Let $T \in \mathcal L(\ell^2(\mathbb C))$, so that $$ T(x_n) = \left(\frac{x_n}{n}\right)\,. $$ Show that the range $\mathcal R(T)$ is a dense subspace, knowing that the spectrum $\sigma(T) = \{1/n \mid n \in \mathbb N\} \cup \{0\}$ and a point spectrum $\sigma_p(T) = \sigma(T) \setminus \{0\}$.
We also know that $$ \mathcal R(T) = S := \left\{(y_n) \in \ell^2(\mathbb C) \mid \sum_n n^2 |y_n|^2 < \infty\right\}\,. $$
Hint: The dense subspace of finite sequences $\ell^{\mathrm{fin}}(\mathbb C)$ might be useful.
An attempt
We know that the range of $T$ is a subspace, as $T0_{\ell^2} = 0_{\ell^2} / n = 0_{\ell^2}$, and for $x, y \in \ell^2(\mathbb C)$ and $\alpha, \beta \in \mathbb C$, we have
\begin{align} \sum_n n^2 \left| T(\alpha x_n + \beta y_n)\right|^2 &= \sum_n n^2 \left| \frac \alpha n x_n + \frac\beta n y_n) \right|^2 \\ &= \sum_n \left| \alpha x_n + \beta y_n\right|^2 < \infty\,, \end{align} as $\ell^2(\mathbb C)$ is its own subspace.
As for density, let $y \in \ell^2(\mathbb C)$, so $$ \sum_k |y_k|^2 < \infty. $$ We know that for a converging series the tail approaches zero starting from some large enough $n$, which leads to $\ell^{\mathrm{fin}}(\mathbb C)$, the set of sequences whose tail is zero, being dense in $\ell^2(\mathbb C)$ and $\mathcal R(T)$. As density is transitive, $\mathcal R(T)$ is a subspace of $\ell^2(\mathbb C)$ and $\ell^{\mathrm{fin}}(\mathbb C)$, then $\mathcal R(T)$ has to be dense in $\ell^2(\mathbb C)$ as well.
Did I get the gist of this right, or am I missing something crucial?
For the fast answer you'll just need to recall some facts about the spectrum of a bounded linear operator on a Hilbert space:
the spectrum of a bounded linear operator on a Hilbert space is comprised of the point spectrum, the continuous spectrum and the residue spectrum.
For a normal operator, the residue spectrum is empty. The operator $T$ is even self adjoint.
Now, you know that $\sigma(T) = \{1/n \vert n \in \mathbb{N}\} \cup \{0\}$ and $\sigma_p(T) = \{1/n \vert n \in \mathbb{N}\}$. In other words, the continuous spectrum is $\{0\}$ and thus you know that $0I - T = T$ is a injective but not surjective and has dense image, which means that $R(T)$ is dense in $\ell^2(\mathbb{C})$.