Higher-Order Approximation of Catalan-Numbers

Question

I have a question considering the higher-order approximations of the Catalan-Numbers, following the book Analytic Combinatorics by Flajolet and Sedgewick. First we set $$ C_n = \frac{1}{n+1} \binom{2n}{n} = \frac{1}{n+1} [z^n] (1-4z)^{-1/2} = \frac{4^n}{n+1} [z^n] (1-z)^{-1/2}, $$ where we use that the generating function of the central binomial coefficient is $(1-4z)^{-1/2}$.
Using Theorem VI.1 from the quoted book, we know that $$ (1-z)^{-\alpha} \sim \frac{n^{\alpha-1}}{\Gamma(\alpha)} \left( 1 + \frac{\alpha(\alpha -1) }{2n} + \frac{\alpha(\alpha -1) (\alpha -2) (3 \alpha -1)}{24 n^2} + O\left( \frac{1}{n^3}\right) \right) $$ (the full approximation can be found in the book, page 382). So we get $$ C_n = \frac{4^n}{n+1} \frac{n^{-1/2}}{\sqrt{\pi}}\left(1-\frac{1}{8n}+\frac{1}{128n^2}+O\left( \frac{1}{n^3}\right) \right). $$ Yet the book states (page 384) $$ C_n = \frac{4^n}{\sqrt{\pi n^3}}\left(1-\frac{9}{8n}+\frac{145}{128n^2}+O\left( \frac{1}{n^3}\right) \right). $$ I can see that we want to add a bit to the numerators, since we slash a bit of the denominators. But I don't understand the particular choice of the factors. Why do we add the sum of the prior fractions (we add $1$ to the $n^{-1}$ term, $9/8$ to the $n^{-2}$ term etc) to each fraction?

Answer 1

Note that $$\dfrac{n}{n+1} = 1 - \dfrac{1}{n} + \dfrac{1}{n^2} + \ldots $$ and $$ \left(1 - \dfrac{1}{n} + \dfrac{1}{n^2} + \ldots\right) \left( 1 - \dfrac{1}{8n} + \dfrac{1}{128 n^2} + \ldots\right) = 1 - \dfrac{9}{8n} + \dfrac{145}{128 n^2} + \ldots $$