Span of standard 'basis vectors' in $l_\infty$

Question

Let $e_i=(0,...,0,1,0,...)\in l_\infty$ be the sequences with $1$ on the $i$-th position and all other entries $0$ for $i\in\mathbb{N}$.

How do I prove that the span of all $e_i$ is not dense in $l_\infty$?

What I thought:
The span of $e_i$ consists of all finite(?) linear combinations of $e_i$. So I thought the element $(1,-1,1,-1,1,-1,...)\in l_\infty$ is not the limit of any sequence of elements in the span, meaning the span is not dense.
How do I make this proof (if it is correct) more formal?

Answer 1

Hint: Show that if $s$ is your sequence (or, more simply, the sequence of $1$s), then for every finite linear combination $v$ of standard basis vectors, you have $\|s - v\|_{\infty} \geq 1$.