Let $(X,\|.\|)$ be a separable Banach space and $A$ be a nonempty weak compact convex subsets of $X$.
Let $n\geq 1$. Can we say that: $$ cl\big(\text{co}\cup_{k\geq n}{\frac{1}{k}A}\big) $$ is weak compact ? with: $\text{co}(C)=\left\{\sum _{i=1}^{n}\lambda _{i}x_{i}:n\in \mathbb {N} ,\,x_{i}\in C,\,\sum _{i=1}^{n}\lambda _{i}= 1\right\}.$
Yes, we can say that. That is even true. Indeed, let $w$ the the weak topology on $X$. Put $A’=\text{co}\cup_{k\geq n}{\frac{1}{k}A}$. Then $0\in\operatorname{cl} A’$ and $A’\cup \{0\}=\operatorname{co} \left(\frac 1n A\cup\{0\}\right)$. The latter set is an image of a compact set $[0,1]\times \left(\frac 1n A\cup\{0\}\right) \subset\Bbb R\times (X,w)$ by a continuous map $(\lambda,x)\mapsto \lambda x$ for each $\lambda\in\Bbb R$ and $x\in X$. Thus $\operatorname{cl} A’= A’\cup \{0\}$ is weakly compact.