Consider $\ell^\infty$. Let
$$A := \overline{\operatorname{conv}}\left(\left\{\dfrac{e_n}{n}\right\}\right)$$
It's not hard to see that this set is compact (using the Banach-Alaoglu theorem). But how can we show that $A \ne \operatorname{conv}(\operatorname{ex}(A))$?
The only way I see it is to find some convex combination of some points $\{a_i\} \in \operatorname{conv}(\operatorname{ex}(A)) \notin A$. Any hints, please?
Observe that $0 \in A$, so every finitely supported vector $x=(x_k)_{k \ge 1}$ with nonnegative entries such that $s(x)=\sum_{k \ge 1} kx_k \le 1$ is in $A$. Indeed, $$x=\sum_k {kx_k}\cdot(e_k/k)+(1-s(x))\cdot 0 \,.$$
Passing to the limit, we deduce that $$A=\{x=(x_k)_{k \ge 1}: \forall k\ge 1 \; \; x_k \ge 0 \; \:\text{and} \; \; \sum_{k \ge 1} kx_k \le 1\}\,.$$ Thus the extreme points of $A$ are $0$ and $e_k/k$ for $k \ge 1$, since if $x \in A$ has $x_k x_\ell>0$ where $k\ne \ell$, then for positive $\epsilon<\min\{k x_k,\ell x_\ell\}$, the vectors
$$y=x+(\epsilon/k)e_k-(\epsilon/\ell) e_\ell \in A$$ and $$z=x-(\epsilon/k)e_k+(\epsilon/\ell) e_\ell \in A$$ satisfy $x=(y+z)/2$. Finally, $$\sum_{k\ge 1} \frac{e_k}{k^2(k+1)} \in A$$ is not a finite convex combination of $0$ and $e_k/k$ for $k \ge 1$.