The ratio of volumes of a cone, half-ellipsoid, and cylinder, in higher dimensions Start with a circular region, and build a right-circular cylinder of any height. Next construct the right-triangular cone with apex the center of one of the faces of the cylinder, and between them construct the half-ellipsoid with a vertex at that same apex. I just learned a well-known fact that the ratios of the volumes of these three solids will be 1 : 2 : 3. 

If you go down a dimension, and place a triangle in a half-ellipse in a rectangle you get a ratio of areas of ${1:\frac{\pi}{2}:2}$. Generalizing this to $n$ dimensional space, starting with an $(n-1)$-ball as your base, the ratio of the volumes of the cone and cylinder will be $1$ to $n$. But how does the hyper-volume of the higher-dimensional ellipsoid fit into this ratio? 
 A: Considering Cavalieri's principle we only need to think about the case of a ball (instead of a general ellipsiod). After seeing the Wikipedia page for the $n$-ball and the closed-form and recursive formulas for a hyper-volume of an $n$-ball of radius $r$, the volume $V_n$ is given by 
$$
V_n = \frac{\sqrt{\pi^n}}{\operatorname{\Gamma}\left(\frac{n}{2}+1\right)}r^n
\quad\text{ and }\quad
V_n = \sqrt{\pi}\frac{\operatorname{\Gamma}\left(\frac{n+1}{2}\right)}{\operatorname{\Gamma}\left(\frac{n}{2}+1\right)}r V_{n-1}
\,.
$$
where $\Gamma$ is the gamma function. The ratio of these three $n$-dimensional volumes will then be
$$
\frac{1}{n}r V_{n-1}
\;:\;
\frac{1}{2}V_n
\;:\;
rV_{n-1}
\\[3ex]\text{or}\\[1ex]
1
\;:\;
\frac{\sqrt{\pi}\operatorname{\Gamma}\left(\frac{n+1}{2}\right)}{2\operatorname{\Gamma}\left(\frac{n}{2}+1\right)}n
\;:\;
n
\,.
$$
For the first few values of $n$, the middle number $B_n$ in this ratio is
$$
\begin{array}{c|cc}
n          & 2&3&4&5&6&7&8&\dotsb&\\\hline
B_n        & \frac{1}{2}\pi & 2 & \frac{3}{4}\pi & \frac{8}{3} & \frac{15}{16}\pi &\frac{16}{5} & \frac{35}{16}\pi & \dotsb\\
\end{array}
\,.
$$
You get the factor of $\pi$ only in the even numbered terms because $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$, so they will have an additional $\sqrt{\pi}$ in the numerator, whereas odd numbered terms will have a $\sqrt{\pi}$ in the denominator to cancel with the $\sqrt{\pi}$ already there.
Now let's consider what happens to this ratio of volumes as $n \to \infty$. Using Gautschi's inequality and the squeeze theorem we have, for $s=\frac{1}{2}$ and $x=\frac{n}{2}$ that 
$$
x^{1-s} < \frac{\Gamma(x+1)}{\Gamma(x+s)} < (x+1)^{1-s}\\[2ex]
\sqrt{\frac{2}{n}}n>
\frac{\Gamma\left(\frac{n}{2}+\frac{1}{2}\right)}{\operatorname{\Gamma}\left(\frac{n}{2}+1\right)}n>
\sqrt{\frac{2}{n+2}}n
\\[2ex]
\infty>
\lim_{n\to \infty}\frac{\Gamma\left(\frac{n}{2}+\frac{1}{2}\right)}{\operatorname{\Gamma}\left(\frac{n}{2}+1\right)}n>
\infty
$$
So as $n \to \infty$, the ratio of the volume of the half-ball to cone approaches infinity! Now considering the volume of the cylinder compared to the volume of the half-ball (so we divide out the factor of $n$):
$$
\sqrt{\frac{2}{n}}>
\frac{\Gamma\left(\frac{n}{2}+\frac{1}{2}\right)}{\operatorname{\Gamma}\left(\frac{n}{2}+1\right)}>
\sqrt{\frac{2}{n+2}}
\\[2ex]
0>
\lim_{n\to \infty}\frac{\Gamma\left(\frac{n}{2}+\frac{1}{2}\right)}{\operatorname{\Gamma}\left(\frac{n}{2}+1\right)}>
0
$$
So the ratio of the volume of the half-ball to the cylinder approaches zero as $n \to \infty$. This is weird, but it's not new: this oddity is basically the same oddity as the known oddity that the volume of the $n$-ball of a fixed radius approaches zero as $n \to \infty$. 
