I don't think the converse of the Weierstrass test holds. However, you can use the following
Fact: A series $\sum\limits_{i=1}^\infty f_i(x)$ of real-valued functions converges uniformly on a set $E$ if and only if it is uniformly Cauchy. That is, if and only if, given $\epsilon>0$, there is an $N$ such that $\Bigl|\sum_{i=n}^m f_i(x)\,\Bigr |<\epsilon$ for all $m\ge n\ge N$ and for all $x\in E$.
Now think about this and the triangle inequality.
$\color{maroon}{\text {Warning! Solution follows:}}$
We now show that your series $\sum\limits_{i=n}^m f_i(x)$ converges uniformly:
Let $\epsilon>0$. Since, $\sum\limits_{i=1}^\infty |f_i(x)|$ converges uniformly, there is an $N$ such that for all $m\ge n\ge N$ and for all $x$: $ \sum_{i=n}^m |f_i(x)| <\epsilon.$
Using this and the triangle inequality, it follows that for all $m\ge n\ge N$ and for all $x$: $ \Bigl|\sum_{i=n}^m f_i(x)\,\Bigr| \le \sum_{i=n}^m |f_i(x)| <\epsilon. $
So $\sum\limits_{i=1}^\infty f_i(x)$ is uniformly Cauchy, and thus uniformly convergent.
For completeness:
The Fact above follows (by looking at the sequence of partial sums of the series) from the following standard theorem:
Theorem: A sequence of real-valued functions $\{f_n\}$ is uniformly convergent on a set $E$ if and only if it is uniformly Cauchy on $E$; that is, given $\epsilon>0$, there is an $N$ so that $ \tag{1}|f_n(x)-f_m(x)|<\epsilon,\quad \text{ for all }n,m\ge N\text{ and for all }x\in E $
Proof:
To prove the forward implication, suppose $\{f_n\}$ converges uniformly to $f$ on the set $E$. Then, given $\epsilon>0$, there is an $N$ so that $ |f_n(x)-f(x)|<\epsilon/2 $ for all $n\ge N$ and for all $x\in E$. Thus, if $m,n\ge N$ and $x\in E$ $ |f_n(x)-f_m(x)|<|f_n(x)-f(x)|+|f_m(x)-f(x)|<{\epsilon\over2}+{\epsilon\over2}=\epsilon. $ From this, it follows that $\{f_n\} $ is uniformly Cauchy on $E$.
To prove the reverse implication, suppose $\{f_n\}$ is uniformly Cauchy on $E$. Then for each $x\in E$, the sequence $\{f_n(x)\}$ is Cauchy and thus converges to some number $f(x)$.
We claim that $\{f_n\}$ converges uniformly to $f$, as defined above, on $E$.
Towards proving the claim, let $\epsilon>0$ and choose $N$ a positive integer that verifies equation (1).
Then if $m\ge N$ is fixed, $n\ge N$, and $x\in E$: $ \tag{2}|f_m(x)-f_n(x)|<\epsilon $ Taking the limit as $n\rightarrow\infty$ in (2) gives $ \tag{3}|f_m(x)-f (x)|\le\epsilon . $ Since $\epsilon$ was an arbitrary positive number, and since (3) holds for all $m\ge N$ and all $x\in E$, it follows that $\{f_n\}$ converges uniformly to $f$ on $E$.