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.

Solution 1:

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 $n$.

Solution 2:

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.