In the top voted answer here, it says
$A_n = n V_n$
But from wikipedia, I see
\begin{align*} A_n(R) &= \frac{2 \pi^{\frac{n+1}{2}}}{\Gamma\left( \frac{n+1}{2} \right)} R^n \end{align*}
and \begin{align*} V_n(R) &= \frac{\pi^{n/2}}{\Gamma(1 + \frac{n}{2})} R^n \end{align*}
But writing this out, I get
\begin{align} \frac{V_n(R)}{A_n(R)} &= \frac{1}{2 \sqrt{\pi}} \frac{\Gamma \left( \frac{n + 1}{2}\right)}{\Gamma \left( \frac{n + 2}{2}\right)} = O\left( \frac{1}{\sqrt{n}} \right) \end{align} which decays as $\frac{1}{\sqrt{n}}$, not $\frac{1}{n}$.
In any case, this shows that $\frac{V_n(R)}{A_n(R)}$ decays with $n$, thereby showing that a thin shell near the surface contains most of the volume in a high-dimensional sphere.
I think there is difference in notation of A_n between wikipedia version and the answer version. $A_n$ in the answer corresponds to $A_{n-1}$ in the wikipedia. And taking that difference in consideration, you easily get the right ratio.