Let $\alpha=\{\alpha_n\}_{n\geq 1}\subseteq \mathbb{R}$ such that $\sum_{n=1}^{\infty} \alpha_n x_n <\infty$, for every $x=\{x_n\}_{n\geq 1}\in\ell_1$. The goal is to prove that $\sup_{n\geq 1} |\alpha_n|<\infty$.
I noticed that this is equivalent to proving that $\lVert \alpha\rVert_{\infty}<\infty$. However, I don't know how to start the proof. What I also noticed is that $\lVert \cdot\rVert_{\infty}$ is the natural norm of $c_0=\{x=\{x_n\}_{n\geq 1}\subseteq\mathbb{R}: x_n\rightarrow 0\}$ and $c_0^* = \ell_1$, where $c_0^*$ denotes the dual space of $c_0$, but I don't know if this is useful for the proof.
Thank you for your help!
Suppose $\sup |\alpha_n|=\infty$. Then there is a sequence $a_{n_k}$ with $|a_{n_k}|>k$. Choose now $x_n\in l_1$ with $x_{n_k}=\operatorname{sign}(\alpha_{n_k})/k^2$ and the rest of its elements equal zero. Then $$ \sum \alpha_n x_n>\sum_k \frac 1{k}=\infty, $$ contradicting your assumption.