Limit $\lim (\frac{n!}{n^n})^{\frac{1}{n}}$

Question

I need to calculate $$\lim_{n\rightarrow \infty} \left(\frac{n!}{n^n}\right)^{\frac{1}{n}}$$ My try:

When $n!$ is large we have $n!\approx(\frac{n}{e})^n\sqrt {2\pi n}$ (Stirling approximation) $$\lim_{n\rightarrow \infty} \left(\frac{n!}{n^n}\right)^{\frac{1}{n}}=\lim_{n\rightarrow \infty}\frac{\left((\frac{n}{e})^n\sqrt {2\pi n}\right)^{\frac{1}{n}}}{n}$$ Simplifying we get, $$\frac{1}{e}\lim_{n\rightarrow \infty} \left(\sqrt{2\pi n}\right)^{\frac{1}{n}}$$

I am stuck here. I don't know how to proceed further.

You have a mistake it should be $ (n/e)^n$ is Stirling approximation. — Mhenni Benghorbal, Nov 29 '13 at 05:52

score 7 · Accepted Answer · answered Nov 29 '13 at 05:23

7

Another way :

Let $$A= \lim_{n \to \infty}\left(\frac{n!}{n^n}\right)^{\frac1n}$$

$$\implies \ln A= \lim_{n \to \infty}\frac1n\sum_{1\le r\le n}\ln \frac rn$$

Now use $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$

answered Nov 29 '13 at 05:23

lab bhattacharjee

274,582

Can you elaborate on how $\ln \frac{n!}{n^n}$ became $\sum_{1\le r\le n}\ln \frac rn$ ? – GTX OC Nov 29 '13 at 05:25
1

@GTXOC, $$\ln \frac{n!}{n^n}=\ln \prod_{1\le r\le n}\frac rn$$ Now, $\ln ab=\ln a+\ln b$ – lab bhattacharjee Nov 29 '13 at 05:27
So I get ln A=$\int_0^1 lnx~dx$ right? – GTX OC Nov 29 '13 at 06:18
@GTXOC, yes. Now can you finish of? Else http://math.stackexchange.com/questions/264887/integral-of-ln-x-from-0-to-1 – lab bhattacharjee Nov 29 '13 at 06:19
Thanks. I learnt a lot from this question. – GTX OC Nov 29 '13 at 06:28
@GTXOC, my pleasure. Nice to hear that – lab bhattacharjee Nov 29 '13 at 06:30

score 2 · Answer 2 · answered Nov 29 '13 at 05:53

2

You are doing fine. You are almost done

$$ \frac{1}{e}\lim_{n\rightarrow \infty} (\sqrt{2\pi n})^{\frac{1}{n}} =\frac{1}{e}\lim_{n\to \infty} (\sqrt{2\pi})^{1/n} \lim_{n\to \infty} (\sqrt{n})^{1/n}=\frac{1}{e}1.1=\frac{1}{e}.$$

Note:

$$ (\sqrt{n})^{1/n} = n^{1/(2n)} = e^{\frac{1}{2n}\ln n}\longrightarrow_{n\to \infty} e^0=1 .$$

answered Nov 29 '13 at 05:53

Mhenni Benghorbal

47,431

@GTXOC: You are welcome. – Mhenni Benghorbal Nov 29 '13 at 06:31

score 0 · Answer 3 · answered Nov 29 '13 at 05:19

0

Instead of using the stirling approximation, try using the definition of $n!$ and distribute the $n^n$ over all terms in $n!$. Then try using the limit of $(1-x/n)^{1/n}$ as n tends to $\infty$.

Can you solve it from here?

answered Nov 29 '13 at 05:19

shyamupa

411

Limit $\lim (\frac{n!}{n^n})^{\frac{1}{n}}$

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