I am trying to find the asymptotic behavior of the sum:


$$ \sum^n\_{i=0} \begin{pmatrix} 2n \\ i \end{pmatrix} x^i y^{2n-i}$$


as $n\rightarrow\infty$. Here $x$, $y$ are complex numbers and I have $|x|\leq1$ and $|y|\leq1$. I also know that $|x+y|\leq1$ and $|4xy|\leq1$.


One of the approaches I thought about was to approximate the sum by an integral. However, as it turns out, this approach runs into trouble since the terms in the sum do not give a sufficiently smooth function (I think: we have complex numbers raised to non-integer powers thus giving phase ambiguities).


On the other hand this sum may be related to Hoeffding's inequality applied to binomial distribution. This inequality gives an exponential upper bound. However, since $x, y$ are complex and $|x+y|\leq1$, I don't know how to apply this inequality.


Is there a way to bound this sum? I need to show it decays to $0$ as $n\rightarrow\infty$.



Bound on sum of complex summands involving binomial coefficients
Assuming that $|x+y|<1$ and $4|xy| \le 1$, here's a proof of the decay. 


First suppose that $|x|> |y|$. The desired sum is 
$$ 
\le \binom{2n}{n} |xy|^n \sum\_{j=0}^{n} |y/x|^j \le 
\binom{2n}{n} |xy|^n \frac{1}{1-|y/x|}. 
$$
Since $\binom{2n}{n}$ is of size about $4^n/\sqrt{n}$, the desired decay follows in this case.


Next suppose $|x| <|y|$. Here we flip the roles of $x$ and $y$ using 
$$ 
\sum\_{k=0}^{n} \binom{2n}{k}x^k y^{2n-k} = (x+y)^{2n} - \sum\_{\ell =0}^{n-1} \binom{2n}{\ell} y^{\ell} x^{2n-\ell},
$$ 
and now use the previous argument. In this step, to get decay, we used that $|x+y|<1$. 


Finally suppose that $|x|=|y|$. If $|x|=|y|<1/2$ then the sum is 
$$ 
\le \binom{2n}{n} |y|^{2n} (n+1) 
$$
which decays exponentially. We are left with $|x|=|y|=1/2$ and $|x+y|<1$. It is enough to treat $x=e^{i\alpha}/2$ and $y=1/2$, with $0<\alpha<2\pi$. We need to bound 
$$
4^{-n} \sum\_{j=0}^{n} \binom{2n}{j}e^{ij\alpha} = \frac{4^{-n}}{2\pi} 
\int\_0^{2\pi} (1+e^{i\alpha-i\theta})^{2n} (1+e^{i\theta}+\ldots +e^{in\theta}) d\theta,
$$ 
by Parseval's identity. The portion of the integral with $|1+e^{i\alpha-i\theta}| \le 2 - 10 (\log n)/n$ contributes 
$$ 
\le 4^{-n} \Big(2-10\frac{\log n}{n}\Big)^{2n} (n+1), 
$$ 
which goes to zero for large $n$. The set of $\theta \in [0,2\pi]$ with $|1+e^{i\alpha-i\theta}| \ge 2-10(\log n)/n$ has measure $O(\frac{\log n}{n})$ and for such $\theta$ we see that $|1+e^{i\theta}+\ldots +e^{in\theta}|$ is bounded (since $0<\alpha <2\pi$ by assumption, and so $\theta$ is bounded away from $0$ and $2\pi$). Therefore this portion of the integral is $O(\frac{\log n}{n})$, completing the proof.


