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Math Help - Help with a limit

  1. #1
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    Help with a limit

    Hi,

    How I show that

     \lim_{n\to \infty} \frac{n!e^n}{n^n}=\infty


    thanks.
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  2. #2
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    Hello,
    Quote Originally Posted by Andres Perez View Post
    Hi,

    How I show that

     \lim_{n\to \infty} \frac{n!e^n}{n^n}=\infty


    thanks.
    Using Stirling's approximation for n!, it is quite straightforward.

    For a large n, n! \sim \sqrt{2\pi n} \cdot \frac{n^n}{e^n}


    See here : http://en.wikipedia.org/wiki/Stirling%27s_approximation
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  3. #3
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    Hello, Andres Perez!

    How do I show that: . \lim_{n\to \infty} \frac{n!e^n}{n^n}\:=\:\infty
    Ratio test . . .

    R \;=\;\frac{a_{n+1}}{a_n} \;=\;\frac{(n+1)!e^{n+1}}{(n+1)^{n+1}}\cdot\frac{n  ^n}{n!e^n} \;=\;\frac{e^{n+1}}{e^n}\cdot\frac{(n+1)!}{n!}\cdo  t\frac{n^n}{(n+1)^{n+1}}

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


    Divide top and botom by n\!:\;\;e\cdot\left( \frac{1}{1+\frac{1}{n}}\right)^n \;=\;e\cdot\frac{1}{(1 + \frac{1}{n})^n}

    Then: . \lim_{n\to\infty} R \;=\;\lim_{n\to\infty}\bigg[e\cdot\frac{1}{(1+\frac{1}{n})^n}\bigg] \;=\;e\cdot\frac{1}{\lim(1+\frac{1}{n})} \;=\;e\cdot\frac{1}{e} \;=\;1
    . . . . . . . . . . . . . . . . . . . . . . . . . . . ^{^{n\to\infty}}


    Since \lim_{n\to\infty}|R| \:\not<\:1, the series diverges.

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  4. #4
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    Quote Originally Posted by Soroban View Post


    Since \lim_{n\to\infty}|R| \:\not<\:1, the series diverges.
    we can't say anything about convergence or divergence of the series because \lim_{n\to\infty} \frac{a_{n+1}}{a_n}=1. even if it diverges, which it does but for another reason, we can't deduce that \lim_{n\to\infty}a_n=\infty.
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  5. #5
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    Can we obtain that result without Stirling's approximation?

    How we show the divergence of the series?
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Angry

    you already asked this and i already gave you a hint. not very polite of you to re-post Andres.
    Quote Originally Posted by Andres Perez View Post
    Can we obtain that result without Stirling's approximation?
    probably.

    How we show the divergence of the series?
    we don't have a series here, we have a limit. doing the ratio test was a beat around the bush way to attack this problem. of course, it yields no fruit since the limit was exactly 1, it gives us no information.

    a variation on the root test may work though.

    note that \lim_{n \to \infty} \sqrt[n]{\frac {n!e^n}{n^n}} \le \lim_{n \to \infty} \frac {n!e^n}{n^n}
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  7. #7
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    on the Jevon's inequality we get that \lim_{n \to \infty}\frac{n!e^n}{n^n}\geq \lim_{n \to \infty}\sqrt[n]{\frac{n!e^n}{n^n}}=1, so it doesn't says anything about \lim_{n \to \infty}\frac{n!e^n}{n^n}=\infty
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Andres Perez View Post
    on the Jevon's inequality we get that \lim_{n \to \infty}\frac{n!e^n}{n^n}\geq \lim_{n \to \infty}\sqrt[n]{\frac{n!e^n}{n^n}}=1, so it doesn't says anything about \lim_{n \to \infty}\frac{n!e^n}{n^n}=\infty
    ah yes, the ratio and root tests are equivalent when both work

    without Stirling's, this is a beast.

    did you try using power series as i suggested before?
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