In an equilateral triangle, line segments connect a vertex to $n$ uniformly distributed points on the opposite side, including points at the ends. Assuming $\lim_{n\to\infty}\text{(product of lengths of line segments)}$ exists and is positive, what is the value of the limit?
My attempt:
Let $x=\text{side length}$.
The limit expression is:
$$L=\lim_{n\to\infty}\prod_{k=0}^n{x\sqrt{\frac{3}{4}+\left(\frac{k}{n}-\frac{1}{2}\right)^2}}$$
The idea is to find $x$ so that the limit exists and is positive, and then evaluate the limit.
I took the log of the product to change it to a series of logs, then used the Maclaurin expansion of $\ln{(1+x)}$ and the fact that $\sum_{k=0}^n{k^p}=\frac{1}{p+1}n^{p+1}+\frac{1}{2}n^p+…$, but the series turned into a mess.
I conjecture that $x=e^{1-\frac{\pi}{2\sqrt{3}}}\approx{1.098}$ and that (surprisingly) $L=x$.
(The basis of my conjecture is that I solved a similar but easier question: In a $45^{\text{o}}$ right triangle, line segments connect a $45^{\text{o}}$ vertex to $n$ uniformly distributed points on the opposite side, including points at the ends. Assuming $\lim_{n\to\infty}\text{(product of lengths of line segments)}$ exists and is positive, what is the value of the limit? Using the method I described above, I found that the length of the leg of the triangle is $(2^{-1/2})e^{1-\frac{\pi}{4}}\approx0.876$ and the limit is $(2^{-1/4})e^{1-\frac{\pi}{4}}\approx1.042$. Then I used desmos to work out my conjecture.)
Context: I was thinking about infinite products of areas and lengths in various shapes, and I came up with this question.
My background: high school math teacher.
Best Answer
The question is how does $\displaystyle f(n)=\prod_{k=0}^n{\sqrt{\frac{3}{4}+\left(\frac{k}{n}-\frac{1}{2}\right)^2}}$ behave as $n\to\infty$.
Indeed, looking at $g(n)=2\times\ln{f(n)}=\displaystyle \sum_{k=0}^n{\ln\left(\frac{3}{4}+\left(\frac{k}{n}-\frac{1}{2}\right)^2\right)}$,
think about this as the approximation of an integral except that you forgot to account for the width of each rectangle, i.e. you only added up the heights of the rectangles.
So $\dfrac{g(n)}{n}\to\displaystyle\int_0^1 \ln\left(\dfrac{3}{4}+\left(x-\dfrac{1}{2}\right)^2\right)\text{d}x=\dfrac{\pi}{\sqrt{3}}-2$, which you can check on Wolfram Alpha.
I'll call this constant $r$, to save space.
It follows easily that $L(x)=\displaystyle\lim_{n\to\infty}\prod_{k=0}^n{x\,\sqrt{\frac{3}{4}+\left(\frac{k}{n}-\frac{1}{2}\right)^2}}=\begin{cases} 0\;\;\;\text{if}\;\;0<x<\exp\dfrac{-r}{2}\\ \infty\;\,\text{if}\;\;x>\exp\dfrac{-r}{2}\\ \end{cases}$.
$\\$
But how about the case $x=\exp\dfrac{-r}{2}$?
Go back to the earlier graph of the integral. Here's the hand-waving version of an argument you can make more precise if you have the time:
Again, I'm talking about $j(x)=\ln\left(\dfrac{3}{4}+\left(x-\dfrac{1}{2}\right)^2\right)$. Draw the graph of $j(x)$ from $c$ to $c+\Delta x$. The area under the curve is roughly $j(c)\Delta x+\dfrac{1}{2}j'(c)(\Delta x)^2+O((\Delta x)^3)$.
I'll leave it to you as an exercise to show that $rn-g(n)=O\left(\dfrac{1}{n}\right)$. Hint: it involves integrating $j'(x)$ from $0$ to $1$.
Therefore, $L\left(\exp\dfrac{-r}{2}\right)=\exp\dfrac{-r}{2}$.