Asymptotic approximation of Catalan Numbers
We can use Stirling's approximation formula \begin{align*} n!=\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(1+O\left(\frac{1}{n}\right)\right) \end{align*} to prove:
The following is valid
\begin{align*} C_n=\frac{1}{n+1}\binom{2n}{n}= \frac {4^n} { \sqrt {\pi} n^{3/2}} (1+ O(1/n)) \tag{1} \end{align*}
We obtain using (1) \begin{align*} \frac{1}{n+1}\binom{2n}{n}&=\frac{1}{n+1}\cdot\frac{(2n)!}{n!n!}\\ &=\frac{1}{n+1}\cdot\sqrt{4\pi n}\left(\frac{2n}{e}\right)^{2n}\left(1+O\left(\frac{1}{n}\right)\right)\\ &\qquad \cdot \left(\frac{1}{\sqrt {2 \pi n} {\left( \frac {n}{e} \right)}^{n} {\left( 1+ O \left(\frac {1} {n}\right) \right)}}\right)^2\\ \\ &=\frac{1}{n+1}\cdot\frac{4^n}{\sqrt{\pi n}}\cdot \frac{\left(1+O\left(\frac{1}{n}\right)\right)}{\left(1+O\left(\frac{1}{n}\right)\right)\left(1+O\left(\frac{1}{n}\right)\right)}\tag{2}\\ &=\frac{1}{n\left(1+O\left(\frac{1}{n}\right)\right)}\cdot\frac{4^n}{\sqrt{\pi n}} \left(1+O\left(\frac{1}{n}\right)\right)^3\tag{3}\\ &=\frac{1}{n}\cdot\frac{4^n}{\sqrt{\pi n}} \left(1+O\left(\frac{1}{n}\right)\right)^4\\ &=\frac{4^n}{\sqrt{\pi}n^{3/2}} \left(1+O\left(\frac{1}{n}\right)\right)\tag{4}\\ \end{align*} and the claim follows.
Comment:
In (2) we do some cancellation
In (3) we use the geometric series expansion \begin{align*} \frac{1}{1+O\left(\frac{1}{n}\right)}=1+O\left(\frac{1}{n}\right) \end{align*}
In (4) we use \begin{align*} \left(1+O\left(\frac{1}{n}\right)\right)\left(1+O\left(\frac{1}{n}\right)\right)=1+O\left(\frac{1}{n}\right) \end{align*}
Use the binomial approximation for $(1+y)^k$:
$$ (1+y)^k=1+ky+\Theta(y^2) $$ as $y \to 0$.
In your case, you can take $k=-1$ to show that any function which is $\frac{1}{1+\Theta(1/n)}$ is also $1+\Theta(\frac{1}{n})$.