Divide a ball of volume $\frac{e^2}{6}n$ into $n$ slices of equal height. What is the product of the volumes of the slices as $n\rightarrow\infty$?

Question

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 $\frac{e^2}{6}$.)

Answer 1

Compute the Radius

If the volume of the sphere is $\frac{e^2}6n$, then we must have $$ r^3=\frac{ne^2}{8\pi}\tag1 $$

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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 $k$

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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 $2$.

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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 $$ P_n\sim2n^{-\frac1{6n}}e^{\frac{Cn+D}{n^2}}\tag7 $$

Answer 2

For simplicity's sake, let's assume $n=2m$ is even and let's assume we don't know the volume constant $V$ and say we have a sphere of radius $R_m = C\cdot m^{1/3}$. With this we can say that the volume of the $k$th slice from the "equator" would be (using the solid of revolution/cylindrical coordinates)

$$v_k = 2\pi \int_{\frac{(k-1)R_m}{m}}^{\frac{kR_m}{m}}\int_{0}^{\sqrt{R_m^2-z^2}}r\:dr\:dz = \pi\int_{\frac{(k-1)R_m}{m}}^{\frac{kR_m}{m}}R_m^2-z^2\:dz$$ $$ = \pi C^3\left[1-\frac{k^2}{m^2}+\frac{k}{m^2}-\frac{1}{3m^2}\right]$$

Now stating our problem in its final form, does the sqaure root of our desired product exist in the limit

$$\sqrt{P} = \lim_{m\to\infty}(\pi C^3)^m\prod_{k=1}^m \left[1-\frac{k^2}{m^2}+\frac{k}{m^2}-\frac{1}{3m^2}\right]$$

Taking the log of the product retrieves the following the sum

$$\lim_{m\to\infty} m \left(\log (\pi C^3) + \sum_{k=1}^m \log\left[1-\frac{k^2}{m^2}+\frac{k}{m^2}-\frac{1}{3m^2}\right]\cdot\frac{1}{m}\right)$$

The term on the right is almost a Riemann sum (courtesy of squeeze theorem), but it diverges at a rate $\theta(m)$. Fortunately for us, the term on the left diverges at exactly the same rate. The Riemann sum evaluates to

$$\int_0^1 \log(1-x^2)\:dx = \log 4 - 2$$

Immediately we get that for the product to converge at all, the constant from the beginning must have been

$$\begin{cases}\log(\pi C^3) = 2 - \log 4 \\ \frac{4\pi}{3}C^3m = V\cdot(2m)\end{cases} \implies V = \frac{e^2}{6}$$

Working back with the product, we arrive at

$$\sqrt{P} = \lim_{m\to\infty}\left(\frac{e}{2m}\right)^{2m}\prod_{k=1}^m m^2-k^2+k-\frac{1}{3}$$

From Stirling's approximation we have

$$\left(\frac{2m}{e}\right)^{2m} \sim \frac{1}{2^{2m}\sqrt{2}}\cdot\frac{(4m)!}{2^{2m}(2m)!}$$

where we recognize the term on the right as the product of the odd integers between $1$ and $4m$. Since we can multiply the integers in whichever order we want, we should be clever about the order and not blindly multiply them in an ascending fashion. This lets us rewrite the limit as

$$\sqrt{P}=\left(\lim_{m\to\infty}\frac{\left(\frac{e}{2m}\right)^{2m}}{2^{2m}\sqrt{2}\cdot\frac{2^{2m}(2m)!}{(4m)!}}\right)\cdot\left( \lim_{m\to\infty}2^{2m}\sqrt{2}\cdot\frac{2^{2m}(2m)!}{(4m)!}\prod_{k=1}^m m^2-k^2+k-\frac{1}{3}\right)$$

$$=(1)\cdot\left(\lim_{m\to\infty}\sqrt{2}\cdot\prod_{k=1}^m \frac{4m^2-4k^2+4k-\frac{4}{3}}{(2m-2k+1)(2m+2k-1)}\right)$$ $$ = \lim_{m\to\infty}\sqrt{2}\cdot\prod_{k=1}^m \frac{m^2-k^2+k-\frac{1}{3}}{m^2-k^2+k-\frac{1}{4}}$$

The product is bounded above by $1$ and numerical investigations show that the product tends towards $1$, but I have not yet found a proper lower bound for squeeze theorem. In the end though, squeeze theorem will show that the limit of the product of the slices will be

$$P = (\sqrt{P})^2 = 2$$