Convex hull of a bounded sequence. Let $B$ be a nonempty set and $\{x_n\}$ a bounded sequence in $l_{\infty} (B)$. 
Assume that for all $A_n \geq 0, n \in \mathbb{N}$, satisfying $\sum_{n=1}^{\infty} A_n = 1$, the vector $\sum_{n=1}^{\infty} A_n x_n$ attains its supremum over $B$. 
Is it true that 
$$\left\{ \sum_{n=1}^{\infty} A_n x_n : A_n \geq 0, \sum_{n=1}^{\infty} A_n = 1 \right\} \subseteq \text{conv} \{x_n : n \geq 1\}?$$

 A: Consider $B=[0,1]\subset\mathbb R$ and
$$
x_n=(0,\ldots,0,\underbrace{1}_{n.th~position},0\ldots).
$$
Then $\{x_n\}_n$ is bounded in $\ell_\infty(B)$.
Further for each sequence $(A_n)_n$ with $A_n\geq 0$ and $\sum_{n=1}^\infty A_n$ you get $A_n\to 0$ for $n\to \infty$. Therefore $(A_n)_n$ attains its maximum at some index $N$ where $A_N\in(0,1)$ and 
$$
\sum_{n=1}^\infty A_nx_n=(A_n)_n
$$
attains its maximum in $B$ as assumed.
Now you can observe that $\text{conv}\{x_n~:~n\geq 1\}$ contains just sequences which has finite nonzero entries. For $A_n=2^{-n}$ you get
$$
(A_n)_n=\sum_{n=1}^\infty A_nx_n\in\left\{\sum_{n=1}^\infty \tilde{A}_nx_n~:~\tilde{A}_n\geq 0,~\sum_{n=1}^\infty \tilde{A}_n=1\right\}
$$
but $(A_n)_n\notin \text{conv}\{x_n~:~n\geq 1\}$.

As uniquesolution stated you go to the closure of the convex hull and the inclusion becomes correct, since you can approximate the infinite convex combination by finite convex combination. See here:
Let be $y=\sum_{n=1}^\infty A_nx_n=(A_n)_n$ for $A_n\geq 0$ and $\sum_{n=1}^\infty A_n=1$ and let be $\varepsilon>0$. Since $\sum_{n=1}^\infty A_n=1$ there exists $N\geq 1$ such that $\sum_{n=N}^\infty A_n<\varepsilon$ hence $A_m<\varepsilon$ for all $m\geq N$. Now you can define $z=\sum_{n=1}^{N-1} A_nx_n\in \text{conv}\{x_n~:~n\geq 1\}$ and you get
$$
\|y-z\|_{\ell_\infty(B)}=\left\|\sum_{n=N}^\infty A_nx_n\right\|_{\ell_\infty(B)}=\sup_{n\geq N}|A_n|<\varepsilon.
$$
This proves that $y$ is in the closure of $\text{conv}\{x_n~:~n\geq 1\}$.
(With small modifications, you can use this argument for each bounded sequence $\{x_n\}_n\subset\ell_\infty(B)$).
