Let us continue this discussion in chat.

I mean $S_k\leq \nu k$ etc.

I'm not sure, though, if what I wrote is formally true; however, if you substitute all $<$'s to $\leq$'s in the statement, then it should hold.

I think it should be true that $$ \frac{1}{n}P[S_n<\nu n] \leq P[S_k<\nu k \text{ for all }k\leq n] \leq P[S_n<\nu n], $$ which is, probably, what you need. The 2nd inequality is evident; as for the 1st one, note that $S_n<\nu n$ implies that there is a cyclic shift of $X_1,\ldots,X_n$ such that $S'_k<\nu k \text{ for all }k\leq n$ for the ``new'' partial sums (just shift to the point where $S_j - \nu j$ is maximized).

For example, if $\sum p_n<\infty$, then a.s. there will be only a finite number of 1's (so that $\sum X_n$ converges), and (since $E X_n=p_n$) $\sum E X_n$ converges as well. The other case is analogous (or consider $Y_n=1-X_n$, and observe that $X_n-EX_n=-(Y_n-EY_n)$).

Ah, seems that he changed chapters' order. In that version, it's Theorem 5.3.1.