## 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 $n$ $\ge$ $k$