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A calculus-free argument suggested by fedja:

The ultimate reason is, of course, that the typical coordinate of a point in the unit ball is just of size $n^{-1/2}\ll \frac{1}{\sqrt{n}}\ll 1$, as several people have already mentioned. This can be turned into a very simple geometric argument (as suggested by fedja) using the fact that an $n$-element set has $2^n$ subsets. :

At least $n/2$ of the coordinates of any a point in the unit ball are at most $\sqrt{2/n}$ or less \sqrt{\frac{2}{n}}$in absolute value, and the rest are at most$1$in absolute value. Thus, the unit ball can be covered by at most$2^n$bricks (meaning right-angled parallelepipeds) of volume$(8/n)^{n/4}\left(2\sqrt{\frac{2}{n}}\right)^{n/2},$$each . Each brick corresponds corresponding to one choice of the a subset of for the small coordinates. So Hence, the volume of the unit ball is at most 2^n 2^n \cdot (8/n)^{n/4} \left(2\sqrt{\frac{2}{n}}\right)^{n/2} = (128/n)^{n/4}, which clearly goes to 0. Indeed\left(\frac{128}{n}\right)^{n/4}\rightarrow0.$$ In fact, the argument shows that the volume of the unit ball decreases faster than any exponential, which is to say that so the volume of the $n$-sphere ball of any fixed radius $r$ also goes to $0$.

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Argument

A calculus-free argument suggested by fedja:

The ultimate reason is, of course, that the typical coordinate of a point in the unit ball is just of size $n^{-1/2}\ll 1$, as several people have already mentioned. This can be turned into a very simple geometric argument using the fact that an $n$-element set has $2^n$ subsets. At least $n/2$ of the coordinates of any point in the unit ball are $\sqrt{2/n}$ or less and the rest are at most $1$ in absolute value. Thus, the ball can be covered by at most $2^n$ bricks (meaning right-angled parallelepipeds) of volume $(8/n)^{n/4}$ each. Each brick corresponds to one choice of the subset of small coordinates. So the volume of the ball is at most $2^n \cdot (8/n)^{n/4} = (128/n)^{n/4}$, which clearly goes to $0$.

Indeed, the volume decreases faster than any exponential, which is to say that the volume of the $n$-sphere of any fixed radius $r$ also goes to $0$.

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Argument suggested by fedja:

The ultimate reason is, of course, that the typical coordinate of a point in the unit ball is just of size $n^{-1/2}\ll 1$, as several people have already mentioned. This can be turned into a very simple geometric argument if we are allowed to use using the simple combinatorial fact that and an $n$-element set has $2^n$ subsets. Indeed, at At least $n/2$ of the coordinates of any point in the unit ball are $\sqrt{2/n}$ or less and the rest are at most $1$ in absolute value. Thus, the ball can be covered by at most $2^n$ bricks (meaning right-angled parallelepipeds) of volume $(8/n)^{n/4}$ each(each parallelepiped . Each brick corresponds to one choice of the subset of small coordinates)coordinates. So the volume of the ball is at most $2^n \cdot (8/n)^{n/4} = (128/n)^{n/4}$, which clearly tends goes to $0$.

Indeed, the volume decreases faster than any exponential, which is to say that the volume of the $n$-sphere of any fixed radius $r$ also goes to $0$.

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