Answer
${A_8} \simeq 4.059$
${A_9} \simeq 3.298$
${A_{10}} \simeq 2.55$
We show that ${C_n} \le 1$ for $n \ge 6$ and that of all unit balls, the five-dimensional ball has the largest volume.
Using the formulas for the volume ${A_n}$ of the unit ball, we show that ${A_n} \to 0$ as $n \to \infty $.
Work Step by Step
We have the formulas for the volume ${A_n}$ of the unit ball in ${\mathbb{R}^n}$ given by
${A_{2m}} = \frac{{{\pi ^m}}}{{m!}}$, ${\ \ \ \ \ }$ ${A_{2m + 1}} = \frac{{{2^{m + 1}}{\pi ^m}}}{{1\cdot3\cdot5\cdot\cdot\cdot\left( {2m + 1} \right)}}$
So,
${A_8} = {A_{2\cdot4}} = \frac{{{\pi ^4}}}{{4!}} \simeq 4.059$
${A_9} = {A_{2\cdot4 + 1}} = \frac{{{2^{4 + 1}}{\pi ^4}}}{{1\cdot3\cdot5\cdot\cdot\cdot\left( {2\cdot4 + 1} \right)}} = \frac{{{2^{4 + 1}}{\pi ^4}}}{{1\cdot3\cdot5\cdot7\cdot9}} \simeq 3.298$
${A_{10}} = {A_{2\cdot5}} = \frac{{{\pi ^5}}}{{5!}} \simeq 2.55$
Using the definition ${C_n} = \mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} {\cos ^n}\theta {\rm{d}}\theta $ we compute ${C_0}$ and ${C_1}$:
${C_0} = \mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} {\cos ^0}\theta {\rm{d}}\theta = \left( {\theta |_{ - \pi /2}^{\pi /2}} \right)$
${C_0} = \pi $
${C_1} = \mathop \smallint \limits_{\theta = - \pi /2}^{\pi /2} {\cos ^1}\theta {\rm{d}}\theta = \left( {\sin \theta |_{ - \pi /2}^{\pi /2}} \right)$
${C_1} = 2$
Using Eq. (6) we compute the volume ${V_1}\left( r \right)$ and ${V_2}\left( r \right)$:
${V_n}\left( r \right) = {A_n}{r^n}$
${V_1}\left( r \right) = {A_1}r$, ${\ \ \ \ \ }$ ${V_2}\left( r \right) = {A_2}{r^2}$
Since ${V_1}\left( r \right) = 2r$ and ${V_2}\left( r \right) = \pi {r^2}$, so ${A_1} = 2$ and ${A_2} = \pi $.
Using the results obtained above: ${C_0} = \pi $, ${C_1} = 2$, ${A_1} = 2$, ${A_2} = \pi $, Eq. (7) and Eq. (8), we generate a table for some values of ${C_n} = \left( {\frac{{n - 1}}{n}} \right){C_{n - 2}}$ and ${A_n} = {A_{n - 1}}{C_n}$:
$\begin{array}{*{20}{c}}
n&0&1&2&3&4&5&6&7&8&9\\
{{C_n}}&\pi &2&{1.57}&{1.33}&{1.18}&{1.07}&{0.98}&{0.91}&{0.86}&{0.81}\\
{{A_n}}& - &2&\pi &{4.19}&{4.93}&{5.26}&{5.17}&{4.72}&{4.059}&{3.299}
\end{array}$
The table above shows that ${C_n} \le 1$ for $n \ge 6$ and that of all unit balls, the five-dimensional ball has the largest volume. Since $\frac{{n - 1}}{n} < 1$, for $n = 1,2,...$, so ${C_n} < {C_{n - 1}}$. Therefore ${C_n} \to 0$ as $n \to \infty $.
Recall from the formulas:
${A_{2m}} = \frac{{{\pi ^m}}}{{m!}}$, ${\ \ \ \ \ }$ ${A_{2m + 1}} = \frac{{{2^{m + 1}}{\pi ^m}}}{{1\cdot3\cdot5\cdot\cdot\cdot\left( {2m + 1} \right)}}$
Because the factorials in the denominators grow faster than the exponentials in the numerators, we conclude that ${A_n} \to 0$ as $n \to \infty $.
