Mysterious Series Rearrangement/Permutation Problem This problem appears in J. Dieudonne's "Foundations of Foundations of Modern Analysis". (chapter 5, section 2):
Let $(x_n)$ be a convergent series in a normed space $E$; let $\sigma$ be a bijection of $N$ onto itself, and let 
$$r(n) = |\sigma(n) - n| \cdot \sup_{m \geq n}\|x_m\|.$$
Show that if $\lim_{n \rightarrow \infty}r(n) = 0$, the series $(x_{\sigma(n)})$ is convergent in $E$ and that $\sum_{n=0}^{\infty}x_n = \sum_{n=0}^{\infty}x_{\sigma(n)}$.
Any ideas about a solution? A similar post from a couple years ago came up dry.
 A: The function $f(k):= |\sigma(k)-k|$ can be seen as the 'local discrepancy' of the permutation. If $f(k)$ is bounded, it is easy to see that the correspondig series converges and has the same limit.
I need additionally that
$$ \widetilde{r}(n) := |\sigma(n)-n| \sup_{k \geq \min(n,\sigma(n))} \|x_i\| \rightarrow 0$$
in order to prove the above result: First, we observe that for all $m \geq M$, where $M$ is chosen later,
$$ \| \sum_{k=1}^m x_k - \sum_{k=1}^m x_{\sigma(k)}\| \leq \max_{i \in B \setminus A} \|x_i\| \# (B \setminus A) + \sup_{i \in A \setminus B} \|x_i\| \# (A \setminus B),$$
where $A:= \{1,\ldots,m\}$ and $B:= \{\sigma(1),\ldots,\sigma(m)\}$.
Fix $N$ with $\widetilde{r}(n) < \varepsilon$ for any $n \geq N$ and take $M$ so large that


*

*$\sigma(1),\ldots,\sigma(N) \leq M$

*$\{1,\ldots,N\} \subset \{\sigma(1),\ldots,\sigma(M)\}$.


Now take $\sigma(k) \in B \setminus A$ with minimal $1 \leq k \leq m$. Then $$\# B \setminus A \leq m -(k-1) \leq \sigma(k)-k.$$
Then we must have $k > N$. Noting that $\sigma(h) > m \geq k$ for any $h \in B \setminus A$, we find 
$$\max_{h \in B \setminus A} \|x_i\| \# (B \setminus A) \leq \max_{i \geq k} \|x_i\| |\sigma(k)-k| = \widetilde{r}(k) \leq \varepsilon.$$
If $h \in A \setminus B$, then $\sigma(l) = h$ with $l > m$ and $h >N$. Thus, taking $h$ minimal, we get again $\# A \setminus B \leq m - (h-1) \leq l - \sigma(l)$. Using this, shows that
$$\sup_{i \in A \setminus B} \|x_i\| \# (A \setminus B) \leq \sup_{i >\sigma(l)} \|x_i\| |l-\sigma(l)| \leq \widetilde{r}(l) \leq \varepsilon$$
because $l > m \geq M  \geq N$.
However, I am not sure, if the additional requirement is necessary.
