 Solution

For the diagonal operator $S$, the equation $Sa=\lambda a$ gives $(j^{-1}-\lambda )a_j=0$ for every $j$. Thus a nonzero eigenvector exists exactly for $\lambda =1/j$, with eigenvector $e_j$; zero is not an eigenvalue because all diagonal entries are nonzero.

If $\lambda$ lies outside $\{0\}\cup \{1/j:j\ge 1\}$, its distance from this closed set is positive. Dividing each coordinate by $j^{-1}-\lambda$ gives a bounded inverse of $S-\lambda I$. Zero lies in the spectrum: $\Vert S{e}_j{\Vert }_1=1/j\to 0$, excluding a bounded inverse. Therefore