I'm trying to prove that $F=\{x=\{x_n\}_{n\in \mathbb{N}}\in l^2(\mathbb{N}):\sum_{n=1}^{\infty} x_n=0\}$ is dense in the sequence space $l^2(\mathbb{N})$.
I think it should be an easy exercise, but I don't see why an arbitrary sequence $x \in l^2(\mathbb{N})$ is the limit of one in F.
Any hint would be of help, thanks!
If $F$ is not dense, then its closure is not the whole space. Therefore, there is a non-zero functional that vanishes on $F$. In a Hilbert space all (continuous) functionals are of the form $(\cdot, y)$ for some $y$.
Assume $y\in\ell^2(\mathbb{N})$ is such that $(x,y)=0$ for all $x\in F$.
Then $y$ is orthogonal to all vectors of the form $e_i-e_j\in F$ where $e_i$ has a $1$ in the $i$-th component and zeros otherwise. Since $0=(e_i-e_j,y)=\overline{y_i}-\overline{y_j}$, it follows that all components of $y$ are equal. This is only possible if $y=0$.