Show that $V_n(R)$ has one maximum and that it decreases for large $n$ I am studying the volume of $n$-balls, and I have deduced the following formula for the volume of a $n$-dimensional ball of radius $R$:
$$V_{2k}(R) = \frac{(\pi R^2)^k}{k!} \hspace{1cm} V_{2k+1}(R) = 2R\frac{(4\pi R^2)^kk!}{(2k+1)!}$$
I now want to show that $V_n(R)$ tends to zero as $n$ tends to infinity, so I have to calculate $\displaystyle\lim_{k\rightarrow\infty} \frac{(\pi R^2)^k}{k!}$ and $\displaystyle\lim_{k\rightarrow\infty} 2R\frac{(4\pi R^2)^kk!}{(2k+1)!}$. However, I am not sure how to calculate the limits because of the factorials in them.
I also what to show that $V_n(R)$ has always one maximum in terms of $n$, that is, that there is a dimension where a ball with radius $R$ has the maximum volume, for example, for $R=1$, it is the 5th dimension ($n = 5$). However, again because of the factorials, I don't know how to calculate the derivative to optimize the function.
I would prefer a solution that does not use the gamma function, Stirling's approximation, or double factorials.
 A: Since $R$ hass some fixed value, there exists $n=m$, for which $k=\pi R^2/m < 1$. Thus, for $n>m$:
$$
\frac{(\pi R^2)^n}{n!} = \left(\frac{(\pi R^2)^m}{m!}\right)\frac{\pi R^2}{m+1}\frac{\pi R^2}{m+2}\ldots\frac{\pi R^2}{n} < \left(\frac{(\pi R^2)^m}{m!}\right)\frac{\pi R^2}{m}\frac{\pi R^2}{m}\ldots\frac{\pi R^2}{m}=\left(\frac{(\pi R^2)^m}{m!}\right)k^{n-m}.
$$
In other words, your sequence is bounded by a geometric sequence with ratio $k<1$.
For the second sequence, you can do a similar trick, but first you need to simplify:
$$
\frac{n!}{(2n+1)!}=\frac{1}{(n+1)(n+2)\ldots(2n+1)}
$$
To address the question with one maximum, you can show that $K_n=V_{n+1}/V_n$ is a strictly decreasing sequence. So as long as $K_n>1$ $V_n$ is increasing and after $K_n<1$ volume is decreasing. However, note that there exist $R$ and $n$ such that $K_n(R) = 1$. Which means that $V_{n+1}(R)=V_n(R)$. In other words, volume can reach maximal value at two neighbour dimensions.
A: WLOG consider $R=1$. First the odd case:
$$V_{\text{odd}}(n)=\left|\frac{\pi^n}{n!}\right|\leq \left|\frac{\pi^n}{4^n}\right|$$
(For large enough $n$.) Obviously,
$$\lim_{n\to\infty}\left|\frac{\pi}{4}\right|^n=0$$
Hence $\lim_{n\to\infty}V_{\text{odd}}(n)=0$. Apply similar reasoning to the even case.
