if you remove enough of the first terms of an infinite product that converges, the product of remaining terms must converge to 1
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).
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.