= Solution
A bounded linear operator is <compact> if it maps the unit ball to a set with compact closure; equivalently, every bounded sequence has a subsequence whose images converge.
If $S,T$ are <compact operators>, select a subsequence on which $Sx_n$ converges and then a further subsequence on which $Tx_n$ converges. This proves <compactness> of any <linear combination>. For norm closure, suppose compact $T_j$ converge to $T$ in <operator norm>. Given $\varepsilon>0$, choose $j$ with $\|T-T_j\|<\varepsilon/3$. A finite $\varepsilon/3$-net for $T_j$ of the unit ball is then a finite $2\varepsilon/3$-net for its image under $T$. This image is totally bounded, and its closure is compact because $X$ is complete. Thus $B_0(X)$ is a closed linear subspace.
For bounded $S$ and compact $T$, the image of a bounded sequence under $S$ is bounded, so $TS$ is compact; and a convergent subsequence of $Tx_n$ remains convergent after applying the continuous map $S$, so $ST$ is compact. Therefore \b[the <compact operators> form a closed two-sided ideal in $B(X)$].
For the weighted backward shift, truncate its output after coordinate $N$, obtaining a finite-rank operator $T_N$. The exact norm estimate is
$$
\|T-T_N\|=\sup_{k>N}\frac1{k+1}=\frac1{N+2}\to0.
$$
Hence $T$ is compact. Its iterates are
$$
(T^mx)_k=\frac{x_{k+m}}{(k+1)(k+2)\cdots(k+m)},\qquad
\|T^m\|=\frac1{(m+1)!}.
$$
Thus for every nonzero complex $\lambda$ the <Neumann series>
$$
(\lambda I-T)^{-1}
=\sum_{m\ge0}\lambda^{-m-1}T^m
$$
converges absolutely in <operator norm>. Zero is an <eigenvalue>, since $Te_1=0$, whereas every nonzero value is in the resolvent set. Consequently
$$
\boxed{T\text{ is compact},\qquad \sigma(T)=\{0\}.}
$$
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