I'm wondering what approach I should take to find the image of a linear operator: $T: \ell_2 \to \ell_2$ with $Tx = T[(x_n)_{n \ge 1}] = (\frac{x_n}{n})_{n \ge 1}$.
Now I know that since $T$ is clearly injective and $T^{-1}: T(\ell_2) \to \ell_2 $ is defined as $T^{-1}x = (n x_n)_{n \ge 1}$ is clearly unbounded, that the image of $T$, $T \ell_2 $ is not closed, but I am interested in trying to actually show that explicitly.
The range of $T$ is simply $\{(y_n)\in l^{2}: \sum_{n=1}^{\infty} n^{2}y_n^{2}<\infty\}$. This is not closed because the sequence $\{(1,\frac 1 {2^{3/4}},\frac 1 {3^{3/4}},...,\frac 1 {n^{3/4}},0,0,...)\}:n=1,2,...\}$ lies in the range and converges to $(1,\frac 1 {2^{3/4}},\frac 1 {3^{3/4}},...)$ which does not belong the range of $T$.