I have been studying functional analysis and I came across this question: Show that $$F := \{(s\_n)\in l^1 : (ns\_n)\notin l^{\infty}\}$$ is dense in $l\_1$.

So here is my approach and I would like some feedback if possible: $(e\_n) \in l^1$ (a sequence with unit value in the nth position) is certainly part of this set as $(ne\_n)$ is unbounded. We know already that the span of $(e\_n)$ is dense in $l^1$. Every other sequence in F can be represented by the span of $(e\_n)$. Hence, $(e\_n)$ is dense in F itself. Hence, F is dense in $l^1$.

Is this correct? I am sorry that I can be very vague in structuring my arguments (I am new to mathematics) and was wondering if anyone could help me with structuring this better or correct me if I am wrong somewhere.