I need help explaining the steps below pulled from a proof.
Solution:
Stirlings Formula: $N! = \sqrt{2\pi N}(\frac N e)^N(1+\frac {1}{12N} + \frac{1}{288N^2}- \frac{139}{51840N^3}+O(\frac{1}{N^4})$
Catalan Numbers: $\frac{1}{N+1} {2N \choose N}$
After substituting $N!$ into Catalan numbers and simplifying, I am stumped on how the following steps are obtained and evaluated:
(Step 1:)
$ = (\frac{1}{N+1})(\frac{4^N}{\sqrt{ N \pi }})(\frac{1+\frac{1}{24N}+\frac{1}{288(4N^2)}-\frac{139}{51840(8N^3)}+O(\frac{1}{N^4})}{1+\frac{1}{6N}+\frac{1}{72N^2}-\frac{31}{6480N^3}+O(\frac{1}{N^4})} $
(Step 2:)
$= (\frac{1}{N(1+\frac 1 N)})(\frac{4^N}{\sqrt{ N \pi }})({1+\frac{1}{24N}+\frac{1}{288(4N^2)}-\frac{139}{51840(8N^3)}+O(\frac{1}{N^4})})*(1-(\frac{1}{6N}+\frac{1}{72N^2}-\frac{31}{6480N^3}+O(\frac{1}{N^4}))+(\frac{1}{6N}+\frac{1}{72N^2}-\frac{31}{6480N^3}+O(\frac{1}{N^4}))^2-(\frac{1}{6N}+\frac{1}{72N^2}-\frac{31}{6480N^3}+O(\frac{1}{N^4}))^3+O(\frac{1}{N^4})) $
(Step 3:)
$= \frac{4^N}{N(1+\frac 1 N)\sqrt{ \pi N }} (1 - \frac{1}{8N} + \frac{1}{128N^2} + \frac{5}{1024N^3}+ O(\frac{1}{N^4}))$
(Step 4:)
$ = \frac{4^N}{\sqrt{\pi N}} (1 - \frac{1}{8N} + \frac{1}{128N^2} + \frac{5}{1024N^3}+ O(\frac{1}{N^4}))(1-\frac{1}{N}+\frac{1}{N^2}-\frac{1}{N^3}+O(\frac{1}{N^4})) $
I am confused about how the author went from steps 2-4. Could anyone explain, please? I know it's algebra, but having trouble with it and tell me where $\frac{1}{1-x}= 1+x+x^2+x^3+O(x^4) $ fits into these steps?
$$C_n=\frac 1{n+1}\binom{2 n}{n}=\frac 1{n+1}\frac{(2 n)!}{(n!)^2}=\frac{(2 n)!}{n! (n+1)!}$$ Now, you could make life simpler writing $$\log(C_n)=\log[(2n)!]-\log[n!]-\log[(n+1)!]$$ Using Stirling three times and continuing with Taylor series to get $$\log(C_n)=n \log (4)-\frac{1}{2} \log \left({\pi n^3}\right)-\frac{9}{8 n}+\frac{1}{2 n^2}-\frac{21}{64 n^3}+\frac{1}{4 n^4}-\frac{129}{640 n^5}+O\left(\frac{1}{n^6}\right)$$ Now, using $$C_n=e^{\log(C_n)}=\frac {4^n}{n\sqrt{\pi n}}\left(1-\frac{9}{8 n}+\frac{145}{128 n^2}-\frac{1155}{1024 n^3}+\frac{36939}{32768 n^4}+O\left(\frac{1}{n^5}\right)\right)$$