Suppose that $\sum_{n = 0}^\infty a_n$ is absolutely convergent and $\{P_1,\ldots, P_r\}$ is a partition of $\mathbb{N}$ (i.e. $\bigcup_{i=1}^r P_i = \mathbb{N}$ and $P_i \cap P_j = \emptyset$ for any $i \ne j$). Then, $\sum_{n \in P_i} a_n$ is absolutely convergent ($\sum_{n \in P_i} |a_n| \le \sum_{n=0}^\infty |a_n| < \infty$) for any $i \in \{1,\ldots,r\}$. Then, is true that $\sum_{n = 0}^\infty a_n = \sum_{i=1}^r \sum_{n \in P_i} a_n$? I think it's true because we can write \begin{align*} \sum_{n = 0}^\infty a_n &= \lim_{k \to \infty} \sum_{n=0}^k a_n \\ &= \lim_{k \to \infty} \sum_{i=1}^r \sum_{n \in P_i \\ n \le k} a_n \\ &= \sum_{i=1}^r(\lim_{k \to \infty} \sum_{n \in P_i \\ n \le k} a_n) \\ &= \sum_{i=1}^r \sum_{n \in P_i} a_n. \end{align*}
- 3$\begingroup$ Yes. Rearranging the terms of an absolutely convergent series does not change the value of the series. $\endgroup$Shou– Shou2025-09-07 21:49:26 +00:00Commented Sep 7 at 21:49
- $\begingroup$ @Shou I've added an edit now. Can you verify if it's ok please? $\endgroup$ProofSeeker– ProofSeeker2025-09-07 21:53:30 +00:00Commented Sep 7 at 21:53
- $\begingroup$ @Show By rearranging I understand a permutation of terms, but there is no permutation which put odd numbers first and even numbers second, so to prove that $\sum_n = \sum_{n = \text{even}} + \sum_{n = \text{odd}}$ I think that this is easier to understand $\endgroup$ProofSeeker– ProofSeeker2025-09-07 21:55:11 +00:00Commented Sep 7 at 21:55
1 Answer
Let $s=\sum_{n=0}^\infty a_n$ and, for each $i\in\{1,2,\ldots,r\}$, let $s_i=\sum_{n\in P_i}a_n$. What you want to prove is that $s=\sum_{i=1}^rs_i$. This is the same thing as asserting that$$(\forall\varepsilon>0):\left|s-\sum_{i=1}^rs_i\right|<\varepsilon.$$So, take $\varepsilon>0\require{cancel}$. There is some $N\in\Bbb N$ such that$$(\forall m\in\Bbb N):m\geqslant N\implies\left|s-\sum_{n=0}^ma_n\right|<\frac\varepsilon2.$$And, for each $i\in\{1,2,\ldots,r\}$, there is some $N_i\in\Bbb N$ such that$$(\forall m\in\Bbb N):m\geqslant N_i\wedge m\in P_i\implies\left|s_i-\sum_{n\in P_i\\n\leqslant m}a_n\right|<\frac\varepsilon{2r}.$$Therefore, if $M=\max\{N,N_1,N_2,\ldots,N_r\}$, then\begin{align}\left|s-\sum_{i=1}^rs_i\right|&=\left|s-\sum_{n=0}^Ma_n+\sum_{n=0}^Ma_n-\sum_{i=1}^rs_i\right|\\&\leqslant\left|s-\sum_{n=0}^Ma_n\right|+\left|\sum_{i=1}^rs_i-\sum_{n=0}^Ma_n\right|\\&<\frac\varepsilon2+\sum_{i=1}^r\left|s_i-\sum_{n\in P_i\\n\leqslant M}a_n\right|\\&<\frac\varepsilon2+\cancel r\frac\varepsilon{2\cancel r}\\&=\varepsilon.\end{align}
Note: At no point this proof uses the assumption that the given series is absolutely convergent. That assumption is only needed to prove that, for each $i\in\{1,2,\ldots,r\}$, the series $\sum_{n\in P_i}a_n$ converges.
