Question: Let $f\_1,f\_2,\ldots,g\colon\mathbb{Z}\rightarrow\mathbb{R}$ be functions such that $|f\_N(n)|\leq g(n)$, $\sum\_{n=-\infty}^\infty g(n)<\infty$, and $\lim\_{N\rightarrow\infty}f\_N(n)=f(n)$. Then show that $$\lim\_{N\rightarrow\infty}\sum\_{n=-\infty}^\infty f\_N(n)=\sum\_{n=-\infty}^\infty f(n).$$

This looks like the dominated convergence theorem, but how can we prove it directly?

**Edit**: As T. Bongers helpfully pointed out, this can be shown using the dominated convergence theorem. Is there a direct way to do it without the theorem?

Answer: 
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).
Since $\sum\limits\_{n\in\mathbb{Z}} g(n)$ converges, for each $k$, $\sum\limits\_{n\in\mathbb{Z}}f\_k(n)$ converges. Furthermore, for any $\epsilon\gt0$, there is an $n\_\epsilon$ so that
$$
\sum\_{|n|\gt n\_\epsilon}g(n)\le\frac\epsilon2\tag1
$$
Thus, for any $\epsilon\gt0$,
$$
\begin{align}
\hspace{-9pt}\limsup\_{k\to\infty}\left|\,\sum\_{n\in\mathbb{Z}} f\_k(n)-\sum\_{n\in\mathbb{Z}}f(n)\,\right|
&\le\lim\_{k\to\infty}\sum\_{|n|\le n\_\epsilon}\!\!|f\_k(n)-f(n)|+\limsup\_{k\to\infty}\!\!\sum\_{|n|\gt n\_\epsilon}\!\!|f\_k(n)-f(n)|\\
&\le0+\epsilon\tag2
\end{align}
$$
Since $(2)$ is true for any $\epsilon\gt0$,
$$
\lim\_{k\to\infty}\sum\_{n\in\mathbb{Z}}f\_k(n)=\sum\_{n\in\mathbb{Z}}f(n)\tag3
$$