# evaluation of limit $\lim_{n\rightarrow \infty}\left(\frac{n!}{n^n}\right)^{\frac{1}{n}}$

Calculate the value of the limit

$$\lim_{n\rightarrow \infty}\left(\frac{n!}{n^n}\right)^{1/n}$$

Can we solve this without using a Riemann sum method? If so, how?

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Use Stirling:

$$n! \sim \sqrt{2 \pi n} n^n e^{-n}$$

Note that

$$\left(n^{1/2}\right)^{1/n} = \exp{\left( \frac{\log{n}}{2 n}\right)} \sim 1$$

as $n \rightarrow \infty$. The limit is then $1/e$.

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Thanks Ron Gordon, one of my friend told me that we can solve it using Stloz Theorem. have any idea about that, Thanks – juantheron Mar 22 '13 at 1:35
Yes Thanks Ron Gordon – juantheron Mar 24 '13 at 1:01

$(1)$ Stirling's Formula: Applying Stirling's formula allows us to conclude that $$\left(\frac{n!}{n^n}\right)^\frac{1}{n}=\frac{1}{e}\left(1+o(1)\right).$$

$(2)$ Riemann Sum: Take the logarithm and notice that this is $$\exp \left(\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^{n}\log\frac{i}{n}\right)=\exp\left(\int_{0}^{1}\log xdx\right)=\frac{1}{e}.$$

Do note that Stirling's formula is usually derived from recognizing the Riemann Sum and using analysis.

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Thanks Ron Gordon, can we solve it using stolz theorem, plz explain here, thanks – juantheron Mar 22 '13 at 1:36

We will use the result

$$\lim_{n \to \infty} a_n^{1/n} = \lim_{n\to \infty}\frac{a_{n+1}}{a_n} .$$

Let $$a_n=\frac{n!}{n^n} \implies \frac{a_{n+1}}{a_n}=\frac{(n+1)!n^n}{(n+1)^{n+1}n!} = \left(\frac{n}{n+1}\right)^n$$

$$\implies \lim_{n\to \infty}\frac{a_{n+1}}{a_n}= \lim_{n\to \infty }\frac{1}{(1+1/n)^n} =e^{-1}.$$

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I assume you want to write $\frac1e$ instead of $1$, since all the remaining steps are correct. – user17762 Mar 22 '13 at 4:06
@Marvis: Yes, you are right. It is getting late. Thanks. – Mhenni Benghorbal Mar 22 '13 at 4:11
Thanks Mhenni Benghorbal – juantheron Mar 24 '13 at 1:01