Prove that the operator $T:\ell^2\rightarrow\ell^2$ defined as \begin{equation} Tx = \bigg(x_2, \frac{x_3}{2}, \frac{x_4}{3}, \dots \bigg) \end{equation}for $x = (x_1, x_2, x_3, \dots) \in \ell^2$ is compact
I am trying to prove this problem be following approach, first let we defined a sequence of operator $T_n$ such that $T_n x = \bigg(x_2, \frac{x_3}{2}, \frac{x_4}{3}, \dots, \frac{x_{n+1}}{n}, 0, 0, 0, \dots \bigg)$, which $T_n x$ only doing the operation for first $n$ element for $x = (x_1, x_2, x_3, \dots) \in \ell^2$. Each $T_n$ is clearly a compact operator since it is finite rank.
Then if I could show that $T_n x$ converge to $Tx$, then there is a theorem guarantee that the limit of $T_n$, which should be $T$, is also an compact operator. I think this is a right approach, but I am sticking on showing that $\|Tx - T_nx\| \to 0$. I cannot bounded $\|Tx - T_nx\|$ by something like $\|x\|/n$. Could someone can help me on this problem?