Thank you for the counterexample. Let me give a little background info, cause it is about weak-mixing and Ergodicity of a transformation T. Apparantly, if we assume T is weakly mixing, and S Ergodic, then $T\times S$ is Ergodic. This amounts to assuming:
$\frac{1}{n}\sum_{i=1}^n|a_k-p|\to 0$ (absolute convergence on one hand) $\frac{1}{n}\sum_{i=1}^n b_k-q\to 0$ (convergence on the other hand)
and showing this implies $\frac{1}{n}\sum_{i=1}^na_kb_k-pq \to 0$
\begin{align*} \frac{1}{n}\sum_{k=1}^n a_kb_k-pq & = \frac{1}{n} \sum_{k=1}^n (a_k-p)(b_k-q) + \frac{1}{n} \sum_{k=1}^np(b_k-q)+ \frac{1}{n} \sum_{k=1}^nq(a_k-p)\\ & \to \lim_{n\to\infty}\frac{1}{n} \sum_{k=1}^n (a_k-p)(b_k-q)+0+0 \end{align*}
Maybe I'm missing something, but what extra conditions do we need to make sure this limit is 0. ?