Here's Prob. 8, Sec. 2.3 in the book Introductory Functional Analysis With Applications by Erwine Kreyszig:
If in a normed space $X$, absolute convergence of any series always implies convergence of that series, show that $X$ is complete.
My effort:
Let $\left( x_n \right)_{n \in \mathbb{N}}$ be a Cauchy sequence in $X$. Then, given a real number $\varepsilon > 0$, we can find a natural number $N$ such that $$ \left\Vert x_m - x_n \right\Vert < \varepsilon$$ for all $m \in \mathbb{N}$ and $n \in \mathbb{N}$ such that $m > N$ and $n > N$. And so, $$ \left\vert \ \left\Vert x_m \right\Vert - \left\Vert x_n \right\Vert \ \right\vert < \varepsilon $$ for all $m \in \mathbb{N}$ and $n \in \mathbb{N}$ such that $m > N$ and $n > N$.
Now we define a sequence $\left( y_n \right)_{n \in \mathbb{N}}$ as follows: Let $y_1 \colon= x_1$, and let $y_n \colon= x_n - x_{n-1}$ for $n = 2, 3, 4, \ldots$. Then we see that $y_1 + \cdots + y_n = x_n$ for all $n \in \mathbb{N}$.
What next?
In fact this is an equivalence.
For the implication you seek, take $n_1<n_2<n_3<\cdots$ such that $$ \|x_{n_k}-x_{n_{k+1}}\|\leq\frac{1}{2^k}\quad\forall k $$ Then $$ \sum_{k\geq1}\|x_{n_k}-x_{n_{k+1}}\|<\infty $$ Hence $\sum_k(x_{n_k}-x_{n_{k+1}})$ converges by hypothesis. This means $x_{n_1}-x_{n_{k+1}}$ tends to a limit (telescoping series). So $x_{n_k}$ tends to a limit $x\in X$. Finally, $$ \|x_n-x\|\leq\|x_n-x_{n_k}\|+\|x_{n_k}-x\|<\epsilon+\epsilon $$ if $n,n_k$ are big enough. This shows that the entire sequence $x_n\to x$.