I do not know how to solve the following problem:

Prove that the volume of the $n$-dimensional unit-sphere converges faster to >$0$ than any sequence of the form $a_n := a^n$ for $0<a<1$.

I know that the volume of the $n$-dimensional unit-sphere is given by

$$V(n)=\frac{\pi^{n/2}}{\Gamma\left(\frac{n}{2}+1\right)} $$

and I know about Stirlings formula

$$lim_{z \rightarrow \infty} \frac{\Gamma(z+1)}{\sqrt{2 \pi z}(\frac{z}{e})^z } = 1$$

which might be useful in the context of this exercise. Could you help me?

  • 1
    $\begingroup$ Using Stirling's formula, the denominator has a term $n^{n+\frac{1}{2}}$ which makes $V(n)$ go to zero very fast. $\endgroup$ – herb steinberg Dec 5 '18 at 19:47
  • $\begingroup$ Could you expand on that please? I do not get it right now. $\endgroup$ – 3nondatur Dec 5 '18 at 19:51
  • $\begingroup$ $V(n)\approx \frac{\pi^{n/2}}{\sqrt{\pi n}(\frac{n}{2e})^{n/2}}$ The essential point is that $\frac{(2\pi e)^{n/2}}{n^{(n+1)/2}}$ goes to zero faster than $a^n$ for any fixed $a$. (Note minor error in previous comment.) $\endgroup$ – herb steinberg Dec 5 '18 at 20:13

Consider $0<a<1.$ We want to prove $\lim\limits_{n\to \infty}\frac{V(n)}{a^n}=0,$ or equivalently $\lim\limits_{n\to \infty}\log\left(\frac{V(n)}{a^n}\right)=-\infty.$

Due to the Stirling formula, for large values of $n$ replace $\Gamma(\frac n2+1)$ by $\sqrt{\pi n} \left(\frac{n}{2 e}\right)^{n/2}.$ Compute $$\begin{aligned}\log\left(\frac{V(n)}{a^n}\right)&=\log V(n)-n\log a\\ &\sim \frac{n}{2}\log{\pi}-\log\left(\sqrt{\pi n} \left(\frac{n}{2 e}\right)^{n/2}\right)-n\log a\\&=\frac{n}{2} \log\frac{{\pi}}{a^2}-\frac{1}{2}\log(\pi n)-\frac{n}{2}\log\frac{n}{2e}\\&=\frac{n}{2}\log\left(\frac{2\pi e}{a^2\cdot n \cdot(\pi n)^{1/n}}\right)\to \infty\times(-\infty)\end{aligned}$$


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