On Tuesday, April 29, 2003, at 04:39 PM, asimovd@aol.com wrote:
I think this is an old problem, almost certainly solved, but I don't think I've ever seen the solution:
Given the harmonic series but with an independently chosen random sign in front of each term:
±1 ± 1/2 ± 1/3 ± ...,
then
a. what is the probability of convergence?
b. when it converges, what is the probability distribution of the sum?
I think it helps to get a conceptual picture by first think of a related series like ±1 ±1/2 ±1/2 ±1/4 ±1/4 ±1/4 ±1/4 ... If you group the equal magnitude terms together, you can apply the central limit theorem to conclude that the sum a_n (binomially distributed) is eventually close to a normal distribution with mean 0 and variance (2^n) * 2^(-2n), so standard deviation 2^(-n/2). The sequence of s_n of standard deviations is a geometric series. Therefore also the sequence n*s_n converges. Obviously, the probability is 1 that eventually, |a_n| < n * s_n, so the series converges with probability 1. This argument easily implies that the harmonic series with random signs also converges with probability 1. I would guess that distribution of the sums is kind of bumpy in some sense, being the convolution of a lot of 2-point distributions, but I haven't thought this through nor tried to plot experimental partial distributions. In any case, my inclination is to think it would be surprising to find any nice formula for it. Bill