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hairyfigment comments on Probabilistic Löb theorem - Less Wrong Discussion
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Let a(n) be any series whose sum converges and let b(n) = exp(a(n)). Then the product of b(n) converges iff the sum of a(n) converges. In particule, we can take b(n) = exp(1/2^n) for n = 0 to infinity, whose product converges to exp(2).
My claim was intended to be limited to infinite products where each term is between 0 and 1.
exp(1/2^0)=exp(1)=e>1
exp(epsilon)>1
b(n)>1 for all positive n.
The addition (or removal) of a finite number of nonzero terms at the beginning of an infinite product does not change whether or not the product converges.
Plain language: if you remove enough of the first terms of an infinite product that converges, the product of remaining terms must converge to 1. I'm rusty on my math and LaTeX, but given an infinite product that converges, we can prove that the first N terms have a product within epsilon of the product of the series (by the definitions of limit and converge), which means that the remaining infinite number of terms must have a product within some different epsilon of 1 (again, simply by definitions).
Given an infinite series of numbers all between 0 and 1 and dropping the first N terms, I can find an epsilon such that the product of the remaining terms is always greater than that epsilon from 1. That epsilon is 1-(the N+1th term).
I've made some error somewhere, since the infinite product of b(n)=.01 converges (to 0). I intuit that I may have proven only that the infinite product does not converge to 1. I suspect that there's a step somewhere around the point where I used the phrase 'some different epsilon'.
This is not sufficiently precise. More precisely: if T(n) is the tail product of all but the first n terms, then for each fixed n, T(n) is a convergent product, and as n approaches infinity, T(n) should to converge to 1 (assuming the initial infinite product converges to a nonzero value).
I think the flaw in your proof is the confusion between convergence of T(n) for a fixed n, as an infinite product, and convergence of T(n), the value of that infinite product, as n goes to infinity.
Also, a more general family of counterexamples is the following: take any nonnegative series a(n) whose sum converges to A, and let b(n) = exp(-a(n)). Then 0<b(n)<1 and the product of b(n) converges to exp(-A). (Obviously this is more or less recycled from the grandparent.) For example, b(n) = exp(-1/2^n).
Concur- the infinite product of 0<b(n)<1 is in [0,1). That makes sense for probabilities.
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.