$\ell^1$ denotes absolutely summable sequences, and $\ell^2$ square summable sequences. It is well-known that $\ell^1$ is dense in $\ell^2$, with the standard argument that for any sequence $(a_1, a_2, \ldots) \in \ell^2$ one can take the finite sequence $(a_1, \ldots, a_N, 0, 0, \ldots) \in \ell^1$ and approximate the original sequence arbitrarily well as $N \to \infty$.
However, consider any $(a_n) \in \ell^2$, where I will assume that $a_n>0$ for simplicity, take any sequence $(b_n) \in \ell^2 \setminus \ell^1$ such as $b_n = \frac{1}{n}$, and define the sequence $(c_n)$ by $c_n = a_n + \epsilon b_n$. Now, $(c_n) \in \ell^2$ by Cauchy-Schwarz, but clearly $(c_n) \notin \ell^1$ because we're adding a positive sequence and a sequence with a divergent series. But since $(c_n) \to (a_n)$ as $\epsilon \to 0$, doesn't this mean that $\ell^2 \setminus \ell^1$ is dense in $\ell^2$? How to reconcile this with the fact that $\ell^1$ is dense in $\ell^2$?
