Finding $a$ and $r$ such that $\lim\limits_{n\to \infty} n^r \cdot \frac12 \cdot \frac34 \cdots \frac{2n-1}{2n}=a$ Find $a,r>0$ such that
$$\lim_{n\to \infty} n^r \cdot \frac12 \cdot \frac34 \cdots \frac{2n-1}{2n}=a$$
I don't have any idea to solve it. How can I solve it?
 A: Hints: (1) Write the product of rational numbers as a single rational number, using only powers of $2$ and factorials. (2) Use Stirling's formula to compute simple equivalents of the numerator and the denominator. The ratio of these should be your $an^{-r}$.
(To help you check your computations, I mention that $r=\frac12$.)
A: I'll start out from a celebre limit, namely Wallis product that states that:
$$ \frac{\pi}{2} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdot \cdot \cdot $$
Without loss of generality, we consider an even factors number of the limit excepting ${n^r}$, and then by applying Wallis product we get that:
$$ \lim_{n\to \infty}\frac{n^r}{\sqrt{2n+1}} \frac{\sqrt{2}}{\sqrt{\pi}}$$ that obviously gives us $L =\frac{1}{\sqrt{\pi}}$ for $r=\frac{1}{2}$
The proof is complete.
A: Here are the details of @did's answer. Write
$$
\frac12 \cdot \frac34 \cdots \frac{2n-1}{2n} = \frac{(2n)!}{(2^n n!)^2}=\frac{1}{4^{n}}{2n \choose n}
$$
We have the following asymptotics for the central binomial coefficient:
$$
{2n \choose n} \sim \frac{4^n}{\sqrt{\pi n}}\text{ as }n\rightarrow\infty
$$
Hence
$$
\frac12 \cdot \frac34 \cdots \frac{2n-1}{2n} \sim \frac{1}{\sqrt{\pi n}}
$$
and so
$$
n^{1/2} \frac12 \cdot \frac34 \cdots \frac{2n-1}{2n} \sim \frac{1}{\sqrt{\pi}}
$$
