# Prove that in Banach space $X$ the following are equivalent for series $\sum_{i\in\mathbb{N}} x_i$

Prove that in Banach space $X$ the following are equivalent for series $\sum_{i\in\mathbb{N}} x_i$:

1) $\forall\varepsilon>0 \hspace{2mm} \exists$ finite $F\subset\mathbb{N}$ such that for every finite $F'\subset \mathbb{N}$ with $F\subseteq F'$, $\lVert\sum_{i\in F'}x_i-x\rVert\ < \varepsilon$

2) For every permutation $\pi$ of $\mathbb{N}$, $\sum_{i\in\mathbb{N}} x_{\pi(i)}$ converges to $x$.

For 1)$\Rightarrow$2) Let $\pi$ be a permutation of the natural numbers and $\varepsilon>0$. By the assumption in 1), there exists F with the properties above. For some $n_0$ we have that $F\subseteq\{\pi(0),\pi(1),\dots,\pi(n_0)\}$ and thus $\lVert\sum_{i=1}^nx_i-x\rVert\ < \varepsilon$ whenever $n\geq n_0$

For 2)$\Rightarrow 1)$ let's assume that 1) does not hold, that is $\exists\varepsilon_0>0 \hspace{2mm} \forall$ finite $F\subset\mathbb{N} \hspace{2mm} \exists F'\subset \mathbb{N}$ with $F\subseteq F'$ and $\lVert\sum_{i\in F'}x_i-x\rVert\ \geq \varepsilon_0$.

We have that there is $n_0$ such that $\lVert\sum_{i=1}^{n_0}x_i-x\rVert\ < \varepsilon_0$ (because the series converjes for the identity permutation). By the assumption there is a finite set $F_1\supset F$ with $\lVert\sum_{i\in F_1}x_i-x\rVert\ \geq \varepsilon_0$. Let's look at the permutation $\pi=F\cup(F_1\setminus F)\cup (N\setminus F)$. Since 2) holds for $\pi$there is a $j_0>\max\{F\}$ such that $\lVert\sum_{i=0}^{\pi(i)=j_0}x_\pi(i)-x\rVert\ < \varepsilon_0$. Now there is a set $F_2$ such that ... and continue by recursion. I'm not 100% sure if the permutation that I get is such that the series doesn't converge to $x$. And is there a simpler construction to get a contradiction?

I have to admit that I didn't manage to digest your final paragraph, but I think I can see a way of generating a permutation $\pi$ such that $\sum_{i \in \mathbb N}x_{\pi(i)}$ does not converge to $x$.
The key is the look at the statement of the negation of (1). This says that there exists an $\epsilon_0 > 0$ such that for all finite $F \subset \mathbb N$, there exists a finite $F' \subset \mathbb N$ with $F \subseteq F'$ and $|| \sum_{i \in F'} x_i - x || \geq \epsilon_0$.
A consequence is that there exists a strictly increasing sequence $$F_1 \subset F_2 \subset F_3 \subset\dots \subset \mathbb N$$ where each $F_j$ is finite and $||\sum_{i \in F_j} x_i -x || \geq \epsilon_0$ for each $j$.
[You can generate this sequence recursively. For example, you can start with $F_1 = \emptyset$. And for each $F_j$, you can take $F_{j+1}$ to be the $F'$ provided by applying the above statement with $F = F_j \cup \{ p_j \}$, where $p_j$ is any integer that isn't in $F_j$. (Including the extra element $p_j$ ensures that the inclusion $F_j \subset F_{j+1}$ obtained in this way is a strict inclusion.)]
Then (to use your notation), we can consider a permutation $\pi$ of the form $F_1 \cup (F_2 \setminus F_1) \cup (F_3 \setminus F_2) \cup \dots$, which clearly violates condition (2).