Let us start with a given function $f_n$ defined on a certain domain $I$ and s.t. $f_n:I\mapsto\text{ some normed vector space}$. We know that:
$$\sum_{n=0}^\infty \|f_n\| := \sum_{n=0}^\infty \sup_I |f_n(x)| < \infty\implies \sum_{n=0}^\infty f_n(x)\text{ is normally convergent}\tag{1}$$
I was thinking that, at this point, it could have been enough to apply the so called Weierstrass M-test to show that $\sum_{n=0}^{\infty}f_n$ is uniformly and absolutely convergent as well.
However, Weierstrass M-test first requires absolute convergence of the series, that is: $$\sum_{n=0}^\infty |f_n(x)|\text{ convergent}$$ and I know that in general it holds true that:
$$\sum_{n=0}^\infty f_n(x)\text{ convergent}\nRightarrow \sum_{n=0}^\infty |f_n(x)|\text{ convergent} $$
Hence, starting from normal convergence of a series, how is it possible to use Weierstrass M-test so as to show that the series converges absolutely and uniformly as well? Which is the missing part above?
For any $x \in I$, $$\sum_{n=0}^\infty |f_n(x)| < \sum_{n=0}^\infty \sup_{x \in I}|f_n(x)| = \sum_{n=0}^\infty \|f_n\| < \infty$$ Then by the $M$-test the series is uniformly and absolutely convergent if the normed vector space is a Banach space.