Fix $0<\varepsilon<1$. For $m\in\mathbb{N}$, let $$c_m=\max\left\{\frac{{m-1\choose s-1}{m\choose k-s}}{{2m\choose k}}:\;k=1,2,\dots,2m(1-\varepsilon)\;\mbox{and}\;s=1,2,\dots,k\right\}$$ Prove that $c_m\to0$ as $m\to\infty$.
I have made an initial attempt as follows:
Note that \begin{align*} \frac{{m-1\choose s-1}{m\choose k-s}}{{2m\choose k}}&=\frac{{2m-1\choose k-1}}{{2m\choose k}}\cdot\frac{{m-1\choose s-1}{m\choose k-s}}{{2m-1\choose k-1}}\\ &=\frac{k}{2m}{k-1\choose s-1}\prod_{j=1}^{s-1}\frac{m-s+j}{2m-s+j}\cdot\prod_{\ell=1}^{k-s}\frac{m-k+s+\ell}{2m-k+\ell}\\ &<\frac{k}{2m}{k-1\choose s-1}\left(\frac{1}{2}\right)^{s-1}\left(\frac{m}{2m-s}\right)^{k-s}\\ &\le\frac{k}{2m}{k-1\choose \frac{k-1}{2}}\left(\frac{1}{2}\right)^{k}\left(\frac{2m}{2m-s}\right)^{k-s}\\ &\sim\frac{k}{2m}\frac{1}{\sqrt{2\pi}(k-1)^{1/2}}\left(\frac{2m}{2m-s}\right)^{k-s}, \end{align*} where the approximation follows by using Stirling's formula.
I have no idea about how to continue the proof. It seems not like it will achieve the goal.
Any help will be greatly appreciated