Problem:
Let $M = \{ x \in \ell^2 : \sum_{k = 1}^\infty x_k = 0 \}$. Prove that the $M$ is a linear subspace of $\ell^2$ and that $\overline M = \ell^2$.
Idea:
It's quite trivial to prove, unless I am mistaken, that the $M$ is a linear subspace of $\ell^2$.
What I'm having difficulties proving is that $\overline M = \ell^2$.
We know that $(M^{\perp})^{\perp} = \overline{span M}$, and as $M$ is a linear subspace of $\ell^2$, then $\overline{span M} = \overline M$.
Thus, it would suffice to show that $(M^\perp)^\perp = \ell^2$.
What I've noticed is that $M = \{\vec 1\}^\perp$, however the element $\vec 1$ is not member of $\ell^2$ – as far as I know.
Furthermore, if I take an arbitrary element $y \in M^\perp$, then $$y \perp (0, ..., 0, 1, -1, 0, ...) \\ \Rightarrow (\forall n \in \mathbb{N}) y_n = y_{n+1} \\ \Rightarrow y = c \vec 1$$ for some $c \in \mathbb{C}$. Thus again we see that $\vec 1 \in M^\perp$, which is not in $\ell^2$.
Is my idea good? What am I missing?