Question

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

Answer 1

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

Answer 2

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:

\(\large \displaystyle \lim_{n \rightarrow \infty}\left[ \sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n~-~n! \right]=-\infty\)

\(\large \displaystyle \lim_{n \rightarrow \infty}\left[ \frac{\sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n~}{~n!} \right]=1\)

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

wolfram said so

Answer 7

And if you subtract Stirling's approximation from factorial:

n! - Stirling's approximation , then the limit as n-> infinity , = infinity.

Answer 8

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

Answer 9

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

Answer 10

@ganeshie8 check this

Answer 11

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.

Answer 12

show*

Answer 13

whats the question

Answer 14

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

\(\large \displaystyle \lim_{n \rightarrow \infty}\left[ \frac{\sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n~}{~n!} \right]=1\)

Answer 15

let me ask you a question

Answer 16

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

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

Answer 17

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

Answer 18

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

Answer 19

right, good question to spend time on !

Answer 20

\(\large \displaystyle \sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n~~~?~~~~n! \)

\(\large \displaystyle \sqrt{2\pi n}\cdot n^n~~~?~~~~e^nn! \)

I guess like this.

\(\large \displaystyle \sqrt{2\pi}\cdot n^{n+0.5}~~~?~~~~e^nn! \)

we can disregard coefficient 2pi for this, perhaps.

\(\large \displaystyle n^{n+0.5}~~~?~~~~e^nn! \)

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

what about e^n ?

Answer 25

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

Answer 26

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...

Answer 27

oh, my bad

Answer 28

wait, I am mixinging something up

Answer 29

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

just this.

Answer 30

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!

Answer 31

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

Answer 32

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.

Answer 33

Tnx for your input ganeshie and ikram