The limit of the product $\prod_{i=1}^n\frac{1 - (2i + 1)a/(2n)}{1 - ia/n}$ as $n\to\infty$ Given that $0 < a < 1$, what is the value of
$$ P = \lim_{n\to \infty} \prod_{i=1}^n \frac{1 - (2i + 1)a/(2n)}{1 - ia/n} $$
Thus, $P_n$, the $n$th term, is
$$ P_n = \frac{1-3a/2n}{1-a/n}\cdot \frac{1-5a/2n}{1-2a/n}\cdots \frac{1-(2n+1)a/2n}{1-a} $$

Motivation
The product describes a real-live physical problem. Put simply, if you have a river or stream in which you're sampling at an upstream site A and a downstream site B and you analyze the chemistry of the stream at A and B, if the concentration has changed, it is because of ground water which has seeped in between A and B. By making some simplifying assumptions, the concentration and amount of ground water which has entered are calculated using the evaluation of this infinite product.
 A: The product can be rewritten as
$$\mathcal{P}_n=\prod_{i=1}^{n}\frac{1 - (2i + 1)a/(2n)}{1 - ia/n}=\prod_{i=1}^{n}\frac{\frac{n}{a}-\frac12-i}{\frac{n}{a}-i}=\frac{\Gamma\left(\frac{n}{a}-n\right)}{\Gamma\left(\frac{n}{a}\right)}\cdot\frac{\Gamma\left(\frac{n}{a}-\frac12\right)}{\Gamma\left(\frac{n}{a}-n-\frac12\right)}$$
Now taking the logarithm and using Stirling formula 
$$\ln\Gamma(x)=x(\ln x-1)+\frac12(\ln2\pi-\ln x)+O\left(x^{-1}\right)\qquad \text{as}\;
x\rightarrow\infty, $$
we arrive at 
$$\ln\mathcal{P}_{\infty}=\frac12\ln(1-a),$$
which reproduces Ron Gordon's answer.
A: Take logs of both sides to get
$$\log{P_n} = \sum_{i=1}^n \left(\log{\left [ 1- a\frac{i}{n} - \frac{a}{2 n}\right]} -\log{\left [ 1- a\frac{i}{n}\right]} \right) $$
Noting that $n$ is large, so we can rewrite the summand as
$$\log{\left [ 1- a\frac{i}{n} - \frac{a}{2 n}\right]} -\log{\left [ 1- a\frac{i}{n}\right]} = \log{\left( 1-\frac{a/(2 n)}{1-a i/n}\right)} \sim -\frac{a}{2} \frac{1}{n} \frac{1}{1-a i/n} $$
In the limit as $n \to \infty$, this is a Riemann sum and becomes the integral
$$\log{P} = -\frac{a}{2} \int_0^1 \frac{dx}{1-a x} = \frac12 \log{(1-a)}$$
Therefore
$$P=\sqrt{1-a}$$
BONUS
Here are a few plots of log base $2$ of the relative error between the above result $P$ and the finite product $P_n$ vs $\log_2{n}$.  The values of $a$ from left to right are $1/4$, $1/3$, $2/3$, $9/10$, and $99/100$.

EDIT
@Did has pointed out that the step in the second line needs more justification, i.e., a quantification of the error term inherent in the $\sim$ symbol.  Along these lines, for sufficiently large $n$, we may write
$$\log{\left( 1-\frac{a/(2 n)}{1-a i/n}\right)} = -\frac{a}{2} \frac{1}{n} \frac{1}{1-a i/n} + \frac12 \left ( \frac{a}{2} \frac{1}{n} \frac{1}{1-a i/n} \right )^2 + \cdots$$
As we did for the first term on the RHS, we may also sum over $i$ for the subsequent terms.  For example, the first error term has behavior, for large $n$
$$\frac{a^2}{8 n}\; \frac1n \sum_{i=1}^n \frac1{(1-a i/n)^2} $$
which, in the limit as $n\to\infty$, is another Riemann sum; the term therefore behaves as
$$\frac{a^2}{8(1-a)} \frac1n$$
It should be clear that higher order terms in the log behave as higher orders of $1/n$.  Thus, the error terms arranged in this way do not contribute to the sum for sufficiently large $n$.
A: $\newcommand{\+}{^{\dagger}}
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$\ds{P\equiv\lim_{n\ \to\ \infty}\prod_{i = 1}^{n}
     {1 - \pars{2i + 1}a/\pars{2n} \over 1 - ia/n}:\ {\large ?}}$

\begin{align}&\color{#c00000}{%
\sum_{i = 1}^{n}\ln\pars{1 - \pars{2i + 1}a/\pars{2n} \over 1 - ia/n}}
=\sum_{i = 1}^{n}\int_{1 - ia/n}^{1 - \pars{2i + 1}a/\pars{2n}}{\dd x \over x}
\end{align}

Set$\ds{\quad x = 1 - {a \over n}\,i - {a \over 2n}\,t}$:

\begin{align}&\color{#c00000}{%
\sum_{i = 1}^{n}\ln\pars{1 - \pars{2i + 1}a/\pars{2n} \over 1 - ia/n}}
=\sum_{i = 1}^{n}\int_{0}^{1}
{-a\,\dd t/\pars{2n} \over 1 - \pars{a/n}\,i - at/\pars{2n}}
\\[3mm]&=\half\int_{0}^{1}\sum_{i = 1}^{n}
{1 \over -n/a + i + t/2}\,\dd t
=\half\int_{0}^{1}\sum_{i = 1}^{n}
\pars{{1 \over i - n/a +  t/2} - {1 \over i + n - n/a +  t/2}}\,\dd t
\\[3mm]&=\half\int_{0}^{1}\sum_{i = 1}^{\infty}
\pars{{1 \over i - n/a +  t/2} - {1 \over i + n - n/a +  t/2}}\,\dd t
\\[3mm]&=\half\,n\int_{0}^{1}\sum_{i = 0}^{\infty}
{1 \over \pars{i + 1 - n/a +  t/2}\pars{i + 1 + n - n/a +  t/2}}\,\dd t
\\[3mm]&=\half\int_{0}^{1}\bracks{%
\Psi\pars{1 + n - {n \over a} +  {t \over 2}}
-\Psi\pars{1 - {n \over a} +  {t \over 2}}}\,\dd t
\end{align}

\begin{align}&\color{#c00000}{%
\lim_{n\ \to\ \infty}\sum_{i = 1}^{n}\ln\pars{1 - \pars{2i + 1}a/\pars{2n} \over 1 - ia/n}}
=\half\lim_{n\ \to\ \infty}\int_{0}^{1}\bracks{%
\ln\pars{n - {n \over a}} - \ln\pars{-\,{n \over a}}}\,\dd t
\\[3mm]&=\half\,\ln\pars{1 - a}=\ln\pars{\root{1 - a}}
\end{align}

$$P\equiv\color{#66f}{\large\lim_{n\ \to\ \infty}\prod_{i = 1}^{n}
{1 - \pars{2i + 1}a/\pars{2n} \over 1 - ia/n}}
=\exp\pars{\ln\pars{\root{1 - a}}}\color{#66f}=\color{#66f}{\large\root{1 - a}}
$$

