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Let $B^n(r)$ be the $n$-ball of radius $r$. A standard (easy) problem for first year calculus students is the following.

$(1)$ Show that $$ \lim_{n\to \infty} \frac{\text{Vol}(B^n(r))}{\text{Vol}(B^n(1)\setminus B^n(r))}=0$$ for $0\leq r<1$.

An equivalent problem is:

$(2)$ Let $r_n$ be such that $2\cdot\text{Vol}(B^n(r_n))=\text{Vol}(B^n(1))$; show that $$\lim_{n\to \infty} r_n=1.$$

There is a synthetic proof of (1) for $r\leq 1/2$, which goes as follows. Let $C$ be a cube of side-length $2r$ circumscribing $B^n(r)$. Then one may place $2n$ disjoint hemispheres of radius $r$ on the faces of $C$ (with the bases of the hemispheres inscribed in the faces of $C$) such that these hemispheres are all contained in $B^n(r)$. But then we must have that $$\text{Vol}(B^n(1)\setminus B^n(r))\geq n \cdot \text{Vol}(B^n(r))$$ which completes the proof. All this really uses is the Pythagorean theorem, to check that the hemispheres are contained in $B^n(1)$.

This proof is unsatisfying for two reasons---first, it only seems to work for $r\leq 1/2$, and second, the rate of convergence one gets is $\sim1/n$, rather than the actual exponential convergence.

So my question is

Is there a synthetic proof of (1) or (2) that gives the correct convergence rate? Is there a synthetic proof of (1) for $r>1/2$?

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${\text{Vol}(B^n(r))}=r^n{\text{Vol}(B^n(1))}$ by the homogeneity of degree $n$ of the Lebegue measure. Consequently, $\frac{\text{Vol}(B^n(r))}{\text{Vol}(B^n(1)\setminus B^n(r))}=\frac{r^n}{1-r^n}\to 0$ exponentialy fast.

But maybe this is too analytic?

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Indeed, much too analytic. But nice. – Daniel Litt Oct 8 '10 at 0:03

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