= 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 $\{0\}\cup\{1/j:j\geq1\}$, 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: $\|Se_j\|_1=1/j\to0$, excluding a <bounded inverse>. Therefore
$$
\boxed{p\sigma(S)=\{1/j:j\geq1\},\qquad\sigma(S)=\{0\}\cup\{1/j:j\geq1\}.}
$$
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