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I know that the volume of the $n$-ball is given by $$a_n:=\frac{\pi^{n/2}}{\Gamma\left(\frac{n}{2}+1\right)}$$ and know, that the volume goes to $0$ for $n$ tending to $\infty$ (I even know that $\sum a_nx^n$ is absolute convergent for all $x\in \Bbb{R}$).

But why is the volume maximal for $n=5$? I can calculate $a_n$ for $n=1,\ldots,6$ and see that this sequence is increasing until $n=5$ and $a_6<a_5$. But how do I know that there isn't another peak, before the $a_n$ are strictly decreasing to $0$?

Edit: My question is not why $a_n\rightarrow 0$ for $n\rightarrow \infty$.

Nathaniel Bubis
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Luke Mathwalker
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2 Answers2

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Unfortunately, I cannot close this question as a duplicate, since the same question appeared on mathoverflow:

https://mathoverflow.net/questions/53119/volumes-of-n-balls-what-is-so-special-about-n-5

To summarize, the answers there show that the choice to measure the dimensionless $V_n(R)/R^n$ is in fact arbitrary, and one can create other dimensionless quantities with different maximum $n$, though all will eventually decrease quickly to zero. Particularly for the case of the volume ratio of the ball to it's enclosing cube, user @Marty points out that:

More interesting geometrically might be the equally dimensionless ratio $V_n(R)/(2R)^n$, which is the ratio of the volume of the $n$-ball to the volume of the smallest $n$-cube containing it. This is monotonic decreasing (for $n≥1$), showing that balls decrease in volume relative to their smallest cubical container, as the dimension increases.

As to proving this arbitrary value, $$V'(n) = 0 \rightarrow (\ln \pi) \pi^{n/2} \Gamma(n/2+1) - \pi^{n/2}\Gamma(n/2+1)\psi_0(n/2+1)=0$$ $$\ln \pi= \psi_0(n/2+1)$$ Since the digamma function is monotonically increasing (when $n>0$), there can only be one solution, which numerically can be found at $n\approx5.26$. You can now check which of $n=5,6$ is the highest integer value.

Nathaniel Bubis
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Consider $\frac{a_{n+2}}{a_n}$. It holds that $$\frac{a_{n+2}}{a_n}=\frac{\pi^{\frac{n+2}{2}}}{\Gamma\left(\frac{n+2}{2}+1\right)}\cdot \frac{\Gamma\left(\frac{n}{2}+1\right)}{\pi^{\frac{n}{2}}}=\pi\frac{\Gamma\left(\frac{n}{2}+1\right)}{\Gamma\left(\left(\frac{n}{2}+1\right)+1\right)}$$ Now use the fact that $\Gamma(x+1)=x\Gamma(x)$ for $x>0$ to obtain that $$\frac{a_{n+2}}{a_n}=\pi\frac{\Gamma\left(\frac{n}{2}+1\right)}{\Gamma\left(\frac{n}{2}+1\right)\left(\frac{n}{2}+1\right)}=\frac{\pi}{\frac{n}{2}+1}$$ so $\frac{a_{n+2}}{a_n}>1$ iff $$n<2(\pi-1)\Leftrightarrow n\leq4$$ Since $a_6<a_5$ this yields that $a_n$ is maximal for $n=5$.

Nightgap
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