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1*3*4*5*… = ?

I suggest ℵ1.

It’s at least 2 to the power of ℵ0 – the number of factors in the title. And by definition, 2 to the power of some aleph is the next aleph. As every finite set with N elements also has its power set of 2^N elements.

This is the lower bound.

The next reason is, that countably many sided cuboid 1 by 2 by 3 by 4 … has a bijective mapping between the real numbers on the interval [0,1] and its consisting hypercubes 1 by 1 by 1 …

And this is also the upper bound.

ℵ1 is therefore the most natural definition for such a product. That the product of all naturals, is equal to the number of all reals.

Standard

One thought on “1*3*4*5*… = ?”

1. Oscar Cunningham says:

We can also give a finite answer using the analytic continuation of the Riemann zeta function:

ζ(s) = 1^-s + 2^-s + 3^-s + …

so

ζ'(s) = – log(1)*1^-s – log(2)*2^-s – log(3)*3^-s – …

then symbolically we ought to have

ζ'(0) = – log(1) – log(2) – log(3) – …

and hence

1*2*3*… = e^-ζ'(0).

The zeta function can be analytically continued to s=0, and it’s derivative there is -log(2π)/2. So

1*2*3*… = e^(log(2π)/2) = sqrt(2π)

whatever that means.