16 Jan
2014
16 Jan
'14
12:01 p.m.
Good point, Joerg. I overlooked the very common case where (assuming 0 < a_k < 1, k=1,2,3,...) oo Product (1 - a_k) > 0 n=1 . Since a_k > 0, this is true if and only if oo Sum a_k < oo, n=1 (which occurs with measure 0 in the countable-cube (0,1)^oo endowed with standard Kolmogorov product measure). --Dan On 2014-01-16, at 12:42 AM, Joerg Arndt wrote:
About that "technical condition":
0 = 1 - 1 = 1 - A = 1 - \sum_{n>=1}{ a_n \prod_{k=1}^{n-1}{ 1 - a_k } = \prod_{n>=1}{ 1 - a_k } (**)
(last equality from 1 + \sum_{n>=1}{ a_n \prod_{k=1}^{n-1}{ 1 + a_k } } = \prod_{n>=1}{ 1 + a_k } and replacing a_k by -a_k )
Now, when is (**) actually zero?