Is this asymptotic equation correct? Is this equation correct?
$$ \frac {1 + \Theta(\frac 1 {2n})} {(1 + \Theta(1/n))^2} = 1 + O(1 / n) $$
I need this equation to prove that 
$$ \binom {2n} n = \frac {2 ^ {2n}} {\sqrt {\pi n}} (1 + O(1 / n)) $$
which could be calculated from Stirling's Approximation:
$$ n! = \sqrt {2\pi n} \left(\frac n e\right)^n \left(1 + \Theta({\frac 1 n})\right).$$
 A: Suppose $Bx\le\Theta(x)\le Cx$ for each $\Theta$.
$$
\begin{align}
\frac{1+\Theta(\frac{1}{2n})}{(1+\Theta(\frac{1}{n}))^2}
&\le\frac{1+\frac{C}{2n}}{(1+\frac{B}{n})^2}\\
&=(1+\frac{C}{2n})(1-2\frac{B}{n}+O(\frac{1}{n^2}))\\
&=(1+(\frac{C}{2}-2B)\frac{1}{n}+O(\frac{1}{n^2}))\\
&=1+O(\frac{1}{n})
\end{align}
$$
$$
\begin{align}
\frac{1+\Theta(\frac{1}{2n})}{(1+\Theta(\frac{1}{n}))^2}
&\ge\frac{1+\frac{B}{2n}}{(1+\frac{C}{n})^2}\\
&=(1+\frac{B}{2n})(1-2\frac{C}{n}+O(\frac{1}{n^2}))\\
&=(1+(\frac{B}{2}-2C)\frac{1}{n}+O(\frac{1}{n^2}))\\
&=1+O(\frac{1}{n})
\end{align}
$$
A: Yes.  One way to see it is to do the long division; i.e., divide the denominator directly into the numerator.  The numerator is $1 + \Theta(\frac{1}{2n}) = 1 + \Theta(\frac{1}{n})$, and the denominator is $\left(1 + \Theta(\frac{1}{n})\right)^2 = 1 + \Theta(\frac{1}{n}) + \Theta(\frac{1}{n^2}) = 1 + \Theta(\frac{1}{n})$.  I'm not going to attempt to typeset the long division on this forum, but try it, and you'll see that you get $1$ with a remainder of $O(\frac{1}{n})$.  So you have 
$$\frac {1 + \Theta(\frac{1}{2n})} {\left(1 + \Theta(\frac{1}{n})\right)^2} = \frac{1 + \Theta(\frac{1}{n})}{1 + \Theta(\frac{1}{n})} = 1 + \frac{O(\frac{1}{n})}{1 + \Theta(\frac{1}{n})} = 1 + O\left(\frac{1}{n}\right),$$
since the denominator is $O(1)$.
(Remember that $\Theta(\frac{1}{2n}) = \Theta(\frac{1}{n})$, since the constant $\frac{1}{2}$ doesn't affect the asymptotic order.)
