Let $(c_n)_n$ a bounded succession of complex numbers, and $T:\ell_2 \to \ell_2$ a operator defined by $T(x)=(c_nx_n)_n$, for all $x = (x_n)_n \in l_2$.
Show that if $\lambda \notin \overline{\{c_n\}}$, then $\lambda \notin \sigma(T)$
I have succeeded in demonstrating that each $(c_n)$ is an eigenvalue of $T$ and considering $\lambda \notin \overline{\{c_n\}}$, $T-\lambda I$ is injective. I would have to prove that it is also surjective, so $\lambda \notin \sigma(T)$, but I can not do it, help please.
Hint: if $\lambda\notin\overline{\{c_n\}}$, then $\delta:=\inf_n|\lambda-c_n|>0$. If $(\lambda I-T)^{-1}$ existed, what would it be?