Problem from the Cormen appendix C 1.13 I am currently working on CLRS 1.13. The idea is to use Stirling's
approximation to prove that
$${2n \choose n} = \frac{2^{2n}}{\sqrt{\pi n}} \left( 1 + O \left( \frac{1}{n} \right)  \right)$$
Now directly replacing Stirling's approximation I arrive to something like:
$${2n \choose n} = \frac{2^{2n} \left( 1 + \Theta(\frac{1}{2n}) \right)}{ \sqrt{\pi n} \left(1 + \Theta( \frac{1}{n}) \right)^2 }$$
Now the quotient is a function that ends up being $O(\frac{1}{n})$ because we
have terms in $O(\frac{1}{n})$ and $O(\frac{1}{2n})$.  The problem is that even if there
is a quotient of two functions that are $O(\frac{1}{n})$.
I don't know if the result is still $O(\frac{1}{n})$.  In fact I haven't seen
any general rule for the quotient of two big $O$ s.
Is there a general rule that I can apply here?
Just simply use the inequality $1 + \Theta(1/n) \geq 1$ in the denominator. You will get the required answer.