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Is it possible to find the following using the squeeze/sandwich theorem? Think of the

$$\lim_{n\rightarrow\infty}\frac{n!}{n^n}$$

Watson
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lost
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  • The inequalities here may help you

    https://en.wikipedia.org/wiki/Stirling%27s_approximation

     – JC574 Oct 23 '16 at 11:04
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    Isn't $0\le \dfrac{n!}{n^n}\le \dfrac 1n?$ – mfl Oct 23 '16 at 11:06
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    How do we know that n!/n^n<=1/n? I'm not disputing that it is, but would that need another proof? What's the justification for using that particular inequality? – lost Oct 23 '16 at 11:10
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    $\dfrac{n!}{n^n}=\dfrac{n}{n}\dfrac{n-1}{n}\dots \dfrac{2}{n}\dfrac{1}{n}\le \dfrac{1}{n}$ because $\dfrac{n-k}{n}\le 1$ for any $k\in{0,\cdots,n-1}.$ – mfl Oct 23 '16 at 11:19
  • Since $n^n$ is the product of $n$ $n$-times, and $n!$ is the product of $1\cdot2\cdots n$, both products have $n$ factors. Pair up the factors. Each factor of $n!$ is less than or equal to the corresponding factor of $n^n$, so you have a product of things less than or equal to $1$. Moreover, since $n!$ has a factor of $1$, which matches with a factor of $n$, the entire product is less than or equal to $1/n$. – Michael Burr Oct 23 '16 at 11:20

1 Answers1

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You can use Stirling's approximation to apply the squeeze theorem. However, you need more than the basic version. In particular, you need the approximation

$$ \sqrt{2\pi}\frac{n^{n+\frac{1}{2}}}{e^n}\leq n!\leq e\frac{n^{n+\frac{1}{2}}}{e^n}. $$

Therefore, $$ \sqrt{2\pi}\frac{\sqrt{n}}{e^n}\leq\sqrt{2\pi}\frac{n^{n+\frac{1}{2}}}{n^ne^n}\leq \frac{n!}{n^n}\leq e\frac{n^{n+\frac{1}{2}}}{n^ne^n}=e\frac{\sqrt{n}}{e^n} $$

Now, study the terms on the end.

Note: mfl's comment is a simpler approach.

Michael Burr
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