Divide a ball of volume $\frac{e^2}{6}n$ into $n$ slices of equal height, as shown below with example $n=8$. What is the limit of the product of the volumes of the slices as $n\rightarrow\infty$?
(If the image doesn't load for you, just imagine $n+1$ equally-spaced horizontal planes, and a ball that is tangent to the top and bottom planes. The planes, between the top and bottom planes, are where you cut the ball.)
I used volume of revolution, and after simplifying I got:
$$\lim_{n\to\infty}\exp{\left(2n-2n\ln{n+\sum_{k=1}^{n}}\ln{\left(k(n+1-k)-\frac{n}{2}-\frac{1}{3}\right)}\right)}$$
Wolfram does not evaluate this limit, but desmos tells me that when $n=10, 100, 1000, 10^6$, the product is approximately $1.847, 1.977, 1.997, 1.99999513$, respectively. So apparently the limit converges and equals $2$, but I do not know how to prove this.
(In case you're wondering how I got the number $\frac{e^2}{6}$: I used trial and error on desmos to hunt for the number that makes the limit converge, assuming such a number exists. I obtained a number like 1.231509. I entered this number into Wolfram and it suggested $\frac{e^2}{6}$.)
Best Answer
Compute the Radius
If the volume of the sphere is $\frac{e^2}6n$, then we must have $$ r^3=\frac{ne^2}{8\pi}\tag1 $$
Compute the Volumes of the Slices
For $1\le k\le n$, we have the volume of slice $k$ to be $$ \begin{align} &\int_{(2k-2-n)r/n}^{(2k-n)r/n}\pi(r^2-x^2)\,\mathrm{d}x\\ &=\frac{2\pi r^3}n-\frac\pi3\frac{r^3}{n^3}\left[(2k-n)^3-(2k-2-n)^3\right]\tag{2a}\\ &=\frac{2\pi r^3}n-\frac{\pi r^3}{3n^3}\left[6(2k-n)^2-12(2k-n)+8\right]\tag{2b}\\ &=\frac{2\pi r^3}n-\frac{2\pi r^3}{3n^3}-\frac{2\pi r^3}{n^3}(2k-n-1)^2\tag{2c}\\ &=\frac{8\pi r^3}{n^3}\left(\frac{n+1}2+\frac12\sqrt{n^2-\tfrac13}-k\right)\left(k-\frac{n+1}2+\frac12\sqrt{n^2-\tfrac13}\right)\tag{2d} \end{align} $$ Explanation:
$\text{(2a)}$: integrate
$\text{(2b)}$: take the difference of the cubes
$\text{(2c)}$: complete the square
$\text{(2d)}$: factor the quadratic polynomial in $k$
Compute the Product of the Volumes of the Slices
Notice that the factor in the left parentheses in $\text{(2d)}$ is the same as the one in the right parentheses under the map $k\mapsto n+1-k$. Therefore, $$ \begin{align} P_n &=\prod_{k=1}^n\int_{(2k-2-n)r/n}^{(2k-n)r/n}\pi(r^2-x^2)\,\mathrm{d}x\\[6pt] &=\left(\frac{8\pi r^3}{n^3}\right)^n\prod_{k=1}^n\left(k-\frac{n+1}2+\frac12\sqrt{n^2-\tfrac13}\right)^2\tag{3a}\\ &=\left(\frac{8\pi r^3}{n^3}\right)^n\frac{\Gamma\left(\frac{n+1}2+\frac12\sqrt{n^2-\frac13}\right)^2}{\Gamma\left(-\frac{n-1}2+\frac12\sqrt{n^2-\frac13}\right)^2}\tag{3b}\\ &=\left(\frac{8\pi r^3}{n^3}\right)^n\frac{\Gamma\left(n+\frac12-\frac1{12n}+O\!\left(\frac1{n^3}\right)\right)^2}{\Gamma\left(\frac12-\frac1{12n}+O\!\left(\frac1{n^3}\right)\right)^2}\tag{3c}\\[6pt] &=\left(\frac{8\pi r^3}{n^3}\right)^n\frac{n!^2}{\pi n}n^{-\frac1{6n}+O\left(\frac1{n\log(n)}\right)}\tag{3d}\\[6pt] &=\left(\frac{e^n}{n^n}\right)^2\frac{2\pi n\left(\frac{n^n}{e^n}\right)^2}{\pi n}n^{-\frac1{6n}+O\left(\frac1{n\log(n)}\right)}\tag{3e}\\[15pt] &=2n^{-\frac1{6n}+O\left(\frac1{n\log(n)}\right)}\tag{3f} \end{align} $$ Explanation:
$\text{(3a)}$: use $\text{(2d)}$
$\text{(3b)}$: write the product as a ratio of Gamma functions
$\text{(3c)}$: $\sqrt{n^2-\frac13}=n-\frac1{6n}+O\!\left(\frac1{n^3}\right)$
$\text{(3d)}$: Gautschi's inequality yields $\Gamma(x+s)=\Gamma(x)x^se^{O(1/x)}$ for $0\le s\le1$
$\text{(3e)}$: Stirling's approximation
$\text{(3f)}$: simplify
$\text{(3f)}$ says that the limit of the product is indeed $2$.
Further Results
If we use the more precise asymptotic approximations $$ \Gamma(x+s)\sim\Gamma(x)x^se^{-\frac{s(1-s)}{2x}}\tag4 $$ and $$ n!\sim\sqrt{2\pi n}\frac{n^n}{e^n}e^{\frac1{12n}}\tag5 $$ we get that the coefficient of the $O\!\left(\frac1{n\log(n)}\right)$ term in $\text{(3f)}$ is $$ \begin{align} C &=\frac16\frac{\Gamma'(1/2)}{\Gamma(1/2)}-\frac1{12}\tag{6a}\\[6pt] &=-0.41058500433690391324\tag{6b} \end{align} $$ According to the table $$ \begin{array}{r|l} n&n^2\log(P_n/2)+\frac16n\log(n)-Cn\\\hline 1&-0.074321645451096396990\\ 10&-0.040105155755794398809\\ 100&-0.035166254783256555024\\ 1000&-0.034391063053238965551\\ 1000000&-0.034269677264259024828\\ 10000000&-0.034269484677864450270 \end{array} $$ if we let $D=-0.034269$, we apparently get $$ P_n\sim2n^{-\frac1{6n}}e^{\frac{Cn+D}{n^2}}\tag7 $$