Dimension that maximizes surface area and volume unit $n$-sphere

Question

Let $S_n(R)$ and $V_n(R)$ denote the surface area and volume of an $n$-sphere with radius $R$ respectively.

It is well known that $$S_n(R) = \frac{n \pi^{\frac{n}{2}}R^{n-1}}{\Gamma\left(\frac{n}{2}+1\right)}\quad\text{and}\quad V_n(R) = \frac{\pi^{\frac{n}{2}}R^{n}}{\Gamma\left(\frac{n}{2}+1\right)}$$ where $\Gamma(z)$ is the gamma function: $$\Gamma(z) = \int_0^\infty e^{-x} x^{z-1}\, dx.$$

Question: Let $v$ and $s$ be positive integers such that $S_s(1)$ and $V_v(1)$ are the maximum. What is $v-s?$

If I am not mistaken, I saw somewhere that $s = 7$ and $v = 5$ (I think) but I couldn't find that source anymore.

While trying to prove the result, I realize that I am not able to do so.

Any hint is appreciated.

Answer 1

Remember that $\Gamma(x+1) = x \Gamma(x).$ You can deal with the odd and even dimensions separately.

Now if you start with the ratio $$V_n = \frac{\pi^{n/2}R^{n}}{\Gamma\left(\frac{n}{2}+1\right)}$$ for some particular value of $n,$ and you add two the dimension repeatedly, you are multiplying the denominator by $x,$ $x+1,$ $x+2, \ldots$ (where $x = \frac n2 + 1$) whereas the numerator is multiplied by $\pi$ each time. First do this sufficiently many times so that the factor $x+k$ in the denominator is greater than $\pi,$ and then a sandwich with $$0 \leq V_i \leq \text{constant} \times \left(\frac\pi{x+k}\right)^i$$ shows that $V_i \to 0.$

For $S_n = nV_n$ you might add two to the dimension enough times so that $n + k > 2\pi$ and $(n+2)/n \leq 2,$ and then you have $$0 \leq S_i \leq \text{constant} \times \left(\frac{2\pi}{x+k}\right)^i.$$

So each of the formulas has a maximum over all odd dimensions and a maximum over all even dimensions, and whichever is greater is the maximum over all dimensions.

Answer 2

Hint: The only tricky bit is the $\Gamma$-function. For $n$ even you need the $\Gamma$-function at integer values, so you can just use $$\Gamma(k)=(k-1)!$$ For $n$ odd you need $\Gamma(k+\frac{1}{2})$ for $k$ an integer. Here you can use the Legendre duplication formula, which will work out to $$\Gamma(k+\frac{1}{2})=\frac{2^{1-2k}\sqrt{\pi}(2k-1)!}{(k-1)!}$$ Now you can just compute the first few terms and see at what point they start decreasing again. $s=7$ and $v=5$ looks roughly correct but I haven't done the computation.

Edit: I will write out the solution for the $S_{2k}(1)$, the other $4$ cases work similar. $$S_{2k}(1)=\frac{2k\pi^{2k/2}}{\Gamma(2k/2+1)}=\frac{2k\pi^{k}}{k!}=2\frac{\pi^k}{(k-1)!}$$ So if we go from $k$ to $k+1$, the volume gets multiplied by $\frac{\pi}{k}$, so until $k=3$ this is bigger than $1$, for $k\ge 4$ this is smaller. The maximum occurs at $k=4$. Hence the highest surface area of an even dimensional sphere is at $k=4$ or $n=8$.