Is there any sufficient condition in terms of moments under which 
$$ \sum\_{n=1}^{\infty} X\_n$$ diverges a.s.?Here $X\_n$ are not independent


I am given that $\sum\_n E[X\_n]$ diverges. Actually, I am interested in the following special case:


Let $\{F\_n\}$ be a filtration, and $\{A\_n\}$ be a sequence of events such that $A\_n \in F\_n$. Suppose $P(A\_n) > 0 \forall n$. Then what is the weakest additional condition that will ensure the following


$$ \sum\_{n=1}^{\infty} P(A\_n | F\_{n-1}) = \infty \mbox{ a.s.}$$


One condition is $P(A\_n i.o) =1$, but I want something verifiable. 



Divergence of general random series and a special case
Here is one simple sufficient condition:
$\sum \mathsf{E} X\_n$ diverges, and $\sum (\mathsf{Var} X\_n)^{1/2} < \infty$.


Indeed, the variance condition ensures that $\sum (X\_n - \mathsf{E} X\_n)$ converges in $L^2$, so if $\sum \mathsf{E} X\_n$ diverges, so does $\sum X\_n$ (say, in probability).


