n-1}{n} \cdot \frac{n}{n}\right) and each limit of this product goes to 0. However, how would you write an \varepsilon-proof for this limit?
    Last edited by akolman; November 29th 2008 at 01:50 PM. Reason: mistake in an equation
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  2. #2
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    Quote Originally Posted by akolman View Post
    I know that  \lim \limits_{n \to \infty } \frac{n!}{{n^n }} = 0
    since  \lim \limits_{n \to \infty } \frac{n!}{{n^n }} =<br />
\lim_{n \to \infty} \left( \frac{1}{n} \cdot \frac{2}{n} \cdot \ \cdots \ \cdot \frac{n-1}{n} \cdot \frac{n}{n}\right) and each limit of this product goes to 0. However, how would you write an \varepsilon-proof for this limit?
    Hint: \frac{n!}{n^n} \leq \frac{1}{n}.
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  3. #3
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    this means  n^{n-1} \geq n!

    How do we know this is true? (for large enough  n of course)
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  4. #4
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    \frac{n!}{n^n} \ = \ \frac{n{\color{red}(n-1)}{\color{blue}(n-2)}{\color{magenta}(n-3)} \cdots (2)}{n^{n-1}} \cdot \frac{1}{n} \ \leq \ \frac{\overbrace{n \cdot {\color{red}n} \cdot {\color{blue}n} \cdot {\color{magenta}n} \cdots n}^{n-1 \ \text{times}}}{n^{n-1}} \cdot \frac{1}{n} \ = \ \frac{n^{n-1}}{n^{n-1}} \cdot \frac{1}{n} \ = \ \frac{1}{n}
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  5. #5
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    Thank you all very much for your help!
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