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

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

  2. 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?

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2 Answers 2

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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.

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  • $\begingroup$ Yes perfect. What I thought was for 2, $(f(n) -n)$ will be bounded, if it satisfies the hypothesis. But is that true? Yes please give some hints for 2. $\endgroup$
    – jerry guna
    Sep 29, 2021 at 5:11
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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?]

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  • $\begingroup$ You wrote that $$\sum_{i\in A_n}a_i - \sum_{i\in A_n}a_{f(i)} = 0$$ But this equality is false. $\endgroup$
    – perroquet
    Oct 21, 2021 at 6:02
  • $\begingroup$ @perroquet you've right, I missed that. Check now after update (defining sets $C_n$ and $D_n$) $\endgroup$
    – jorlinski
    Oct 22, 2021 at 20:27
  • $\begingroup$ you wrote that $$|h(n)-n|+1 \leqslant |f(h(n))-h(n)|$$ But we don't know if $f(h(n))>n$. $\endgroup$
    – perroquet
    Oct 22, 2021 at 22:06
  • $\begingroup$ @perroquet You're right again. I give up. The solution is wrong. $\endgroup$
    – jorlinski
    Oct 23, 2021 at 7:24
  • $\begingroup$ @perroquet If you can take a critical look at the new solution, I would appreciate it. $\endgroup$
    – jorlinski
    Oct 29, 2021 at 20:37

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