If $\sum f_n$ is absolutely convergent and uniformly convergent $\Longrightarrow$ $\sum \|f_n\|$ is uniformly convergent.

Is the following proposition true?

Let $f_n \colon X\subset\mathbb{R}^2\longrightarrow \mathbb{R}^2$ be a sequence of functions such that:

$\bullet$ $\sum f_n$ is absolutely convergent

$\bullet$ $\sum f_n$ is uniformly convergent

Then $\sum \|f_n\|$ is uniformly convergent.

$\|\cdot\|:$ Euclidean norm.

Any help would be appreciated.

Define $f_n(x,y)$ as
$$f_n(x,y) = \left(\frac{(-1)^nx^n}{n}\chi_{[0,1)}(x),0\right).$$
Then $\sum f_n$ is uniformly convergent to $(0,0)$ (Dirichlet test) and absolutely convergent to $0$ (comparison with $x^n$), but $\sum||f_n||$ is not uniformly convergent.
Note that $||f_n(x,y)||= |x^n/n|\chi_{[0,1)}(x)$ and
$$\sup_{x \in [0,1)} \sum_{k=n+1}^{2n} \frac{x^k}{k}\geq \sup_{x \in [0,1)}\frac{nx^{2n}}{2n}= \frac1{2},$$ which does not converge to $0$ as $n \rightarrow \infty$.