Let $A : \ell^1 \to \ell^1$ be an operator where for any $x \in \ell^1$, $Ax = \left(\frac{x_n}{n}\right)_{n \in \mathbb{N}}$.
Question. I'm trying to show that $\text{range}(A)$ is dense subspace of $\ell^1$.
I know that essentially we want to show that the closure of the range is equivalent to $\ell^1$, i.e.
$$\overline{\text{range}}(A) = \{x \in \ell_1 : \exists x_n \in \text{range}(A)~\text{such that}~x_n \to x\} = \ell^1.$$
But I don't really have an idea of how to formally go about showing the above using this definition?