= Solution
Let $e_j$ be the <coordinate vectors> and define finitely supported vectors
$$
v_k=2^{-k}\sum_{j=2^k}^{2^{k+1}-1}e_j,\qquad k=0,1,\ldots.
$$
Each $v_k$ belongs to $\ell^1$, and $\|v_k\|_\infty=2^{-k}$. Hence $\sum_kv_k$ is absolutely convergent for the <supremum norm>. Its <partial sums> converge uniformly in $\ell^\infty$ to the sequence $v$ whose entries on the $k$th block all equal $2^{-k}$. But every block contributes one to its $\ell^1$ sum:
$$
\sum_j|v_j|=\sum_{k=0}^{\infty}2^k2^{-k}=\infty.
$$
Thus $v\notin\ell^1$. Any supremum-<norm> limit in $\ell^1$ would have the same coordinates as $v$, an impossibility. The <partial sums> are Cauchy and have no limit in the space. Therefore \b[$(\ell^1,\|\cdot\|_\infty)$ is not Banach], and the displayed <series> is the required absolutely convergent but nonconvergent <series> in that space.
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