# Convergence of a rearrangement of a series

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?

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.

• 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. Sep 29, 2021 at 5:11

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?]

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