I just read the Wikipidia article "n-sphere" where I quote : "Interestingly, given the radius R, the volume and the surface area of the n-sphere reaches a maximum and then decrease towards zero as the dimension n increases". Can you explain it in an intuitive way ? I think the math behind the formula of the surface or volume of an n-sphere is too difficult for me too understand. I just would like to understand why is the volume/surface decrease and reach 0. That seems unbelievable for me.
Thank you!
Let $V_n(r)=A_nr^n$ be the volume of the $n$-sphere of radius $r$. Then $$A_n=V_n(1)=\int_{-1}^1V_{n-1}(\sqrt {1-x^2}\;)\;dx=$$ $$=2\int_0^1V_{n-1}(\sqrt { 1-x^2}\;)\;dx=$$ $$=2\int_0^1A_{n-1}\cdot (1-x^2)^{(n-1)/2}\;dx.$$
For very large $n$ and for $x\in [0,1]$ the value of $(1-x^2)^{(n-1)/2}$ is small except when $x$ is near $0$.
For example when $n\geq 405$ and $x\in [0.1,1]$ we have $$(1-x^2)^{(n-1)/2}\leq (1-.01)^{202}<1/7.$$ And for $x\in [0,0.01]$ we have (obviously) $(1-x^2)^{(n-1)/2}\leq 1.$
So for $n\geq 405$ we have $$A_n/A_{n-1} =2\int_0^1 (1-x^2)^{(n-1)/2}\;dx\leq$$ $$\leq 2\int_0^{0.01}1\;dx+2\int_{0.01}^1(1/7)\;dx<1/3.$$ So $A_n <(A_{n-1})/3$ for all $ n\geq 405,$ so $A_n$ must converge to $0.$