I'm trying to find the exact asymptotics of a sum:
$$A = \sum^n_{i=0} \begin{pmatrix} 2n \\ i \end{pmatrix} x^{i} y^{2n-i} $$
as $n\rightarrow\infty$. Here $x,y$ are complex numbers, $|x|\leq1, |y|\leq1$. I also have $|x+y|<1$ and $|4xy|\leq1$. My strategy is to write the sum as an integral. Rewrite this sum as: $$ A = (x+y)^{2n} - \sum^{n-1}_{i=0} \begin{pmatrix} 2n \\ i \end{pmatrix} x^{2n-i} y^i \quad\quad\quad\quad (1) $$
Note that this is a remainder for the $(n-1)$-order Taylor expansion so can use the integral formula for the remainder:
$$ A = \begin{pmatrix} 2n \\ n \end{pmatrix} y^n \int^1_0 (x+(1-t)y)^n nt^{n-1} dt $$
Changing variables and introducing $z=x/y$ get:
$$ A = \begin{pmatrix} 2n \\ n \end{pmatrix} (xy)^n \int^1_0 (1+(1-u^{\frac{1}{n}})z)^n du $$
In this integral the integrand converges to $u^{-z}$. Then for $Re(z)<1$ the integral converges and I get (using Stirling's formula) the following asymptotic expression:
$$ A \sim \frac{(4xy)^n}{\sqrt{n}}\frac{1}{1-z} $$
On the other hand, I can approximate the second term in $(1)$ as the Taylor remainder thus interchanging the roles of $x,y$ and then get for $Re(1/z)<1$ that
$$ A \sim (x+y)^{2n} - \frac{(4xy)^n}{\sqrt{n}}\frac{1}{1-1/z} $$
However, now I have something very strange: when both $Re(z)<1$ and $Re(1/z)<1$ are valid, both asymptotic expressions have to agree, which means that it must be $|x+y|^2 < |4xy|$. But it's clearly not true - take for example $x=-1/2$ and $y=1/100$ which satisfy the conditions on $Re(z)$ and $Re(1/z)$. Where have I made a mistake?
