So, a first term greater than 1, and the remainder converges to 1/b(0)?
I did only show that the infinite product of 0<n<1 cannot be one. It can be 1-epsilon for any epsilon.
I don’t think that particular a(n) converges, but that doesn’t invalidate your point that b(n) selected such that the product from 0 to n equals the sum of a(n) from 0 to n must have a product that converges to the sum of a(n). But the sum of a(n) would have to be decreasing for the terms of b(n) to be between 0 and 1.
I wasn’t summing a(n). It’s the sequence a(n) that converges, not its sum, and the partial products of the b(n) are equal to the a(n), not the partial sums of the a(n).
Certainly an infinite product of 0<n<1 can’t be one. Nobody’s disputing that.
Oh, then it sounds like we are in perfect agreement that my initial claim was wrong; however, we can now generate an infinite series of probabilities less than one whose product remains higher than 1-epsilon for any epsilon. If 1-epsilon is used as the determinator of what the system proves true, Löb’s theorem holds.
Here’s a counterexample to your claim:
Let a(n) be a decreasing series which tends to a nonzero limit (for example, a(n) = 1 + 1/n)
Then let b(0) = a(0), b(n) = a(n) / a(n-1)
a(n) is decreasing so for all n>0, b(n) < 1
But the product of the first N b(n)s is a(N), so the infinite product of b(n) converges to the same thing as a(n), which is nonzero.
So, a first term greater than 1, and the remainder converges to 1/b(0)?
I did only show that the infinite product of 0<n<1 cannot be one. It can be 1-epsilon for any epsilon.
I don’t think that particular a(n) converges, but that doesn’t invalidate your point that b(n) selected such that the product from 0 to n equals the sum of a(n) from 0 to n must have a product that converges to the sum of a(n). But the sum of a(n) would have to be decreasing for the terms of b(n) to be between 0 and 1.
I wasn’t summing a(n). It’s the sequence a(n) that converges, not its sum, and the partial products of the b(n) are equal to the a(n), not the partial sums of the a(n).
Certainly an infinite product of 0<n<1 can’t be one. Nobody’s disputing that.
Oh, then it sounds like we are in perfect agreement that my initial claim was wrong; however, we can now generate an infinite series of probabilities less than one whose product remains higher than 1-epsilon for any epsilon. If 1-epsilon is used as the determinator of what the system proves true, Löb’s theorem holds.