Suppose that
$$E\left[|S_j - S_i|^\gamma |S_k - S_j|^\gamma \right] \leq \left(\sum_{i < l \leq j} u_l\right)^\alpha \left(\sum_{j < l \leq k} u_l\right)^\alpha$$
for $0 \leq i \leq j \leq k \leq m$ with $S_k = \sum_{i \leq k} \xi_i$ ($S_0 = 0$) and $\gamma, \alpha$ postitive, and $u_l \geq 0$ for every $l$.
It follows from a theorem in Billingsley's book Convergence of Probability Measures (first edition) that
$$P\left(\max_{n \leq i \leq j \leq k \leq m} \min \{|S_i - S_j|, |S_k - S_j|\} \geq \lambda\right) \leq \frac{K}{\lambda^{2\gamma}} \left(u_{n+1} + \ldots + u_{m}\right)^{2\alpha} \times \min_{n + 1 \leq h \leq m}\left(1 - \frac{u_h}{u_{n + 1} + \ldots + u_m}\right)$$
where $K$ depends only on $\lambda$ and $\gamma$.
Assume additionally that $\sum u_l < \infty$. Can we use the above bounded probability to show that $\sum \xi_i < \infty$ almost surely?
I think we can simplify things by saying that since $\min_{n + 1 \leq h \leq m}\left(1 - \frac{u_h}{u_{n + 1} + \ldots + u_m}\right) \leq 1$ we actually only need to work with
$$P\left(\max_{n \leq i \leq j \leq k \leq m} \min \{|S_i - S_j|, |S_k - S_j|\} \geq \lambda\right) \leq \frac{K}{\lambda^{2\gamma}} \left(u_{n+1} + \ldots + u_{m}\right)^{2\alpha}$$
This question is related to a question I asked before and I think a similar approach to the solution given there could be used here, but I don't know how to do it. I want to relate $\max_{n \leq i, j \leq m} |S_i - S_j|$ to $\max_{n \leq i \leq j \leq k \leq m} \min \{|S_i - S_j|, |S_k - S_j|\}$ but I don't know if I can do so in a useful way.