# Is it possible that $\sum a_{\sigma(n)}$ converges iff $\sum b_{\sigma(n)}$ diverges for every permutation $\sigma :\mathbb N\to \mathbb N$?

Here is my question: Is it possible to find a pair of series $\sum a_n, \sum b_n,$ each having rearrangements that converge conditionally, such that $\sum a_{\sigma(n)}$ converges iff $\sum b_{\sigma(n)}$ diverges for every permutation $\sigma :\mathbb N\to \mathbb N$?

What if we consider more than just two sets (in finite case rest will probably follow from induction)? I intentionally do not divide in cases like "B is divergent but has limit point on extended real number line" since I think they will be probably similar, but please correct me if I am wrong.

As always, feel free to retag.

• @Max Divergent and convergent should not match: "that for every permutation of natural numbers series corresponding to set A is** divergent** if and only if series corresponding to set B is convergent" – Clement C. Jun 30 '17 at 16:50
• Oh my bad, it seems I can't read – Max Jun 30 '17 at 17:45
• I edited your question. I feel it is now simpler, but perhaps I cut out too much. Please re edit if you like. – zhw. Jun 30 '17 at 18:27
• @Robert I don't see how it would follow directly from Riemann's theorem. – 6005 Jun 30 '17 at 18:33
• @Robert For me it is also unclear how it follows from Riemann Rearrangement Theorem (idea presented by zhw. is a little bit different). – Shingle Jun 30 '17 at 19:10

There is no such example. Suppose $\sum a_n, \sum b_n$ each have rearrangements that converge conditionally. Then there is a permutation $\sigma :\mathbb N\to \mathbb N$ such that both $\sum a_{\sigma(n)}, \sum b_{\sigma(n)}$ diverge.

Proof: Let $A^+ = \{n: a_n\ge 0\}, A^-=\{n: a_n< 0\},$ $B^+ = \{n: b_n\ge 0\}, B^-\{n: b_n< 0\}.$ Recall that

$$\sum_{n\in A^+}a_n = \infty,\, \sum_{n\in B^+}b_n = \infty\,\,$$

$$\sum_{n\in A^-}a_n = -\infty,\, \sum_{n\in B^-}b_n = -\infty.$$

We want to think of the sets $A^+, A^-,$ $B^+, B^-$ in their natural order. I'll be inductively choosing finite subsets of these sets. At each step I'll be choosing an initial segment, in order, of whatever remains of $A^+, A^-,$ $B^+, B^-$ as we move along. That will guarantee a permutation arises from these choices.

Choose $A_1 \subset A^+$ such that $\sum_{n\in A_1}a_n > 1.$ Then choose $B_1 \subset B^+\setminus A_1$ such that $\sum_{n\in B_1}b_n > 1.$ Next choose $A_2\subset A^- \setminus (A_1\cup B_1)$ such that $\sum_{n\in A_2}a_n < -1.$ And then $B_2\subset B^- \setminus (A_1\cup B_1\cup A_2)$ such that $\sum_{n\in B_2}b_n < -1.$

We continue choosing sets $A_k,B_k$ in this way. For odd $k,$ $\sum_{n\in A_k} a_n$ is a positive sum, for even $k$ it's a negative sum. Same for the $b_n$'s. We don't know anything about the sums $\sum_{n\in B_k} a_n, \sum_{n\in A_k} b_n ,$ but we don't have to.

Now the sets $A_1,B_1, A_2, B_2, \dots$ define a permutation $\sigma$ of $\mathbb N.$ It follows that both $\sum a_{\sigma(n)}, \sum b_{\sigma(n)}$ diverge. Why? Because neither sum is Cauchy. No matter how far you go out in these sums, you'll find finite strings $A_k,B_k$ beyond that point such that $\sum_{n\in A_k} a_n >1, \sum_{n\in B_k} b_n >1.$ That violates the Cauchy criterion.

• (+1) Great answer, exactly the thing I was looking for. Thanks. – Shingle Jun 30 '17 at 19:05
• Perhaps you should edit, the notation $\displaystyle\sum_{n\in A_n}$ being quite unfortunate – Max Jun 30 '17 at 19:21
• @Max Strange comment. It's absolutely standard notation. – zhw. Jun 30 '17 at 20:24
• @zhw. With the $n$ being at the same time the indexing of the set $A_n$ and the variable $n$ "living" in $A_n$? Shouldn't it rather be something like $k\in A_n$ ? – Max Jun 30 '17 at 20:32
• @Max I see, yes you're right, I was suffereing from author's blindness. Thanks, now edited. – zhw. Jun 30 '17 at 20:42