matt grime said:
(after doing a simple ratio test, I'm actually coming round to the |x|<1/e argument, but it shouldn't be deduced from Stirling which is merely an asymptotic tool - n! can be a long way from the estimate, although I am considering reconsidering that position, as it happens - the asymptotic nature might be precisely the thing that implies the ratio test: the ratio of consecutive terms is x(1+1/n)^n, which converges to xe, and thus for n large enough |x|<e would imply x(1+1/n)^n<1). At |x|=e we can say nothing, and it diverges for |x|>e
I was in big hurry last time and didn't elaborate my hint & answer.
(Living has been hectic here lately).
Since I calculated without pencil and a scrap of paper I left possibility
I made mistake open.However,the reason wasn't I suspected something was
wrong with a the principal usage of of Stirling's approximative formula,
but my imperfect mind pics.I (ab)used the
weakest version
of Stirling to quickly obtain the result:
n!\approx \sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n
Of course,much more preceise formula is:
n!=\sqrt{2\pi n} \left(\frac{n}{e}\right)^n e^{\frac{1}{12n}+O(\frac{1}{n^2})}
And there is much more about Stirling.
Did you know there is also a full asymptotic series version of Stirling
which involves
Gamma function and summation over
Bernouli numbers?
It is derived by the
Mean value theorem for
Gamma function.
It is associated with tracking
n! like nobody's buisness.
If you are interested I can post that monster.
Application of the most powerful version may give an exact value for n!,not
just an aproximation.
For convenience I list some known results:
n!=\sqrt{2\pi}n^{n-\frac{1}{2}}e^{-n}\left(1+\frac{1}{12n}+\frac{1}{288n^2}+O(\frac{1}{n^2})\right)
n!=\sqrt{2\pi n}\left(\frac{n}{e}\right)^n e^{\frac{\theta(n)}{12n}};0<\theta(n)<1
For n>1 an interesting inequality holds:
\sqrt{2\pi}n^{n+1/2}e^{-n+\frac{1}{12n}}>n!>\sqrt{2\pi}n^{n+1/2}e^{-n+\frac{1}{12n+1}}
But such level of preceison is completely unnecessary for the problem we have.
If you still wonder how mathematical criminal tehno may (ab)use the weakest version
of Stirling and get the correct result anyway,my plausible explanation is:
The problem requires only the dominant multiplication term and one doesn't have to
run error estimates for large n! in evaluating the limit.IOW,the limits of the structure
like given where fast multiplication functions ratios are packed are insensitive
to the differences associated up to polynomial factors.
You may find an interesting exercise to rigorously prove why I'm correct in using
weak Stirling approximation when dealing with that limit.
At last ,few comments should be made regarding the simple "ratio" convergence test
you are talking about.Isn't that test also call "d'Alambert" test?
d'Alambert test says that for 0<x<
1/e the series converges,and by Leibnitz
principle for alternating series one concludes that for
-1/e<x<0 must
converge as well.
But what about point
x=1/e?
d'Alambert fails to decide on that point.
However,the series diverges for it!
Use Raabe's test to show that the following sum diverges:
\sum_{n=1}^{\infty}\frac{1}{n!}\left(\frac{n}{e}\right)^n
OTOH,for
x=-1/e the series converges!
Without elaboration why ,I will remind that the situation has much in common
with harmonic series:
\sum_{n=1}^{\infty}\frac{1}{n}
which diverges,but
1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...=ln\ 2
I used no of tests above to reach the conclusion.
Solution for the convergence interval is:
x\in (-1/e,+1/e>
Only the *nifty* weak Stirling approach.
As you can see it is very useful.
zoki85 said:
The proofs weren't so advanced and complicated but longish .
And I don't think they mentioned Stirling's formula .
I recall exactly the results becouse they were surprising.
For example,for n>> it is:
(2^n)*n!<n^n
and
(4^n)*n!>n^n
How advanced and how long ,what are you talking about?
The stronger claim than yours holds:
3^n n!>n^n
And it can be proven in an elementar way ,and isn't long at all!
