how to prove this ?

[distant dove]

I have no idea about this, help me, please. Thanks!

[small ice]

Stirling formula, says n! nearly $\sqrt 2 \pi n (\frac{n}{2})^n} $

[rigid oriole]

remove the space from the last dollar sign and the } bracket

[distant dove]

Stirling formula, says n! nearly $\sqrt 2 \pi n (\frac{n}{2})^n}$

[warm spire]

Nya~
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[distant dove]

oh my god...
why is n! nearly (√2)πn(n/2)^n ?
is that uses the Squeeze theorem?

[rigid oriole]

I believe you meant, that a in R

[rigid oriole]

Yes, you could use it as an upper bound

and lower bound 1/n!

[iron merlin]

No need to use Stirling's formula. Just note that for each k>2a we have a/k<1/2. So, a^n/n! < const/2^n which tends to 0.

[small ice]

sry, some typo error when deal this. Striling formula is $ \sqrt{2 \pi n} (\frac{n}{e})^n$

[small ice]

use Euler-Maclaurin fomula can get this one, but there are other ways to get it

[distant dove]

we haven't start teaching Differentiation and Integration, so i hope there is the way which just uses the concept in limits. i just a freshman.

[distant dove]

i think i need the explanation about why it is feasible

[iron merlin]

I hope this helps.

[distant dove]

if a also approaching ∞
is that still feasible?

[rigid oriole]

a is constant

[rigid oriole]

I honestly dont get how you derived this equation

[iron merlin]

Suppose a is an integer. What exactly is unclear? We group together the first 2a factors in a^n and in n!. So we put aside a^(2a) and (2a)! and are left with n-2a factors in numerator and denominator. Each of the fractions formed by these n-2a factors is less then 1/2.

[iron merlin]

Oh, I see, it should be n-2a. Not 2n-a. Sorry.

Here is the correct version.

[distant dove]

i think i understand that...maybe
if i need the another case that <a^n/n! , like squeeze theorem

[iron merlin]

If a>0 then the whole thing is bounded by 0 from below. So, yes, by squeeze theorem the limit is 0. If a<0, then ok, take the absolute value and apply upper bound for it.

[distant dove]

so, i just take the absolute value, then this proof can end perfectly

[iron merlin]

If a is not an integer you can take a=[|a|]+1, where [...] is a floor function and apply this proof.

[distant dove]

thank you very much !

/solved

i think this doesn't work again

[rigid oriole]

I think it works now, after observation, I am still astonished how you derived the denominator n! = (2a)! * (2a+1) * ... * (2a+(n-2a))

[iron merlin]

Ok, say n=10, a=3. so $10!=6!\cdot 7\cdot 8\cdot 9\cdot 10=6!*(6+1)(6+2)(6+3)(6+4)$.
Btw n-2a=4.

[warm spire]

QD2718