# Product of areas in a disk

geometryinfinite-productlimits

A disk of area $$2n$$ is divided into $$n$$ regions by drawing $$n$$ evenly spaced points on the perimeter and then drawing line segments joining one fixed point with all the other points. An example is shown with $$n=8$$.

What is the limit of the product of the areas of the regions, as $$n$$ approaches $$\infty$$ ?

Using basic trigonometry, I've got:

$$\lim_{n\to\infty}\prod_{k=1}^{n}\left(2-\frac{n}{\pi}\sin\left(\frac{2k\pi}{n}\right)+\frac{n}{\pi}\sin\left(\frac{2(k-1)\pi}{n}\right)\right)$$

I do not know how to evaluate this limit. Wolfram does not evaluate the limit, but tells me that when $$n=10000$$ the product is approximately $$8.3$$.

UPDATE1:

I am fairly confident that the limit is $$4\cosh^2\left({\frac{\pi}{2\sqrt{3}}}\right)=8.29674…$$

Here's why. I was trying to answer a similar question: A ball is divided into $$n$$ concentric shells of equal thickness. Can the average volume of the shells be fixed so that the product of the volumes converges to a positive number as $$n\to\infty$$? The answer turns out to be yes. If we fix the average volume of the shells to be $$\frac{e^2}{3}$$ then the product of the volumes converges to $$2\cosh\left({\frac{\pi}{2\sqrt{3}}}\right)=2.8804…$$ I noticed that this number seemed to be (to many decimal places) the square root of the answer to the question here. I think there must be a connection.

UPDATE2:

Another similar question is this. A disk is divided into $$n$$ regions by equally spaced parallel lines, with the perimeter of the disk being tangent to two of the lines. Can the average area of the regions be fixed so that the product of the areas converges to a positive number as $$n\to\infty$$? The answer seems to be yes. If we fix the average area of the regions to be $$\frac{{\pi}e}{8}$$ then the product of the areas seems to converge to $$2\cos\left(\frac{\pi}{2\sqrt{3}}\right)$$ (notice this is cos, not cosh). Again, I think there must be a connection.

(As for the source of the original question, I thought of the question by myself. It was inspired by an IB (high school) May 2021 exam question, which goes like this. A unit circle has $$n$$ evenly distributed points. Line segments are drawn joining one point with all the other points (like a seashell). The exam question led students to find the product of the lengths of the line segments, using complex numbers. The answer turns out to be $$n$$. Then I wondered, what is the product of the areas of the enclosed regions? Obviously the product approaches $$0$$ as $$n\to\infty$$ (since all the areas approach $$0$$). But could the average area be fixed so that the product of the areas converges to a positive number as $$n\to\infty$$? Definitely beyond the course syllabus. After experimenting on desmos, I found that the answer seems to be yes: if we fix the average area to be $$2$$, then the product of the areas seems to converge to a positive number. The question in this post is, what does it converge to?)

The limit is, as suspected, $$4\cosh^2\left(\frac{\pi}{2\sqrt 3}\right).$$ First, define $$\epsilon_n=1-\frac n\pi\sin(\pi/n)$$, so that $$\epsilon_n=\frac{\pi^2}{6n^2}-O(n^{-3})$$ and \begin{align*} a_n &:=\sum_{k=1}^n\log\left(2-\frac n\pi\sin\left(\frac{2k\pi}n\right)+\frac n\pi\sin\left(\frac{2(k-1)\pi}n\right)\right)\\ &=\sum_{k=1}^n\log\left(2-\frac n\pi\left[\sin\left(\frac{2k\pi}n\right)-\sin\left(\frac{2(k-1)\pi}n\right)\right]\right)\\ &=\sum_{k=1}^n\log\left(2-2\frac n\pi\sin\left(\frac\pi n\right)\cos\left(\frac{(2k-1)\pi}n\right)\right)\\ &=\sum_{k=1}^n\log\left(2-2\cos\left(\frac{(2k-1)\pi}n\right)+2\cos\left(\frac{(2k-1)\pi}n\right)\epsilon_n\right). \end{align*} Define $$b_n:=\sum_{k=1}^n\log\left(2-2\cos\left(\frac{(2k-1)\pi}n\right)\right),$$ so that $$c_n:=a_n-b_n=\sum_{k=1}^n\log\left(1+\frac{\epsilon_n\cos\left(\frac{(2k-1)\pi}n\right)}{1-\cos\left(\frac{(2k-1)\pi}n\right)}\right).$$ The summand here is symmetric about $$k=\frac{n+1}2$$, and is $$\log(1-\epsilon_n/2)=o(1)$$ at $$k=\frac{n+1}2$$, so $$c_n=o(1)+2\sum_{k=1}^{\lfloor n/2\rfloor}\log\left(1+\frac{\epsilon_n\cos\left(\frac{(2k-1)\pi}n\right)}{1-\cos\left(\frac{(2k-1)\pi}n\right)}\right).$$ Write $$c_{n,K}=\sum_{k=1}^K\log\left(1+\frac{\epsilon_n\cos\left(\frac{(2k-1)\pi}n\right)}{1-\cos\left(\frac{(2k-1)\pi}n\right)}\right).$$ We have, for large $$n$$ (being liberal with constant factors), that \begin{align*} |c_n-2c_{n,K}| &\leq 2\sum_{k=K+1}^{\lfloor n/2\rfloor}\left|\log\left(1+\frac{\epsilon_n\cos\left(\frac{(2k-1)\pi}n\right)}{1-\cos\left(\frac{(2k-1)\pi}n\right)}\right)\right|\\ &\leq 4\sum_{k=K+1}^{\lfloor n/2\rfloor}\frac{\epsilon_n}{1-\cos\left(\frac{(2k-1)\pi}n\right)}\\ &=2\epsilon_n\sum_{k=K+1}^{\lfloor n/2\rfloor}\frac1{\sin^2\left(\frac{(2k-1)\pi}{2n}\right)}\\ &\leq 4\epsilon_n\sum_{k=K+1}^{\lfloor n/2\rfloor}\frac1{\left(\frac{(2k-1)\pi}{2n}\right)^2}\\ &=\frac{16\epsilon_nn^2}{\pi^2}\sum_{k=K+1}^{\lfloor n/2\rfloor}\frac1{(2k-1)^2}\\ &\leq\frac 83\sum_{k=K+1}^\infty \frac1{(2k-1)^2}\leq \frac2{3K}. \end{align*} On the other hand, for $$n$$ large and $$k, \begin{align*} \log\left(1+\frac{\epsilon_n\cos\left(\frac{(2k-1)\pi}n\right)}{1-\cos\left(\frac{(2k-1)\pi}n\right)}\right) &=\log\left(1+\frac{\left(\frac{\pi^2}{6n^2}+O(n^{-3})\right)\left(1-O(n^{-1})\right)}{2\sin^2\left(\frac{(2k-1)\pi}{2n}\right)}\right)\\ &=\log\left(1+\frac{\left(\frac{\pi^2}{6n^2}+O(n^{-3})\right)\left(1-O(n^{-3/2})\right)}{\frac{(2k-1)^2\pi^2}{2n^2}+O(n^{-3})}\right)\\ &=\log\left(1+\frac1{3(2k-1)^2}+O(n^{-1})\right)\\ &=\log\left(1+\frac1{3(2k-1)^2}\right)+O\left(\frac 1n\right). \end{align*} This means that, setting $$K=n^{1/4}$$, $$c_n=2c_{n,K}+o(1)=o(1)+2\sum_{k=1}^K\log\left(1+\frac1{3(2k-1)^2}\right),$$ and thus $$\lim_{n\to\infty} c_n=2\sum_{k=1}^\infty \log\left(1+\frac1{3(2k-1)^2}\right).$$ Now, \begin{align*} e^{b_n} &=\prod_{k=1}^n\left(2-2\cos\left(\frac{(2k-1)\pi}n\right)\right)\\ &=\prod_{k=1}^n\left|1-e^{\frac{(2k-1)\pi i}n}\right|^2\\ &=\prod_{z^n=-1}\left|1-z\right|^2\\ &=(1-(-1))^2=4. \end{align*} This shows that $$\lim_{n\to\infty} a_n=2\ln 2+\lim_{n\to\infty}c_n=2\ln 2+2\sum_{k=1}^\infty \log\left(1+\frac1{3(2k-1)^2}\right).$$ So, we need only evaluate this infinite series, which we call $$S$$. We first expand the $$\log$$ to get $$S=\sum_{k=1}^\infty\sum_{m=1}^\infty \frac{(-1)^{m-1}}{m3^m(2k-1)^{2m}}=\sum_{m=1}^\infty \frac{(-1)^{m-1}}{m3^m}\sum_{k=1}^\infty \frac1{(2k-1)^{2m}},$$ where the interchange of summation is justified since the sum is absolutely convergent. Using that $$\sum_{k=1}^\infty \frac1{(2k-1)^{2m}}=\left(1-2^{-2m}\right)\zeta(2m),$$ we get $$S=\sum_{m=1}^\infty \frac{(-1)^{m-1}}m\left(3^{-m}-12^{-m}\right)\zeta(2m).$$ Write $$f(x)=\sum_{m=1}^\infty\frac{(-1)^{m-1}x^m\zeta(2m)}m.$$ We compute $$f'(x)=\sum_{m=1}^\infty(-x)^{m-1}\zeta(2m).$$ An identity here gives that $$\sum_{m=1}^\infty t^{2m}\zeta(2m)=\frac{1-\pi t\cot(\pi t)}{2},$$ and so $$f'(x)=-\frac{1-\pi i\sqrt{x}\cot(\pi i\sqrt x)}{2x}=\frac{\pi\sqrt x\coth(\pi\sqrt x)-1}{2x}.$$ This gives \begin{align*} S &=f\left(\frac13\right)-f\left(\frac1{12}\right)\\ &=\int_{1/12}^{1/3}\frac{\pi\sqrt x\coth(\pi\sqrt x)-1}{2x}dx\\ &=\int_{\frac1{2\sqrt3}}^{\frac1{\sqrt3}}\frac{\pi y\coth(\pi y)-1}ydy\\ &=\int_{\frac1{2\sqrt3}}^{\frac1{\sqrt3}}\pi\coth(\pi y)dy-\log 2\\ &=\log(\sinh(\pi y))\bigg|_{\frac1{2\sqrt3}}^{\frac1{\sqrt3}}-\log 2\\ &=\log\left(\frac{\sinh\left(\frac{\pi}{\sqrt 3}\right)}{2\sinh\left(\frac{\pi}{2\sqrt 3}\right)}\right)\\ &=\log\cosh\frac{\pi}{2\sqrt 3}. \end{align*} So, the limit is $$e^{2\ln 2+2S}=4\cosh^2\left(\frac{\pi}{2\sqrt 3}\right),$$ as desired.