I think that the answer is no
I thought quite a bit about this problem. My idea was to build a sequence $(y_n)_{n \in \mathbb{N}} \subset A$ such that given a $x \in \ell^2$ we pick the first N components of $y \in A$ as $(y_1,y_2, \dots, y_N)=(x_1, \dots, x_N)$ and try to make $\|x-y\| < \varepsilon$. $$ \|x-y\|^2=\sum_{n=1}^\infty |x_n-y_n|^2=\sum_{n=N+1}^\infty |x_n-y_n|^2 \leq 2\sum_{n=N+1}^\infty x_n^2+y_n^2$$
Now as $y \in A$ it also has to satisfy the condition $\sum_{n=1}^\infty \frac{y_n}{n}=0$ But from Cauchy-Schwarz inequality (as both $y \in \ell^2$ and $\frac{1}{n} \in \ell^2$ we have that $\sum_{n=1}^\infty \frac{y_n}{n} \leq (\sum_{n=1}^\infty y_n^2)^{\frac{1}{2}} (\sum_{n=1}^\infty \frac{1}{n^2})^{\frac{1}{2}}$ and therefore if $\sum_{n=1}^{N}\frac{y_n} {n}=-a$ $\implies$ $\sum_{n=N+1}^\infty \frac{y_n} {n}=a$
And it therefore follows that $a \leq (\sum_{n=N+1}^\infty y_n^2)^{\frac{1}{2}} \pi/{\sqrt{6}}$
Now as $y \in \ell^2 \implies \lim_{n\ \to \infty} y_n =0$ and if $a$ is large enough we cannot approximate $x$ by $y$ because I need both $\sum_{n=N+1}^{\infty} x_n^2 <\epsilon/4$ and $\sum_{n=N+1}^{\infty} y_n^2 <\epsilon/4$. In the case of $(x_n)_n$ we can always choose an $N$ large enough but because of the aforementioned argument , we cannot ensure that such a $y \in A$ exists.
Is the idea of the proof correct? What is bothering me is that there could be another way of constructing the sequence $(y_n)_n$ which could work. Thanks
It is certainly not dense.
The linear functional
$$ T(x_n) = \sum_{n=0}^{\infty} \frac{x_n}{n} $$
Is continuous on $l^2$. Thus, the set $T^{-1}(0)$ must be a closed subspace of $l^2$. If it were dense, we'd have $l^2 = T^{-1}(0)$. But the sequence ${\frac{1}{n}}$ obviously isn't in this space.