= Solution
Suppose $a^k\rightharpoonup a$ in $\ell^1$ but $a^k$ does not converge to $a$ in the <norm topology>. There are $\varepsilon>0$ and a <subsequence> $(a^{m_k})$ such that
$$
\lVert a^{m_k}-a\rVert_1\geq\varepsilon
\qquad\text{for every }k.
$$
Set $x^k=a^{m_k}-a$. Then
$$
x^k\rightharpoonup0,
\qquad \lVert x^k\rVert_1\geq\varepsilon,
$$
which proves the first assertion.
We now use a <gliding hump argument>. Put $b_0=0$. Having chosen $b_{n-1}$ and $k_{n-1}$, coordinatewise convergence $x_i^k\to0$ lets us choose $k_n>k_{n-1}$ so that
$$
A_n:=\sum_{i\leq b_{n-1}}|x_i^{k_n}|<\frac\varepsilon6.
$$
For this fixed element of $\ell^1$, choose $b_n>b_{n-1}$ so far out that
$$
C_n:=\sum_{i>b_n}|x_i^{k_n}|<\frac\varepsilon6.
$$
Define one sequence $y$ by
$$
y_i=\operatorname{sgn}(x_i^{k_n})
\qquad\text{when }b_{n-1}<i\leq b_n.
$$
The blocks partition the positive integers and $|y_i|\leq1$, so $y\in\ell^\infty$. On the $n$th assigned block, the signs agree; outside it, use $|y_i|\leq1$. The <duality of l1 and l infinity> gives
$$
\begin{aligned}
\langle x^{k_n},y\rangle
&\geq
\sum_{b_{n-1}<i\leq b_n}|x_i^{k_n}|-A_n-C_n\\
&=\lVert x^{k_n}\rVert_1-2A_n-2C_n\\
&>\varepsilon-\frac{2\varepsilon}{6}-\frac{2\varepsilon}{6}
=\frac\varepsilon3.
\end{aligned}
$$
But $x^{k_n}\rightharpoonup0$ requires $\langle x^{k_n},y\rangle\to0$ for this fixed $y\in(\ell^1)^*=\ell^\infty$, a contradiction. Therefore every weakly convergent sequence in $\ell^1$ converges in norm. This is the <Schur property of l1>.
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