HMMT February

Solution:
Answer: $\frac{216 \pi^{3}-2187 \sqrt{3}}{8 \pi^{2}}$

First, we find the total amount of juice consumed. We can simply subtract the amount of juice remaining at infinity from the initial amount of juice in the cup, which of course is simply the volume of the cup; we'll denote this value by $V$.

Since volume in the cup varies as the cube of height, the amount of juice remaining in the cup after $m$ minutes is
$$V \cdot \prod_{n=1}^{m}\left(\frac{9-\frac{1}{n^{2}}}{9}\right)^{3}=V \cdot\left(\prod_{n=1}^{m}\left(1-\frac{1}{9 n^{2}}\right)\right)^{3}$$
We can now factor the term inside the product to find
$$V\left(\prod_{n=1}^{m} \frac{(3 n+1)(3 n-1)}{9 n^{2}}\right)^{3}=V\left(\frac{(3 m+1)!}{3^{3 m}(m!)^{3}}\right)^{3}$$
It remains to evaluate the limit of this expression as $m$ goes to infinity.

However, by Stirling's approximation, we have
$$\begin{aligned} \lim _{m \rightarrow \infty} \frac{(3 m+1)!}{3^{3 m}(m!)^{3}} & =\lim _{m \rightarrow \infty} \frac{\left(\frac{3 n+1}{e}\right)^{3 n+1} \cdot \sqrt{2 \pi(3 n+1)}}{\left(\frac{3 n}{e}\right)^{3 n} \sqrt{(2 \pi n)^{3}}} \\ & =\lim _{m \rightarrow \infty} \frac{(3 n+1) \sqrt{3}}{2 \pi n e}\left(\frac{3 n+1}{3 n}\right)^{3 n} \\ & =\frac{3 \sqrt{3}}{2 \pi} . \end{aligned}$$
Therefore the total amount of juice the king consumes is
$$V-V\left(\frac{3 \sqrt{3}}{2 \pi}\right)^{3}=\left(\frac{3^{2} \cdot \pi \cdot 9}{3}\right)\left(\frac{8 \pi^{3}-81 \sqrt{3}}{8 \pi^{3}}\right)=\frac{216 \pi^{3}-2187 \sqrt{3}}{8 \pi^{2}}$$

Remark. We present another way to calculate the limit at $m \rightarrow \infty$ of $f(m)=\frac{(3 m+1)!}{3^{3 m}(m!)^{3}}$. We have
$$f(m+1)=\frac{(3 m+4)!}{3^{3 m+3}(m+1)!^{3}}=f(m) \frac{\left(m+\frac{2}{3}\right)\left(m+\frac{4}{3}\right)}{(m+1)^{2}}$$
whence we can write
$$f(m)=\frac{c \Gamma\left(m+\frac{2}{3}\right) \Gamma\left(m+\frac{4}{3}\right)}{\Gamma(m+1)^{2}}$$
for some constant $c$. We can find $c$ by equating the expressions at $m=0$; we have
$$1=f(0)=\frac{c \Gamma\left(\frac{2}{3}\right) \Gamma\left(\frac{4}{3}\right)}{\Gamma(1)^{2}}$$
so that $c=\Gamma(1)^{2} / \Gamma\left(\frac{2}{3}\right) \Gamma\left(\frac{4}{3}\right)$.
Of course, $\Gamma(1)=0!=1$. We can evaluate the other product as follows:
$$\Gamma\left(\frac{2}{3}\right) \Gamma\left(\frac{4}{3}\right)=\frac{1}{3} \Gamma\left(\frac{2}{3}\right) \Gamma\left(\frac{1}{3}\right)=\frac{1}{3} \cdot \frac{\pi}{\sin \pi / 3}=\frac{2 \pi}{3 \sqrt{3}}$$
Here the first step follows from $\Gamma(n+1)=n \Gamma(n)$, while the second follows from Euler's reflection formula. Thus $c=3 \sqrt{3} / 2 \pi$. We can now compute
$$\lim _{m \rightarrow \infty} f(m)=\lim _{m \rightarrow \infty} \frac{c \Gamma\left(m+\frac{2}{3}\right) \Gamma\left(m+\frac{4}{3}\right)}{\Gamma(m+1)^{2}}=\frac{3 \sqrt{3}}{2 \pi} \lim _{m \rightarrow \infty} \frac{\Gamma\left(m+\frac{2}{3}\right) \Gamma\left(m+\frac{4}{3}\right)}{\Gamma(m+1)^{2}}$$

Since $\lim _{n \rightarrow \infty} \Gamma(n+\alpha) /\left[\Gamma(n) n^{\alpha}\right]=1$, this final limit is $1$ and $f(m) \rightarrow 3 \sqrt{3} / 2 \pi$ as $m \rightarrow \infty$.