# Asymptotics of partial sum of binomial coefficients

For some fixed $$0, with $$q=1-p$$, let $$np\leq c\leq np+\sqrt{2npq-2n\log\log n}$$ and $$2np\leq x\leq 2np+2\sqrt{2npq-2n\log\log 2n}$$. I’m trying to get asymptotics of the partial sum $$\sum_{k=x-c}^c\binom{n}{k}\binom{n}{x-k}$$ or equivalently if $$c=n\lambda_1$$ and $$x=2n\lambda_2$$, for constants $$p\leq\lambda_2\leq\lambda_1<1$$ $$\sum_{k=2n\lambda_2-n\lambda_1}^{n\lambda_1}\binom{n}{k}\binom{n}{2n\lambda_2-k}$$

My initial attempt was to adapt @robjohn’s solution in this post.

First we focus on $$a_k=\binom{n}{k}\binom{n}{2n\lambda_2-k}$$ Then letting $$k=n\lambda_2+j$$, $$\log\left(\frac{a_{k+1}}{a_k}\right)=-\frac{2j}{n\lambda_2(1-\lambda_2)}+O(n^{-1})$$ Thus, $$a_k=a_{n\lambda_2}\exp\left(-\frac{2j^2}{n\lambda_2(1-\lambda_2)}+O(j/n)\right)$$ By Stirling’s approximation, we have $$a_{n\lambda_2}\sim \frac{1}{2\pi n\lambda_2(1-\lambda_2)}(1-\lambda_2)^{-2n}\left(\frac{1-\lambda_2}{\lambda_2}\right)^{2n\lambda_2}=C(\lambda_2)$$ Then using the Riemann integral for the exponential, $$\sum_{j=-n(\lambda_1-\lambda_2)}^{n(\lambda_1-\lambda_2)}\exp\left(-\frac{2j^2}{n\lambda_2(1-\lambda_2)}+O(j/n)\right)=\sqrt{n\lambda_2(1-\lambda_2)}\int_{-\infty}^{\infty}\exp\left(-2t^2\right)dt(1+O(1/n))$$ we have $$\begin{eqnarray} \sum_{k=2n\lambda_2-n\lambda_1}^{n\lambda_1}\binom{n}{k}\binom{n}{2n\lambda_2-k}&\sim& C(\lambda_2)\sqrt{n\lambda_2(1-\lambda_2)}\sqrt{\pi/2}\ &=&\frac{1}{2\sqrt{2\pi n\lambda_2(1-\lambda_2)}}(1-\lambda_2)^{-2n}\left(\frac{1-\lambda_2}{\lambda_2}\right)^{2n\lambda_2} \end{eqnarray}$$ Substituting back $$c=n\lambda_1$$ and $$x=2n\lambda_2$$, and noticing Stirling’s formula for $$\binom{2n}{x}$$, we get $$\sum_{k=x-c}^c\binom{n}{k}\binom{n}{x-k}\sim\frac{1}{\sqrt{2}}\sqrt{\frac{2n}{2\pi x(2n-x)}}\left(\frac{2n}{2n-x}\right)^{2n}\left(\frac{2n-x}{x}\right)^x\sim \frac{1}{\sqrt{2}}\binom{2n}{x}$$

However, I’m not entirely convinced that this asymptotic holds for the ranges given for $$c$$ and $$x$$. Is there a refinement for this result?