= Solution
A <bounded operator> on a complex <Hilbert space> is <self-adjoint> when $T=T^*$, equivalently $\langle Tx,y\rangle=\langle x,Ty\rangle$. The <spectral theorem for compact self-adjoint operators> says its nonzero spectrum consists of real <eigenvalues> of finite multiplicity, with at most countably many and with accumulation possible only at zero. Their <eigenspaces> are mutually orthogonal; their closed span is the <orthogonal complement> of the kernel. On that span the operator has the norm-convergent spectral action $Tx=\sum_j\lambda_j\langle x,e_j\rangle e_j$.
Take $H=\ell^2(\mathbb N)$ and $Te_n=e_n/n$. It is <self-adjoint> and has infinite-dimensional range. Truncation to the first $N$ coordinates is a <finite-rank operator> and differs from $T$ in norm by $1/(N+1)$, so $T$ is compact.
The <resolvent set> comprises those $\lambda$ for which $T-\lambda I$ has a bounded everywhere-defined inverse; its complement is the <spectrum>, and the <point spectrum> consists of <eigenvalues>. In this example
$$
\boxed{\sigma_p(T)=\{1/n:n\ge1\},\quad
\sigma(T)=\{0\}\cup\{1/n:n\ge1\},\quad
\rho(T)=\mathbb C\setminus\sigma(T)}.
$$
For $\lambda$ outside the displayed closed set the inverse is diagonal with entries $(1/n-\lambda)^{-1}$ uniformly bounded, proving membership of the resolvent. Each $1/n$ has <eigenvector> $e_n$. Zero is not an <eigenvalue>, but $\|Te_n\|\to0$ prevents a bounded inverse; also $(1/n)_n\in\ell^2$ has no preimage since that would be the non-square-summable constant sequence. This justifies every asserted spectral case.
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