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I am trying to prove that the volume of an $n$-sphere goes to zero as dimension increases to infinity (I know this to be true). Here is my work:

\begin{align} \lim_{n\to\infty} \left( V_n \right) & \\ \lim_{n\to\infty} \left( \frac{\pi^{n/2} R^n} {\Gamma \left(\frac{n+2}2 \right)} \right) & = \frac \infty \infty \\ \text{Subsitute } \Gamma(z) & = \int_0^\infty x^{z-1} e^{-x} dx \\ \lim_{n\to\infty} \left( \frac{\pi^{n/2} R^n} {\int_0^\infty x^{n/2} e^{-x} \, dx} \right) \\ \text{By Leibniz integral rule, } \frac{d}{dx} \int^b_a f(x,y) \, dx & =\int^b_a f_x(x,y)\, dx \\ \text{Looking at denominator, } \int_0^\infty x^{n/2} e^{-x} \, dx \\ \frac d {dn} \int_0^\infty x^{n/2} e^{-x} \, dx & = \int_0^\infty \frac{\partial}{\partial n} x^{n/2} e^{-x} \, dx \\ & =\int_0^\infty x^{n/2} e^{-x} \ln \left(\frac n 2 \right) \, dx \\ & =\ln \left(\frac n 2 \right)\Gamma \left(\frac{n+2} 2 \right) \\ \text{Using L'Hospital's rule, } \lim_{n\to\infty} \left( \frac{\pi^{n/2} R^n} {\int_0^\infty x^{n/2} e^{-x} \, dx } \right) & = \lim_{n\to\infty} \left( \frac{\ln (n) \pi^{n/2} R^n + \ln \left(\frac n 2 \right) \pi^{n/2} R^n }{\ln \left(\frac n 2 \right)\Gamma \left(\frac{n+2} 2 \right)} \right) \end{align} Since the limit starts as the indefinite $\frac \infty \infty$, I am trying to use L'Hospital's rule. However, after using the rule once I am left with another indefinite $\frac{\infty}{\infty}$. I don't believe repeated uses of L'Hospital's Rule will work since iterations don't change the indefinite state.

Any help or insight would be greatly appreciated.

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  • $\begingroup$ The $\Gamma$ function is definitely going to outrun that numerator, right? $\endgroup$ – Tim kinsella Dec 10 '17 at 17:59
  • $\begingroup$ @Timkinsella The Gamma function definitely outruns the numerator, and that is clear intuitively. However, I am unsure how to prove this rigorously. $\endgroup$ – Adamisthename Dec 10 '17 at 18:02
  • $\begingroup$ You could use Stirling's approximation if you wanted to be super careful. It gives the asymptotic growth of $\Gamma$. $\endgroup$ – Tim kinsella Dec 10 '17 at 18:02
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – Tim kinsella Dec 10 '17 at 18:03
  • $\begingroup$ Wait, your $n$ only takes integer values. Have you considered using $\Gamma(n) = (n-1)!$? $\endgroup$ – Tim kinsella Dec 10 '17 at 18:07
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We can get the result without too much fancy stuff. Let $B_n$ be the open unit ball in $\mathbb R^n,$ and let $v_n$ denote volume measure on $\mathbb R^n.$ Then by Fubini,

$$v_{n+1}(B_{n+1}) = \int_{-1}^1 v_{n}((1-x^2)^{1/2}B_{n})\, dx = \int_{-1}^1 (1-x^2)^{n/2}v_{n}(B_{n})\, dx$$ $$ = v_{n}(B_{n})\int_{-1}^1 (1-x^2)^{n/2}\, dx.$$

Since the last integral $\to 0,$ we have the ratio

$$\frac{v_{n+1}(B_{n+1})}{v_{n}(B_{n})} \to 0,$$

and that implies $v_{n}(B_{n}) \to 0$ fast.

Now

$$\frac{v_{n+1}(RB_{n+1})}{v_{n}(RB_{n})} = R\frac{v_{n+1}(B_{n+1})}{v_{n}(B_{n})}$$

for any $R>0,$ and the full result follows.

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Write the points of ${\mathbb R}^n$ in the form $(x,y,{\bf z})$ with ${\bf z}\in{\mathbb R}^{n-2}$, and put $\sqrt{x^2+y^2}=:r$. Denote the volume of the $n$-dimensional unit sphere $B_{n,1}$ by $\kappa_n$. Then $$\eqalign{\kappa_n&=\int_{B_{n,1}}1\>{\rm d}(x,y,{\bf z})=\int_{B_{2,1}}\int_{B_{n-2,\>\sqrt{1-r^2}}} \>{\rm d}({\bf z})\>{\rm d}(x,y)= \kappa_{n-2}\int_0^1(1-r^2)^{(n-2)/2}\>2\pi r\>dr\cr &={2\pi\over n-2}\kappa_{n-2}\ .\cr}$$ It is then obvious that $\lim_{n\to\infty}\kappa_n=0$.

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