In a Banach space $X$, the representation $$\overline{co}(A)=\left\{\sum_{n=1}^\infty t_n a_n: t_n\ge 0, \sum_{n=1}^\infty t_n\le 1\right\}$$ holds for $A=\{a_n:n\in\mathbb N\}$ if $a_n\to 0$ (if $X$ is only normed, you can apply this to the completion).

To prove this, consider the sequence space $\ell^1$ as the dual of the space $c_0$ of all null sequences and $T:\ell^1\to X$, $(t_n)_{n\in\mathbb N}\mapsto \sum_{n=1}^\infty t_n a_n$ (which clearly converges because $X$ is complete). Then $T$ is linear and $\sigma(\ell^1,c_0)-\sigma(X,X')$-continuous. Indeed, for every $f\in X'$ we need the continuity of $f\circ T$ which is the map $(t_n)_n\mapsto \sum_n t_n f(x_n)$ which is $\sigma(\ell^1,c_0)$ continuous because $f(x_n)\to 0$.

The dual unit ball of $\ell^1$ is $\sigma(\ell^1,c_0)$-compact by Alaoglu and hence so is $B=\{(t_n)_n: t_n\ge 0, \sum_n t_n\le 1\}$ because of the continuity of the projections $(t_k)_k \mapsto t_n$. Therefore, the image $T(B)$ (which is the right hand side in the displayed formula) is weakly compact and hence norm closed in $X$. This proves the inclusion $\overline{co}(\{a_n:n\in\mathbb N\})\subseteq T(B)$ whereas the reverse one is elementary (truncating the series).

As in Gerry Myerson's comment, $a_n=1/n$ shows that one cannot replace the condition $\sum_n t_n\le 1$ by $\sum_n t_n=1$ (unless, of course, one term $a_k=0$) . In the proof above, the problem is that $\tilde B=\{(t_n)_n: t_n\ge 0,\, \sum_nt_n=1\}$ is **not** $\sigma(\ell^1,c_0)$-compact because $(t_n)_n\mapsto \sum_n t_n$ is not $\sigma(\ell^1,c_0)$-continuous. The idea to replace $c_0$ by the space $c$ of all convergent sequences fails because then the duality between $c$ and $\ell^1$ is not of the form $\langle (s_n)_n,(t_n)_n \rangle=\sum_n s_nt_n$ because the functional $(s_n)_n\mapsto \lim s_n$ is not representable in that way.

There are more sophisticated representations of elements of the closed convex hull of more general sets $A$ considered in *Choquet theory*. The question when the closed convex hull of a compact or weakly compact set is again (weakly) compact is also very relevant. The compact case is due to Mazur and the weakly compact case is a theorem of Krein.