# Closure of compact sets in Banach space

Let $(X,\vert\vert\cdot\vert\vert)$ be a Banach space. For each $k\in\mathbb{N}$ let $A_k\subseteq X$ be compact and $r_k\in\mathbb{R},r_k>0$, such that $$A_{k+1}\subseteq \{x+u\vert x\in A_k \text{ and } u\in X \text{ with } \vert\vert u\vert\vert\leq r_k\}$$ for every $k\in\mathbb{N}$ and $$\sum\limits_{k=1}^{\infty}r_k<\infty$$ Show that the closure of $\bigcup\limits_{k=1}^{\infty}A_k$ is compact.

I just don't know where to start, any idea or tip would be helpful, thanks!

Let $A:=\bigcup_{k=1}^\infty A_k$. Since $A$ is contained in a complete normed space, it suffices to show that $A$ is totally bounded. So fix $\varepsilon\gt 0$ and $k_0$ such that $\sum_{k\geqslant k_0}r_k\lt\varepsilon$.
For $m\geqslant 1$, $$A_{k_0+m}\subset\left\{x=a_{k_0}+\sum_{l=1}^mx_k, x_k\in A_{k_0+k},\lVert x_k\rVert\leqslant r_k\right\}.$$ This can be shown by induction on $m$.
Therefore, $$A_{k_0+m}\subset\left\{x, x=x'+x'', x'\in A_{k_0}, \lVert x''\rVert\leqslant \sum_{k=k_0+1}^mr_k\right\}\subset A_{k_0}+B(0,\varepsilon),$$ from which we deduce $$A\subset \bigcup_{j=1}^{k_0}A_j\cup(A_{k_0}+B(0,\varepsilon)).$$ If $S_j$ is a finite subset of $A_j$ such that $A_j\subset\bigcup_{x\in S_j}B(x,\varepsilon)$, then $$A\subset\bigcup_{j=1}^{k_0}\bigcup_{x\in S_j}B(x,\varepsilon)\cup \bigcup_{x\in S_{k_0}}B(x,2\varepsilon),$$ hence $A$ is contained in a finite union of ball of radius smaller than $2\varepsilon$.