I'm interested in finding the asymptotics of the following (for $p \in [0,1]$) $$\sum_{k=1}^{\lfloor (n-1)/2 \rfloor} \frac{k {n-1 \choose 2k} {2k \choose k}} {4^{k}p^{k}}.$$
The central binomial coefficient ${2k \choose k}$ is asymptotic to $\frac{4^k}{\sqrt{\pi n}}$. Also the numerator contains a product of two binomial coefficients which combined are a single multinomial coefficient. Perhaps this is useful? I am unsure if there is a single dominant term.
Thanks for the help!
If we replace $n-1$ with $m$ (just for the sake of readability) we have: $$S=\sum_{k=1}^{\lfloor m/2\rfloor}k\binom{m}{2k}\frac{1}{p^k 4^k}\binom{2k}{k}=\frac{1}{2\pi}\sum_{k=0}^{\lfloor m/2\rfloor}2k\binom{m}{2k}\frac{1}{p^k}\int_{0}^{\pi}\sin^{2k}\theta\,d\theta$$ but since: $$\sum_{j\equiv 0\!\!\pmod{\!\!2}}j\binom{m}{j}\eta^j = \frac{1}{2}\sum_{j=0}^{m}\binom{m}{j}(1+(-1)^j)\,j\,\eta^j=\frac{1}{2}m\eta\left((1+\eta)^{m-1}-(1-\eta)^{m-1}\right)$$ it follows that: $$ S = \frac{m}{4\pi\sqrt{p}}\int_{0}^{\pi}\left(\left(1+\frac{\sin\theta}{\sqrt{p}}\right)^{m-1}-\left(1-\frac{\sin\theta}{\sqrt{p}}\right)^{m-1}\right)\sin\theta\,d\theta.$$ The last integral (it can be computed through hypergeometric function) is well suited to be estimated with the saddle point method. Just evaluate the integrand function and its second derivative in $\theta=\frac{\pi}{2}$ to get that $S$ behaves like: $$\sqrt{\frac{m}{2\pi p}}\left(1+\frac{1}{\sqrt{p}}\right)^{\frac{m}{2}}.$$