Do the notions of weak and weak* convergence coincide for $\ell^1(\mathbb{N})$? As my friends and I were studying for our real analysis final exam yesterday, we were playing with various examples and found ourselves asking this question:

The space $\ell^1(\mathbb{N})$ is the dual of $c_0(\mathbb{N})$, and the dual of $\ell^1(\mathbb{N})$ is $\ell^\infty(\mathbb{N})$.
Is it possible to have a sequence $\{b_n\}\in\ell^1(\mathbb{N})$ converge to $b\in\ell^1(\mathbb{N})$ weakly*, but not weakly?

We knew that the weak and weak* topologies agree on a reflexive space, and because $\ell^1(\mathbb{N})$ is non-reflexive, we were wondering whether the weak and weak* topologies would agree.
Parsing the definitions, this is just asking whether there is a sequence $\{b_n\}\in\ell^1(\mathbb{N})$ such that, for any $r\in c_0(\mathbb{N})$, we have
$$\sum_{k=1}^\infty (b_n)_kr_k\to\sum_{k=1}^\infty b_kr_k\quad \text{ as }n\to\infty,$$
but for some $s\in \ell^\infty(\mathbb{N})$,
$$\sum_{k=1}^\infty (b_n)_ks_k\nrightarrow\sum_{k=1}^\infty b_ks_k\quad \text{ as }n\to\infty.$$
WLOG, we can let take $b=0$ (just subtract $b$ from all the $b_n$), so this becomes:

Is there a sequence $\{b_n\}\in\ell^1(\mathbb{N})$ such that, for any $r\in c_0(\mathbb{N})$, we have
$$\sum_{k=1}^\infty (b_n)_kr_k\to 0\quad \text{ as }n\to\infty,$$
but for some $s\in \ell^\infty(\mathbb{N})$,
$$\sum_{k=1}^\infty (b_n)_ks_k\nrightarrow 0\quad \text{ as }n\to\infty ?$$

We weren't able to come up with any examples, but of course that doesn't mean there aren't any. Also, just to double-check, were we correct in assuming that it would suffice to check whether weak and weak* convergence of sequences agreed in order to determine whether the weak and weak* topologies agreed?
 A: The unit vectors  in $\ell_1$ converge weak* to 0 (this is easy to see, with $c_0$ as the pre-dual) but not weakly to 0 (look at the action of $(1,1,\ldots)\in\ell_\infty$ on them; this can also be seen since no sequence of convex combinations of the unit vectors in $\ell_1$ can coverge in norm to 0).  
In fact, the weak and weak* topologies agree if and only if  X is reflexive. One direction of this follows from the definitions, the other can be duduced from the fact that a Banach space is reflexive if and only if its unit ball is weakly compact.
I should also mention that there are non-reflexive Banach Spaces in which weak* convergent sequences weakly converge (the converse of this always holds). So, just considering sequences does not suffice to show the two (non-metrizable) topologies are the same.
One space where this happens is in $\ell_\infty^*$ (c.f., Joseph Diestel, Sequences and Series in Banach Spaces, Theorem 15, page 103).  This result is attributed to A. Grothendieck   from his 1953 paper:  Sur Les applications linéaires faiblement compactes d'espaces du type C(K), Canadian J. Math., 5, 129-173.
