# Volume of the unit sphere converges to $0$ faster than any $a_n := a^n$ with $0<a<1$.

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.

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?

• Using Stirling's formula, the denominator has a term $n^{n+\frac{1}{2}}$ which makes $V(n)$ go to zero very fast. – herb steinberg Dec 5 '18 at 19:47
• Could you expand on that please? I do not get it right now. – 3nondatur Dec 5 '18 at 19:51
• $V(n)\approx \frac{\pi^{n/2}}{\sqrt{\pi n}(\frac{n}{2e})^{n/2}}$ 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.) – herb steinberg Dec 5 '18 at 20:13

Consider $$0 We want to prove $$\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}