I am was doing some problems from my text book and one of them says
" let the volume of of the nth dimensional ball be denoted by
$vol(B^n(R))$ where $R$ is the radius of the ball. Show that $\lim_{n\to\infty} vol(B^n(R))$ "
I have seen explanations of this, but all of them use the gamma function. I was wondering if there was any way to do it without using it. So far I have shown that $vol(B^n(R)=vol(B^{n-2}(R))\frac{2\pi R^2}{n}$.
Using this I was thinking you could keep iterating and you would get $n(n-2)(n-4)$ in the denominator and so on which is less than $n!$ and whatever is above will not depend on n so we could treat it as a constant, thus you would have something like $\frac{T}{n!}<\frac{T}{n(n-2)(n-4)..}$ so then by taking limits the RHS crushes the left hand side so they both go to 0. Is this right ? I am not sure if this is even rigorous enough.
Any help is appreciated :).
Thank you.
The volume of a $2k$-dimensional unit ball is given by $\pi^k/k!$, and it's easy to see this goes to zero as $k\to\infty$. The odd-dimensional ($2k-1$) version is more annoying, but the principle of the same.