- idku

What is going on here? (Stirling's approximation of n!)

- chestercat

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- e.mccormick

Inquiring minds want to know.... what is the rest of this question?

- idku

Ok, there Stirling's approximation of the factorial
\(\large \displaystyle n! {\LARGE\text{ ~}} ~\sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n\)
and I take two limits to test how reasonable it is, and this is what I get:
\(\large \displaystyle \lim_{n \rightarrow \infty}\left[ \sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n~-~n! \right]=-\infty\)
\(\large \displaystyle \lim_{n \rightarrow \infty}\left[ \frac{\sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n~}{~n!} \right]=1\)

- idku

I can see Stirling's is smaller from the first limit, but why is then 2nd limit =1?
(btw, the source is wolfram)

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## More answers

- anonymous

i'm not sure you're first limit is correct.

- idku

http://www.wolframalpha.com/input/?i=Lim+n%E2%86%92%E2%88%9E+%28%E2%88%9A%282%CF%80n%29%E2%80%A2%28%28n%2Fe%29%5En%29-%28n%21%29%29

- idku

wolfram said so

- idku

And if you subtract Stirling's approximation from factorial:
n! - Stirling's approximation , then the limit as n-> infinity , = infinity.

- idku

http://www.wolframalpha.com/input/?i=Lim+n%E2%86%92%E2%88%9E+%28%28n%21%29-%E2%88%9A%282%CF%80n%29%E2%80%A2%28%28n%2Fe%29%5En%29%29

- ikram002p

so basically n! is increasing faster than (n)^{n+1/2} :)

- ikram002p

@ganeshie8 check this

- idku

Yes, basically plain factorial is increasing faster than stirlings.
I guess for limits as n-> infinity, I can only use stirling in a denominator if I want to shjow convergence.

- idku

show*

- ganeshie8

whats the question

- idku

My question is why this limit is (- infinity)
\(\large \displaystyle \lim_{n \rightarrow \infty}\left[ \sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n~-~n! \right]=-\infty\)
and this limit is 1 for some reason
\(\large \displaystyle \lim_{n \rightarrow \infty}\left[ \frac{\sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n~}{~n!} \right]=1\)

- ganeshie8

let me ask you a question

- ganeshie8

\(\lim\limits_{n\to\infty} (2n-n) = ?\)
\(\lim\limits_{n\to\infty} \dfrac{2n}{n} = ?\)

- idku

Yes, yes I think I found why
1000-100
10000-1000
100000-10000
1×10^(n+1)-1×10^n = ∞
but if you divide it is just 10 for every one of them

- ikram002p

this what u need to explain
http://prntscr.com/8977k3

- ganeshie8

right, good question to spend time on !

- idku

\(\large \displaystyle \sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n~~~?~~~~n! \)
\(\large \displaystyle \sqrt{2\pi n}\cdot n^n~~~?~~~~e^nn! \)
I guess like this.
\(\large \displaystyle \sqrt{2\pi}\cdot n^{n+0.5}~~~?~~~~e^nn! \)
we can disregard coefficient 2pi for this, perhaps.
\(\large \displaystyle n^{n+0.5}~~~?~~~~e^nn! \)

- idku

so two growths verses one, but still kind of strange I thought that n^n is the boss

- ikram002p

well in the long time no one can beats factorial :3

- idku

No, we do know that n^n>n!

- ikram002p

what about e^n ?

- idku

yes, so with e^n together factorial is greater, and my logic of n^n being the boss doesn't apply

- idku

Well,
\[\frac{e}{n}\times\frac{2e}{n}\times\frac{3e}{n}\times\frac{4e}{n}\times...\frac{(n-1)e^{n-1}}{n}\times\frac{e^nn}{n}\]
so that e^n * n! can be in a sense intuitively larger if you think about it...

- idku

oh, my bad

- idku

wait, I am mixinging something up

- idku

\[\frac{n!~e^n}{n^n}=\frac{e}{n}\times\frac{2e}{n}\times\frac{3e}{n}\times\frac{4e}{n}\times...\frac{(n-1)e}{n}\times\frac{en}{n}\]
just this.

- idku

i don't know intuitively n^n still rules, but it will take me a while to digest that and actually think about the truth and make some sense of it.
Thanks guys!

- ganeshie8

|dw:1440615166212:dw|

- idku

well, that is very beginning tho...... and the difference will grow infinitely...

- idku

in any case.... I guess will just take the fact and some reasoning without complete comprehension. THere are times when we humans have to do that.

- idku

Tnx for your input ganeshie and ikram

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