Let $D$ be an unbounded densely defined operator on a separable Hilbert space $H$. If $D$ is diagonalisable with all eigenvalues having finite multiplicity and growing towards infinity, does it follow that the spectrum of $D$ contains only eigenvalues?

EDIT: Diagonalisable means the existence of a countable ONB $e_n$ (orthogonal set with dense span) such that $D(e_n) = \alpha_n e_n$, for some complex number $\alpha_n$.

Growing towards infinity means that $\alpha_n \to \infty$ as $n \to \infty$.