Is the set \begin{align} A=\left\{a=(a_1,a_2,\dots)\in\ell^2 \ \ \lvert \ \ \sum_{k=1}^\infty \frac{a_n}{n}=0 \right\}\subset\ell^2 \end{align} dense in $\ell^2$
Is the following argument correct? Let $x=(1,0,0,\dots)$ and assume $\forall \epsilon>0\ \ \exists a\in A$ such that $\lVert x-a\lVert_2<\epsilon $. This implies $\lvert\sum_{n\geq1}\frac{x_n-a_n}{n}\lvert\leq \frac{\pi}{\sqrt{6}}\epsilon$, i.e. $\lvert\sum_{n\geq 1}\frac{a_n}{n}\lvert\geq1-\frac{\pi}{\sqrt{6}}\epsilon$. Choosing $\epsilon$ sufficiently small therefore implies $a\not\in A$