Weighed cointossing

Assume S = \{s_1, s_2, .. \} is a set with 1 \le s_1 \le s_2 \le .. , such that \displaystyle \sum \dfrac{1}{s_i} diverges.  Let a_i = \pm 1, each with probability \dfrac{1}{2}. What can we say about \displaystyle \sum_{i=1}^n \dfrac{a_i}{s_i} when n goes to infinity? Must there, for example, be infinitely many sign changes with probability 1? It can’t be hard to show this for bounded s_i, but what about S = \aleph for example? I honestly have no clue whatsoever.

EDIT 22 Februari 2011
I can show that there is a positive probability that \left|\displaystyle \sum_{i=1}^n \dfrac{a_i}{s_i}\right| > c for any c, for n large enough; Let k = k(c) be the smallest integer such that \displaystyle \sum_{i=1}^k \dfrac{1}{s_i} > c (note that this k exists since the sum of \dfrac{1}{s_i} is assumed to go to infinity).
Then either the probability that \displaystyle \sum_{i=k+1}^n \dfrac{a_i}{s_i} \ge 0 is at least \dfrac{1}{2} or the probability that \displaystyle \sum_{i=k+1}^n \dfrac{a_i}{s_i} \le 0 is at least \dfrac{1}{2} (*). Assume the first case (second case goes analogous), then we have: P\left(\displaystyle \sum_{i=1}^n \dfrac{a_i}{s_i} > c\right) \ge P\left(\displaystyle \sum_{i=k+1}^n \dfrac{a_i}{s_i} \ge 0\right) \displaystyle \prod_{j=1}^k P(a_j = 1) \ge \dfrac{1}{2}\dfrac{1}{2^k} = \dfrac{1}{2^{k+1}} > 0

(*) It is fun to note that both are the case, independent of S. This can be used to show that P\left(\left|\displaystyle \sum_{i=1}^n \dfrac{a_i}{s_i}\right| > c\right) \ge \dfrac{1}{2^k} for all n \ge k


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