$\{a_n\}$ be a (complex) sequence such that $\sum a_nb_n$ converges for all sequence $b_n\to0$. Show that, $\sum |a_n|<\infty$
I define $\phi_i:c_0\to\Bbb{C}$ by $\phi_i(\{b_n\})=\sum\limits_{n=1}^i |a_n|b_n$
Then $|\phi_i(\{b_n\})|\le \lVert \{b_n\}\rVert_\infty (\sum\limits_{n=1}^i |a_n|)$. This implies $\lVert\phi_i\rVert_{op}\le \sum\limits_{n=1}^i |a_n|$.
Also $\phi_i(1,1,\ldots,1,0,0,0,\ldots)=\sum\limits_{n=1}^i |a_n|$. Hence $\lVert\phi_i\rVert_{op}= \sum\limits_{n=1}^i |a_n|$.
I want to apply Uniform Boundedness principal here.
$\text{sup}_i|\phi_i(\{b_n\})|\le\text{sup}_i\sum\limits_{n=1}^i |a_n||b_n|$.
But I don't know the above quantity is finite or not. I only have $\sum a_nb_n$ converges, but it may not imply $\sum |a_n||b_n|$ is finite.
Consider $\mathcal{c}_0$ the space of all sequences that converge to $0$. $c_0$ is a closed linear subspace of $(\ell_\infty,\|\;\|_\infty)$ and so, $(c_0,]|\;\|_\infty)$ is itself a Banach space
Suppose $a:\mathbb{N}\rightarrow\mathbb{C}$ is such that for all $x\in\mathcal{c}_0$, $\sum_na_nx_n$ converges
Following the idea of the OP, for each $m\in\mathbb{N}$ define the linear functional $\Lambda_m:x\mapsto \sum^m_{j=1}a_jx_j$. It is obvious that $\Lambda_m\in\mathcal{c}^*_0$. For any $x\in\mathcal{c}_0$, the orbit $\{\Lambda_mx:m\in\mathbb{N}\}$ is a bounded set in $\mathbb{C}$ since $\lim_m\sum^m_{j=1}a_jx_j=\sum^\infty_{j=1}a_jx_j$ exists and so, there is $c_x>0$ such that
$$|\Lambda_mx|=\Big|\sum^m_{j=1}a_jx_j\Big|\leq c_x$$ for all $m$. An application of the Banach-Steinhaus theorem (a.k.a. uniform boundedness principle) shows that $c:=\sup_m\|\Lambda_m\|<\infty$.
Hence, for any $y\in\mathcal{c}_0$ $$\Big|\sum^\infty_{j=1}a_jy_j\Big|=\lim_{m\rightarrow\infty}\Big|\sum^m_{j=1}a_jy_j\Big|=\lim_m|\Lambda_m y|\leq c\|y\|_\infty$$ that is, the linear map $\Lambda:y\mapsto\sum^\infty_{j=1}a_jy_j$ is an element of $\mathcal{c}^*_0$ which is know to be $\ell_1$. Hence, there is $\alpha\in\ell_1$ such that $\Lambda y=\sum^\infty_{j=1}\alpha_jy_j$. Applying $\Lambda$ to $e_i=\mathbb{1}_{\{i\}}$, we obtain that $\alpha_i=a_i$ for all $i\in\mathbb{N}$.