Inequality from Stirling's formula

[neginf]

In this link is a part of a book on approximations of functions.

http://books.google.com/books/about/An_Introduction_to_the_Approximation_of.html?id=VTW2cmjC43YC

I'd be thankful if someone would explain how the inequality near the top of page 17 was gotten.

 

[Stephen Tashi]

I don't see how to navigate to page 17 in that link. (I'm not enthusiastic about how the interface to Google books behaves on my browser and internet connection. It seems darn slow and tedious.) You'd best type out your question.

 

[neginf]

Thank you for writing back.

The inequality is

n is even
C(n,n/2)/2^(n+1) > 1/(2*sqrt(n)).

"Sharp form fo Stirling's inequality" is

sqrt(2*pi*k) * k^k * e^-k < k! < sqrt(2*pi*k) * k^k * e^-k * (1+1/(4*k))

Is it right? Tried with 4.

With Google books, by clicking on the book in the upper left hand corner, it will appear big on the screen and you can click on the big image and page up and down.
Think I'm missing something here.

 

[Stephen Tashi]

With Google books, by clicking on the book in the upper left hand corner, it will appear big on the screen and you can click on the big image and page up and down.
Think I'm missing something here.

I'm missing all pages after page 15. It does say "some pages are omitted from the preview".

The inequality is
n is even

$$\frac{ \binom{n}{n/2}}{2^{n+1}} > \frac{1}{2 \sqrt{n}}$$

"Sharp form fo Stirling's inequality" is

$$(\sqrt{2 \pi k}) k^k e^{-k} < k! < (\sqrt{2 \pi k}) k^k e^{-k } (1+\frac{1}{ 4k} )$$

Is it right? Tried with 4.

Is what right? Do you mean that you used k = 4 or n = 4 ?

 

[neginf]

Sorry, n.
I tried it with 4 and it seemed not to hold, the inequality. Tried with 2 and same problem.

 

[Stephen Tashi]

This inequality is not correct:
$$\frac{ \binom{n}{n/2}}{2^{n+1}} > \frac{1}{2 \sqrt{n}}$$


Assuming the inequality
$$(\sqrt{2 \pi k}) k^k e^{-k} < k! < (\sqrt{2 \pi k}) k^k e^{-k } (1+\frac{1}{ 4k} )$$

the only similar inequality that I see is:

$$\frac{ \binom{n}{n/2}}{(1 + \frac{1}{2n})^2} > \frac{\sqrt{2}}{\sqrt{\pi n}} > \frac{1}{\sqrt{2n}}$$