I want to prove if the following statement is true:
Let a sequence of real numbers $x_{m}\left(j\right)\to x\left(j\right)$ as $m \to \infty$, $ \forall j\in \mathbb{N}$. Then: \begin{equation*} \sum_{j \in \mathbb{N}} |x_{m}\left(j\right) - x\left(j\right)| \to 0 \text{ as } m\to \infty \end{equation*}
Proof: Since $x_{m}\left(j\right) \to x\left(j\right)$ as $m\to \infty$, this means that $\forall \epsilon>0$ $\exists m_{0} \in \mathbb{N}$ such that $|x_{m}\left(j\right) - x\left(j\right)|<\epsilon$ for all $m\ge m_{0}$.
For each $j\in \mathbb{N}$ we chose $\frac{\epsilon}{2^{j}}$. Then:
\begin{equation*} 0 \le \sum_{j \in \mathbb{N}} |x_{m}\left(j\right) - x\left(j\right)| < \sum_{j \in \mathbb{N}} \frac{\epsilon}{2^{j}} = \epsilon \end{equation*} which is the thesis.
I have big doubs about the last line: I don't know if it is logically correct or not. I quite sure that I can bound each term of the sum for a $j$ fixed and for each of them there exists an $m_{0}$ such that the bound holds. Is this still true when the term becomes infinite? Could anyone explain me the concept? Does it follow from the definition of convergence of series of real numbers?
You are trying to prove something that is not even true. Here is a counter-example: Let $x_m(j)=1$ if $m=j$ and $0$ if $m \neq j$. Let $x_j=0$ for all $j$. Can you verify that this is a counter-example?