A rearrangement of an absolutely convergent series converges absolutely to the same limit I just completed the following proof.  Is it valid?


*

*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 $\sum_{k=1}^{\infty} |a_k| \in \mathbb{R}$.

*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.

*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.

*First let $\epsilon > 0$.

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

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

*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.

*Then it follows that $\sum_{k=1}^{\infty} b_k$ is absolutely convergent (and hence also just convergent).
$\square$
 A: You showed correctly that every rearrangement of an absolutely convergent sequence of complex numbers is also absolutely convergent (and hence convergent). However, there is an additional thing to prove (which you are not asked perhaps). This the fact that:
Every rearrangement of an absolutely convergent series converges to the same limit.
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 $\sum_{n=1}^\infty b_n=a$.
