# A rearrangement of an absolutely convergent complex series is also absolutely convergent

I just completed the following proof. Is it valid?

1. Let $\sum_{k=1}^{\infty} a_k$ be an arbitrary convergent series that also converges absolutely. Then $\sum_{k=1}^{\infty} a_k \in \mathbb{C}$ and $\sum_{k=1}^{\infty} |a_k| \in \mathbb{R}$.

2. Let $\{b_k\}$ denote a rearrangement of the $\{a_k\}$. That is, we have that $\{a_k\} = \{b_k\}$ yet $a_i = b_i$ needn't hold.

3. We will first show that $\sum_{k=1}^{\infty} b_k$ converges absolutely (immediately implying it converges non-absolutely). We will use the Cauchy Criterion on the partial sums of $\sum_{k=1}^{\infty} |b_k|$ to demonstrate this.

4. First let $\epsilon > 0$.

5. Consider that since $\sum_{k=1}^{\infty} |a_k| \in \mathbb{R}$, we have that $\exists N_1 \in \mathbb{N}$ s.t. for all $N_1 < m < n$ we have

$$\sum_{k=m}^n |a_k| \le \epsilon.$$

6. Now let $0 < N_2$ be large enough s.t $\{a_1, \ldots , a_{N_1}\} \subseteq \{b_1, \ldots , b_{N_2}\}$. Let $M > max\{N_1, N_2\}$. Let $M < m' < n'$. Then since $\{b_k\} - \{b_1, \ldots , b_{m'}\} \subseteq \{a_{N_1}, a_{N_{1}+1}, a_{N_{1}+2} \ldots\}$, we will have that

$$\sum_{k=m}^n |a_k| \le \epsilon \implies \sum_{k={m'}}^{n'} |b_k| < \epsilon.$$

7. Then we have that

$$\left| \sum_{k={m'}}^{n'} |b_k| \right| = \sum_{k={m'}}^{n'} |b_k| < \epsilon$$

so that since $\epsilon$ was arbitrary, we have obtained that

$$\left| \sum_{k=1}^n |b_k| - \sum_{k=1}^m |b_k| \right| = \left| \sum_{k=m}^n |b_k| \right| \rightarrow 0$$

so that via the Cauchy Criterion for Convergence we have that $\sum_{k=1}^{\infty} |b_k|$ converges to some value in $\mathbb{R}$ as desired.

8. Then it follows that $\sum_{k=1}^{\infty} b_k$ is absolutely convergent (and hence also just convergent).

$\square$

The proof of this is along the lines of your proof. Assume $\sum a_n=a$, $\{b_n\}$ a rearrangement, and $\varepsilon>0$. Then there is an $N\in\mathbb N$, such that for $n\ge N$ $$\Big|\sum_{k=1}^n a_k-a\,\Big| <\varepsilon \quad \text{and} \quad \sum_{k\ge N}\lvert a_k\rvert<\varepsilon.$$ Clearly, there is a $N_1\ge N$, such the terms $a_1,\ldots,a_N$ are included among the $b_1,\ldots,b_{N_1}$. Hence for $m\ge N_1$ $$b_1+\cdots+b_{m}=a_1+\cdots+a_{N}+\Big(m\!-\!N\,\,\text{terms of \{a_n\} with index \ge N}\Big),$$ and thus $$\Big|\sum_{k=1}^m b_k-a\,\Big|\le \Big|\sum_{k=1}^N a_k-a\,\Big|+\sum_{k\ge N}|a_k|< 2\varepsilon.$$ Therefore $\sum_{n=1}^\infty b_n=a$.