Limit of factorials: $\lim_{n\to \infty}\frac {(2n-1)!}{2^{2n-1}n!(n-1)!}$

Question

I'm failing go figure out how to calculate the limit where I have one factorial divided by two at about half its size. The specific limit I'm trying to find is this: $$\lim_{n\to \infty}\frac {(2n-1)!}{2^{2n-1}n!(n-1)!}$$

Answer 1

The ratio of two consecutive terms is $$\frac{x_{n+1}}{x_n}=\frac{2n+1}{2n+2}=1+\frac{\color{red}{a}}n+o\left(\frac1n\right),$$ with $$\color{red}{a}=-\frac12.$$ Note that $\color{red}{a}\lt0$ hence the sequence $(x_n)$ is eventually decreasing and its limit is $0$. Furthermore (but this is not needed for the limit), at least in a loose sense of $\approx$, $$x_n\approx n^{\color{red}{a}}=\frac1{\sqrt{n}}.$$ A result which confirms this general approach, and that one could want to memorize because of the ubiquity of its use, is that the mid-coefficient of the $n$th line of the Pascal triangle takes roughly $$\frac1{\sqrt{n}}$$ of the sum of the whole $n$th line, in the sense that $${n\choose \lfloor n/2\rfloor}\approx\frac{2^{n}}{\sqrt{n}}.$$ (More precise asymptotics are available but already this crude one would have solved your exercise right away.)