Consider the sequence space $$S_C=\left\{(x_n)_{n\in\mathbb{N}}:x_n\in\mathbb{F}, \left(\sum_{k=1}^n x_k\right)_{n\in\mathbb{N}} \text{ is convergent}\right\}$$ of summable sequences. We consider $\lVert(x_n)_{n\in\mathbb{N}}\rVert =\sup_n | \sum_{k=1}^n x_k |$ to be the norm on $S_C$. In particular $(S_c, \lVert \cdot \rVert)$ is a Banach space. Furthemore consider the Banach space of absolute summable $\mathbb{F}$ sequences $\ell^1$.
(This is a continuation of a post that I made about the norm of this space: Showing that a norm in the space of summable series is well defined).
Prove that the inclusion $i:\ell^1 \to S_C$ has dense range.
I already showed that $i$ is well defined, that is: $(x_n)\in\ell$, then also $(x_n)\in S_C$. We can also see that $\lVert i \rVert = 1$.
Now to show that $i$ has dense range I think that we need to show that $\overline{\text{Im}(i)}=S_C$. In particular (and this is where I might be wrong) for all $\varepsilon>0$ and for any $(x_n)\in S_C$ there is a $(y_n)\in \ell^1$ such that $\lVert (x_n)-(y_n)\rVert < \varepsilon$. I'm not so sure about that last part. But I think that instead of proving this directly there's a more straightforward way. For example we know that because we are working with Banach spaces and $i\in B(l^1, S_C)$ that: $i$ has dense range and is bounded below (which it is I think by $\lVert i \rVert =1$) iff $i$ is invertible, but I don't know if this would be better or not.
I appreciate any help :)
As suggested in the comments, it is convenient to work with sequences that have finite support, which means sequences where only finitely many terms in the sequence are non-zero. Let $(x_{n})_{n\in\mathbb{N}}\in S_{C}$. Let $\varepsilon \in (0, \infty )$. As $(\sum_{k=1}^{n} x_{k})_{n\in\mathbb{N}}$ is convergent, it is a Cauchy sequence. So there exists some $N\in\mathbb{N}$ such that if $m,n\in\mathbb{N}$ with $N\leq m$, $N\leq n$ and $m<n$, it follows that $|\sum_{k=m+1}^{n}x_{k}| = | \sum_{k=1}^{n}x_{k} - \sum_{k=1}^{m}x_{k}| < \varepsilon$. In particular, it follows that $|\sum_{k=N+1}^{n}x_{k}| < \varepsilon$ for all $n\in\mathbb{N}$ with $N < n$.
For each $n\in\mathbb{N}$, define $y_{n} := \begin{cases} x_{n} & \text{ if } n\in \{1, \ldots , N\}, \\ 0 & \text{ if } n\in \mathbb{N} \setminus \{1, \ldots , N\}. \end{cases}$
Consider the sequence $(y_{n})_{n\in\mathbb{N}}$. It is clear that $(y_{n})_{n\in\mathbb{N}} \in \ell^{1}$. Given $n\in\mathbb{N}$, observe that
$$\left| \sum_{k=1}^{n}x_{k} - \sum_{k=1}^{n}y_{k} \right| = \begin{cases} 0 & \text{ if } n\in \{1, \ldots , N\}, \\ |\sum_{k=N+1}^{n}x_{k}| & \text{ if } n\in \mathbb{N} \setminus \{1, \ldots , N\}. \end{cases}$$
So it follows that $|\sum_{k=1}^{n}(x_{k} - y_{k})| < \varepsilon$ for all $n\in\mathbb{N}$. This shows that $\|(x_{n})_{n\in\mathbb{N}} - (y_{n})_{n\in\mathbb{N}}\| \leq \varepsilon$, which proves the desired result.
In fact, the same proof obtains a stronger result. If $c_{00}$ denotes the vector space of $\mathbb{F}$-valued sequences with finite support equipped with the supremum norm, then $c_{00}$ is contained in $\ell^{1}$ and the proof above show that the inclusion map of $c_{00}$ into $S_{C}$ has dense range.