〈 Berger | Dillon 〉

Perhaps @johncarlosbaez has some geometric intuition

(I'm not sure the 'perhaps' will prove srictly necessary there ;) ).

I tried to think geometrically about this sum and didn't get very far, since he was adding thing with different units. Now I feel pressured to continue. I think it'd be better to put in the radius r of the spheres instead of setting it to 1. Then the sum is exp(r pi). 1/n

But how to prove it's exp(r pi)? One way is to prove it's some function f(r) with f(0) = 1 and f'(r) = pi f(r). The first part is easy: the 0-dimensional sphere of radius 0 has 0-volume equal to 1, but all the rest have n-volume equal to 0. So why is f'(r) = pi f(r)? 2/n

There could be a good way to understand this if we really understood the meaning of the sum f(r). But I don't, so for now I can only see f'(r) = pi f(r) this as saying the volume of the unit 2n-sphere is pi/n times the volume of the unit (2n-2)-sphere. 3/n

That fact - volume of the unit 2n-sphere is pi/n times the volume of the unit (2n-2)-sphere - is proved quite nicely here: https://en.wikipedia.org/wiki/Volume_of_an_n-ball#The_two-dimension_recursion_formula … 4/n (n=4)