Discrete version of dominated convergence thm

The direct proof would still follow the proof of DCT (with simplifications). It suffices to show that $$\liminf_{N\to\infty} \sum_{n\in\mathbb Z} f_N(n) \ge \sum_{n\in\mathbb Z} f(n) \tag{1}$$ because by applying (1) to $-f_N$ and $-f$ you get the reverse inequality which completes the proof. In order to show (1), add $\sum_{n\in\mathbb Z} g(n)$ to both sides. This reduces (1) to the case when all functions are nonnegative. Given $\epsilon>0$, pick $K$ such that
$$ \sum_{|n|\le K} f(n)> \sum_{n\in\mathbb Z} f(n)-\epsilon$$ Due to pointwise convergence, for all sufficiently large $N$ you have $f_N(n) > f(n)-\epsilon/K$, for every $n=-K,\dots,K$. Hence, for such $N$ $$ \sum_{n\in\mathbb Z} f_N(n) \ge \sum_{|n|\le K} f_N(n) >\sum_{|n|\le K} f(n)-3\epsilon > \sum_{n\in\mathbb Z} f(n)-4\epsilon$$ which proves (1).