= Solution
A linear operator between Banach spaces is a <compact operator> when it maps the <unit ball> into a relatively <compact set>, equivalently when every bounded sequence of inputs has an image subsequence converging in <norm>. Such an operator is bounded. For a <bounded linear operator> on a complex <Banach space>, the <resolvent set> consists of the complex $\lambda$ for which $T-\lambda I$ is bijective with bounded inverse; its complement is the <spectrum>. The <point spectrum> consists of those $\lambda$ for which $T-\lambda I$ has a nonzero <kernel>, namely the <eigenvalues>.
On a <Hilbert space>, $T$ is <self-adjoint> if $T=T^*$, or equivalently $\langle Tx,y\rangle=\langle x,Ty\rangle$ for all vectors. The given prescription extends uniquely to
$$
Tx=\sum_{j\geq1}\frac{x_j}{j}e_j,\qquad x=\sum_jx_je_j.
$$
It is bounded with <norm> one. The real diagonal entries prove self-adjointness by the inner-product identity. Define the finite-rank truncation $T_mx=\sum_{j\leq m}x_je_j/j$. Then $\|T-T_m\|=1/(m+1)\to0$. To see compactness directly, extract successively convergent subsequences of the finitely many first coordinates from a bounded input sequence; the uniform tail bound just displayed makes the resulting image subsequence Cauchy. Hence \b[$T$ is compact and <self-adjoint>].
Each $1/j$ is an <eigenvalue> with <eigenspace> $\mathbb Ce_j$. If $Tx=\lambda x$, each coordinate satisfies $(1/j-\lambda)x_j=0$, so there are no other <eigenvalues>; in particular zero is not an <eigenvalue>. But $\|Te_j\|=1/j\to0$, which contradicts a bounded inverse at zero. Thus zero is in the <spectrum>.
For $\lambda\notin\{0,1,1/2,1/3,\ldots\}$, its distance from this <closed set> is positive. The diagonal formula
$$
(T-\lambda I)^{-1}y=\sum_{j\geq1}\frac{y_j}{1/j-\lambda}e_j
$$
then defines a bounded inverse. This proves the complete description
$$
\boxed{\sigma(T)=\{0\}\cup\{1/j:j\geq1\},\quad
\sigma_p(T)=\{1/j:j\geq1\},\quad
\rho(T)=\mathbb C\setminus\sigma(T).}
$$
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