What is the value of the infinite product $$ \frac{1}{2} \frac{3}{4} \frac{5}{6}\cdots = \prod_{n=1}^{\infty} \frac{2n-1}{2n} = \prod_{n=1}^{\infty} \left(1-\frac{1}{2n}\right) = \lim_{n\to\infty}\frac{(2n-1)!!}{(2 n)!!}$$
What is the asymptotic of the product as a function of $n$ for large $n$?
On
Without Stirling, but only comparison theorems for series with non-negative terms:
Letting $a_n \stackrel{\rm def}{=} 1-\frac{1}{2n}$, $$\begin{align*} \ln \prod_{n=1}^N a_n &= \sum_{n=1}^N \ln a_n = \sum_{n=1}^N \ln \left(1-\frac{1}{2n}\right) \operatorname*{=}_{N\to\infty} - \sum_{n=1}^N \frac{1}{2n}- \sum_{n=1}^N \underbrace{\left( \frac{1}{4n^2}+o\left(\frac{1}{n^2}\right)\right)}_{\stackrel{\rm def}{=} b_n} \\&= -\frac{1}{2}H_N - \sum_{n=1}^N b_n \tag{1} \end{align*}$$ using the fact that $\ln (1+x) \operatorname*{=}_{x\to0} x - \frac{x^2}{2}+o(x^2)$. Now, recall that $H_N \operatorname*{=}_{N\to\infty} \ln N + \gamma + o(1)$ to get $$\begin{align*} \ln \prod_{n=1}^N a_n &= -\frac{1}{2}\ln N - \frac{1}{2}\gamma +o(1) - \sum_{n=1}^N b_n\tag{2} \end{align*}$$ and, since $\sum_{n=1}^N b_n$ converges (by comparison with $\sum_n \frac{1}{n^2}$) to some real value (call it $c$), $$\begin{align*} \ln \prod_{n=1}^N a_n &= -\frac{1}{2}\ln N - \frac{1}{2}\gamma - c + o(1)\,.\tag{3} \end{align*}$$ By continuity of exponential, this gives" $$\begin{align*} \prod_{n=1}^N a_n &= \frac{1}{\sqrt{N}}\cdot e^{-\frac{1}{2}\gamma - c}\cdot e^{o(1)}\,.\tag{4} \end{align*}$$ and, since $\lim_{N\to\infty}e^o(1) = 1$, we get $$\begin{align*} \boxed{ \prod_{n=1}^N a_n \operatorname*{\sim}_{N\to\infty} \frac{C}{\sqrt{N}} } \end{align*}$$ for some positive constant $C\stackrel{\rm def}{=} e^{-\frac{1}{2}\gamma - c}$.
Apply Stirling's Formula to $$\frac{(2n)!}{(2^nn!)^2}$$