# Permutations don't affect summation in complex series implies absolute convergence?

Let $\sum a_{n}$ be a complex series that converges. Now let $\sum a'_{n}$ be a rearrangement of that series. If we have $$\sum a_{n}=\sum a'_{n}$$ for all rearrangements, is it true that $\sum a_{n}$ converges absolutely?

On a similar note, for all of you who own Rudin's Principles of Mathematical Analysis, can you check if there is a Theorem 3.56 in your book? Rudin cites it in Real and Complex Analysis, and I can't tell if it's a typo or was the theorem added in later printings.

• In my edition this theorem is presented for real-valued series. But it can be easily generalized to complex-valued series. – Norbert Apr 17 '12 at 22:42
• Neither of my copies of Rudin have a theorem 3.56, but Rudin does have two texts, one with a blue cover that only does real analysis and one with a green cover that does both complex and real analysis. Are you sure you're checking the right one? I only have access to the blue book. – chris Apr 17 '12 at 23:30

Yes, it is true. Suppose $\sum a_n$ does not converge absolutely, i.e. $\sum |a_n| = +\infty$. Write $a_n = b_n - c_n + i d_n - i e_n$ where $b_n = \max(\text{Re}(a_n),0)$, $c_n = \max(-\text{Re}(a_n),0)$, $d_n = \max(\text{Im}(a_n),0)$, $e_n = \max(-\text{Im}(a_n),0)$ are all nonnegative. Since $|a_n| \le b_n + c_n + d_n + e_n$, at least one of $\sum b_n$, $\sum c_n$, $\sum d_n$, $\sum e_n$ is $+\infty$. Let's say it is $\sum b_n$. Let $A$ be the set of positive integers $n$ for which $b_n > 0$ (i.e. $\text{Re}(a_n) > 0$), and $B$ those for which $\text{Re}(a_n) \le 0$. Split up $A$ into infinitely many disjoint finite blocks $A_1, A_2, \ldots$, so that $\sum_{n \in A_k} \text{Re}(a_n) \ge 1$. Then rearrange the series as follows: take $a_n$ for $n \in A_1$, then the first member of $B$, then $A_2$, then the second member of $B$, etc. The rearranged series diverges, because summing over a block $A_k$ always changes the partial sum by at least $1$.