A trouble-some limit.. For a unique value $r$, the value of $$\lim_{n\rightarrow \infty} \ n^r \times \frac{1}{2} \times \frac{3}{4} \times\dots\times\frac{2n-1}{2n}$$ exists and is non-zero. For this value of $r$, what is the limit?
I've tried a few values but I haven't made much progress. Any help would be appreciated.
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$\ds{%
\lim_{n \to\infty}\pars{n^{r} \times \half \times {3 \over 4} \times\ldots\times
{2n - 1 \over 2n}}.\quad}$ $ r = ?$.

\begin{align}
&\ln\pars{n^{r} \times \half \times {3 \over 4} \times\ldots\times
{2n - 1 \over 2n}}
=r\ln\pars{n} + \sum_{k = 1}^{n}\ln\pars{2k - 1} - \sum_{k = 1}^{n}\ln\pars{2k}
\\[3mm]&=r\ln\pars{n} + \sum_{k = 1}^{n}\ln\pars{1 - {1 \over 2k}}
\\[3mm]&=r\ln\pars{n}
+ \sum_{k = 1}^{n}\bracks{\ln\pars{1 - {1 \over 2k}} + {1 \over 2k}}
- \half\bracks{\sum_{k = 1}^{n}{1 \over k} - \ln\pars{n}} - \half\,\ln\pars{n}
\\[3mm]&=
\pars{r - \half}\ln\pars{n}
+ \overbrace{\sum_{k = 1}^{n}\bracks{\ln\pars{1 - {1 \over 2k}} + {1 \over 2k}}}
^{\ds{\mbox{converges when}\ n \to \infty}}
- \half
\overbrace{\bracks{\sum_{k = 1}^{n}{1 \over k} - \ln\pars{n}}}
^{\ds{\to \gamma\ \pars{~\mbox{Euler constant}~} \atop \mbox{when}\ n \to \infty}}
\end{align}

This expresión converges when $\color{#0000ff}{\Large r = \half}$
In particular, $\ds{\sum_{k = 1}^{\infty}\bracks{\ln\pars{1 - {1 \over 2k}} + {1 \over 2k}} = \half\,\gamma - \half\,\ln\pars{\pi}}$. Then,
$$\color{#0000ff}{\large%
\lim_{n \to\infty}\pars{n^{1/2} \times \half \times {3 \over 4} \times\ldots\times
{2n - 1 \over 2n}}
=\exp\pars{-\,\half\,\ln\pars{\pi}} = {1 \over \root{\pi}}}
$$
A: $$n^r\times\frac{\frac{(2n-1)!}{2^{n-1}\cdot (n-1)!}}{2^n\cdot n!}=n^r\cdot\frac{(2n-1)!}{2^{2n-1}\cdot n!\cdot (n-1)!}$$
Now use Stirling's formula
$$n!\approx\frac{\sqrt{2\pi n}\cdot n^n}{e^n}$$
