Convergence of a rearrangement of a series Let $\sum a_n$ be a convergent series and $f$ is a bijection on $\mathbb{N}$.

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*Suppose $(f(n) -n)$ is a bounded sequence the rearrangement series $\sum a_{f(n)}$ converges to same limit.


*Suppose $m_n=\sup\{|a_k| : k> n\}$. If the sequence $(m_n |f(n) -n|) $ converges to $0$, then the rearrangement series converges to same limit.
Since $(f(n) -n)$ is bounded sequence, the terms can be grouped in some way such that the limit does not change. But how it can be done. $2$ seems consequence of $1$. How to do it?
 A: You mentioned you can see how $2$ is a consequence of $1$, so here is the proof for $1$.  Let me know if you need some ideas for $2$ as well.
Since $(f(n)-n)$ is bounded, let $M \in \mathbb{N}$ be such that $|f(n)-n| \le M$ for all $n \in \mathbb{N}$.  This implies $n-M \le f(n) \le n + M$ for all $n$ as well.
Let $L:= \sum_{n=1}^\infty a_n$.  By definition, given $\varepsilon > 0$, there exists $N \in \mathbb{N}$ such that for $n \ge N$ we have $|\sum_{k=1}^{n-M} a_k - L | < \frac{\varepsilon}{2}$.  Since $\sum a_n$ is convergent, we also have $(a_n) \rightarrow 0$, so by taking a potentially large $N$ we also have $|a_n| < \frac{\varepsilon}{4M}$ for all $n \ge N$.
Since $n-M \le f(n) \le n+M$ for all $n$, we have $f(\{1,...,n\}) = \{1,...,n-M\} \cup S$ for some $S \subset \{n-M,...,n+M\}$.  Note that $S$ has less than $2M$ elements.  Therefore, if $n \ge N$, we have
\begin{align*}
\left| \sum_{k=1}^n a_{f(k)} - L \right| &= \left| \sum_{k=1}^{n-M} a_{f(k)} + \sum_{k \in S} a_{f(k)} - L \right| \\
&\le \left| \sum_{k=1}^{n-M} a_{k} - L \right| + \left| \sum_{k \in S} a_{k} \right| \\
&< \frac{\varepsilon}{2} + \sum_{k \in S} |a_{k}| \\
&< \frac{\varepsilon}{2} + \sum_{k \in S} \frac{\varepsilon}{4M} \\
&< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.
\end{align*}
Therefore we conclude $\sum_{k=1}^\infty a_{f(k)} = L$ as desired.
A: 2 Let
$$\begin{align*}
x_n= |S_n - S'_n| &= |a_1+\cdots+ a_n -(a_{f(1)}+\cdots+a_{f(n)})| \\
&= \left| \sum_{i\in C_n}a_i + \sum_{i\in D_n}a_i - \sum_{i\in A_n}a_{f(i)} - \sum_{i\in B_n}a_{f(i)} \right|
\end{align*},$$
where $A_n=\{i\in\mathbb N: f(i)\leq n, 1\leq i\leq n\}$, $B_n=\{i\in\mathbb N: f(i)> n, 1\leq i\leq n\}$, $C_n =\{i\in\mathbb N: i=f(j), j\in A_n \}$ and $D_n=\{1,\dots, n \} \setminus  C_n$. Then we have
$$\sum_{i\in C_n}a_i - \sum_{i\in A_n}a_{f(i)} = 0. \quad\quad(1)$$
Now, becouse for each $k\in\mathbb N$ there is $l\in\mathbb N$ such that
$$\forall_{i\leq k} \exists_{j\leq l} \quad i=f(j)$$
thus $h(n):=\min\{i:i\in D_n\} \to \infty$ when $n\to\infty$.
We denote $p(n) :=\max\{|f(n)-n|: n\in B_n\}=|f(k_n) - k_n|$ for some $k_n\in\mathbb N:h(n)\leq k_n\leq n$ and then we have
$$N(D_n)=N(B_n)\leq p(n) \quad\quad (2)$$
where $N(X)$ is the number of elements of $X$.
Let $\epsilon > 0$. Then... [what then?]
