Let's define the subset of $\ell^2(\mathbb C)$ $$\mathcal D(A) = \left\{ {z \in {\ell ^2}\left( C \right),\sum\limits_{k = 1}^\infty {k^2{{\left| {{z_k}} \right|}^2} < \infty } } \right\},$$
I want to show the completeness of this subset.
my attempt:
We take a Cauchy sequence $\left\{ {z_{}^{\left( m \right)}} \right\}$ in $\mathcal{D}(A)$ and verify that the limit point is in $\mathcal{D}(A)$. Let \begin{eqnarray} {z_n} = \mathop {\lim }\limits_{m \to \infty } z_n^{\left( m \right)} \end{eqnarray} This implies component wise convergence, i.e, $\left| {z_n^{\left( m \right)} - {z_n}} \right| \to 0$ since ${\left\| {z - {z^{(m)}}} \right\|_2} \to 0$ in the $\ell_2$ sense, i.e., convergence in the $\ell_2$ sense implies convergence component wise. However, since $\sum\limits_{k = 1}^\infty {{k^2}{{\left| {z_k^{\left( m \right)}} \right|}^2}} < \infty$ for all $m$ then it holds that \begin{eqnarray} \left| {\sum\limits_{k = 1}^\infty {{k^2}{{\left| {{z_k}} \right|}^2}} - \sum\limits_{k = 1}^\infty {{k^2}{{\left| {z_k^{\left( m \right)}} \right|}^2}} } \right| \to 0 \end{eqnarray} as $m\to \infty$. The fact that $z$ is in $\ell^2$ is tivial since $\ell^2$ is a complete normed space (it is both Hilbert and Banach). Hence we showed that $z \in \mathcal D(A)$.
Are there any mistakes in my approach?
Possibly you are not asking the question you'd really want to ask.
First, no, the closure of that set is the whole $\ell^2$, because it includes all finite linear combinations of standard basis elements.
But it might be (as in some real-life applications, e.g., Sobolev spaces) that you want to define a new norm by (the square root of) the expression that you wrote. This gives a new inner-product-space structure, and (as with $\ell^2$ itself) gives a complete space.