When we try to prove the hypergeometric series $F(a,b;c;z)=\sum_{n=0}^\infty\frac{a_{(n)} b_{(n)}}{c_{(n)}n!}z^n$ converge absolutely when $R(a+b-c)<0$, we use that $ \frac{n}{c+n-1}=1-\frac{c-1}{n}+O(\frac{ 1}{n^2})$. (Without that we need to prove the condition of absolute convergence by subtracting module of complex number $(1+(a-1)/n) (1+(b-1)/n)$ and that of $(1+(c-1)/n)$ and then remove $\sqrt{}$ by $\sqrt{x}-\sqrt{y} =(x-y)/(\sqrt{x}+\sqrt{y})$, which’s quite complicated.)
The equation is valid because $ \frac{n}{c+n-1} = \frac{1}{1+\frac{c-1} {n}}$ and $ \frac{1}{1+x}=1-x+O({x^2})$, right?
More generally, can we say $\frac{1}{1+\frac{A}{n}+ O(\frac{1}{n^2})}=1-\frac{A}{n}+ O(\frac{1}{n^2}) $?
I guess it’s yes since $\{1+\frac{A}{n}+ O(\frac{1}{n^2})\} \{1-\frac{A}{n}+ O(\frac{1}{n^2})\}= 1+ O(\frac{1}{n^2}) $, but then how to eliminate $O(\frac{1}{n^2}) $ on the right side? (I guess basically we need to prove there is an expression of $O(\frac{1}{n^2})$ in the 2nd bracket {} so that the right side is exactly 1.)
Solved: let $\frac{A}{n}+ O(\frac{1}{n^2})$ be $x$.
