# Infinite product, arranging of terms

1. Nov 1, 2008

### Ookke

Consider a set that has integers p and inverses 1/q, infinitely many of both. We are about to form a product of all numbers in this set, with two different arrangements.

Arrange terms, way 1:
For each p (countably many), select 1/q so that q>p, the resulting product term is (p * 1/q) < 1. The supply of p's are exhausted this way, and in addition there are infinitely many left-over 1/q's. As far as I understand (and possibly not), the product of all should be 0.

Arrange terms, way 2:
For each 1/q select p>q, the resulting terms are (1/q * p) > 1 with infinitely many left-over p's. This seems an infinite product.

2. Nov 1, 2008

### HallsofIvy

Doesn't it seem possible that such an "infinite product" will depend on the order in which the products is done?

3. Nov 2, 2008

### Ookke

I find it hard to believe that the product (or sum of a series) would depend on arranging of terms. It's easier to think that product or sum does not exist in that case.

Also this kind of examples cast some doubt over our ability to choose infinitely many items from a set, though this is somewhat common in mathematics. There is a big difference between hand-picked set and a set that is assumed to exist by some axiom.

4. Nov 2, 2008

### HallsofIvy

Any conditionally convergent sum can be rearranged to converge to any given value. That was why I asked.

5. Nov 3, 2008

### EternalVortex

For a product to converge, the terms have to go to 1. This is not the case in your example.