Let $\sum a_n$ be a convergent series and $f$ is a bijection on $\mathbb{N}$.
-
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?
Best Answer
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.