# Dense convex proper subset of the Hilbert space $l_2$: $\{x|\sum x_i=0\}$ [duplicate]

Let's consider the space $l_2$ (all sequences $x$ with $\sum x_i^2 < +\infty$) and its subset $Z = \{x|\sum x_i = 0\}$.
I want to prove that the closure of $Z$ is $l_2$, but I can't. I tried to build some constructions to get $Z$ elements $\epsilon$-close to an arbitrary $l_2$ element, but nothing comes out. I ask for some help. Is there any theorem or a method which might help me?

Given $x\in\ell^2$ and $\varepsilon>0$, you can use the fact that $\sum\limits_{n=k+1}^\infty |x_n|^2\to 0$ as $k\to \infty$ to show that there is a $z\in Z$ of the form $(x_1,x_2,x_3,\ldots,x_k,a,a,a,\ldots,a,a,0,0,0,\ldots)$, with $m$ copies of the number $a$, such that $\|x-z\|_2<\varepsilon$. First you choose $k$ big enough, then choose $m$ big enough ($k$ will depend on $\varepsilon$, $m$ will depend on $k$ and $\varepsilon$, $a$ is determined by the definition of $Z$).