Ratio of volumes of $n$-dimensional unit balls

Question

Prove that $$\left( \frac{nS_n}{S_{n-1}}\right)^{1/n} \le 2$$ where $S_i$ is the volume of the $i$th dimensional unit ball and $n\ge 2$.

I think we can use the fact that an $n$-dimensional unit ball is contained in a hypercube whose edge measures $2$ units, and the ball contains a hypercube whose edge measures $\sqrt{2}$ units. So we have $\sqrt{2}^n\le S_n \le 2^n$ and thus $$\left( \frac{nS_n}{S_{n-1}}\right)^{1/n} \le \left( \frac{n2^n}{\sqrt{2}^{n-1}} \right)^{1/n} = n^{1/n}2^{\frac{n+1}{2n}} = n^{1/n}\sqrt{2}\sqrt{2}^{1/n}.$$

So if we take $n^{1/n} \le 2$ and $\sqrt{2}^{1/n}\le \sqrt{2}^{1/2}$, then an upper bound to my problem is $2^{1+3/4}$ which isn't good enough.

Do we need another approach or do we get better bounds on $n^{1/n}$?

Answer 1

It is enough to show that $S_n\leq 2\,S_{n-1}$, which is trivial, since

$$\{(x_1,\ldots,x_{n-1}):x_1^2+\ldots+x_{n-1}^2\leq 1\}\times[-1,1]\supset \{(x_1,\ldots,x_{n}):x_1^2+\ldots+x_{n}^2\leq 1\}.$$ Over $n\geq 2$, $(2n)^{1/n}$ attains its maximum, $2$, at $n=2$.

Answer 2

Since the volume of a $n$-dimensional unit ball is (see volume of an $n$- ball)$$ S_n = \frac{π^{\frac{n}{2}}}{Γ\left( n + \frac{1}{2} \right)}, $$ then\begin{align*} \left( \frac{nS_n}{S_{n - 1}} \right)^{\frac{1}{n}} \leqslant 2 &\Longleftrightarrow \frac{n \sqrt{π} \cdot Γ\left( n - \frac{1}{2} \right)}{Γ\left( n + \frac{1}{2} \right)} \leqslant 2^n\\ &\Longleftrightarrow \frac{n \sqrt{π}}{n - \frac{1}{2}} \leqslant 2^n. \end{align*} Because$$ 2^n \geqslant 4 > 2 \sqrt{π} \geqslant \frac{n \sqrt{π}}{n - \frac{1}{2}}, $$ then $\displaystyle \left( \frac{nS_n}{S_{n - 1}} \right)^{\frac{1}{n}} \leqslant 2$.