= Solution
Let $Q_N$ be truncation to the first $N$ coordinates in $\ell^1$. For $x\in\ell^1$, boundedness of $T$ and absolute convergence give
$$
(T-TQ_N)x=\sum_{n>N}x_nTe_n,\qquad
\|(T-TQ_N)x\|_1\leq\sup_{n>N}\|Te_n\|_1\,\|x\|_1.
$$
Applying the difference to each $e_n$ with $n>N$ gives the reverse bound on the <operator norm>. Hence
$$
\boxed{\|T-TQ_N\|=\sup_{n>N}\|Te_n\|_1.}
$$
The stated condition makes this tend to zero. Each $TQ_N$ has finite rank, so part (a)(ii) proves <compactness>. Thus the condition is \b[sufficient] on $\ell^1$.
For failure of necessity, set $Tx=(\sum_{n\geq1}x_n)e_1$. The sum defines a bounded <linear functional> on $\ell^1$, so $T$ is a bounded rank-one operator and therefore <compact>. But $Te_n=e_1$ for every $n$, so $\|Te_n\|_1=1$. Thus \b[<compactness> need not make the basis images tend to zero on $\ell^1$].
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