If a series $\sum a_{n}$ converges but not $\sum |a_{n}|$, then $\sum a_{n}$ can be rearranged so that the newly rearranged sum takes a value different from the original one.

This is a well-known fact and I do remember having learnt about it some long years ago.

But recently, I have been thinking about this. "There is a new rearrangement", but what sort of rearrangement is it?

Suppose that $\sum_{n \geq 1}a_{n}$ converges and $\mathbb{N} = A_{1} \cup A_{2} \cup \cdots$ with $A_{i}$ disjoint from one another.

Then what are the conditions on those subsets $A_{i}$'s so that \[ \sum_{j \geq 1} \sum_{n \in A_{j}}a_{n} \not = \sum_{n \geq 1} a_{n}. \]

The limit of its partial sums $\lim_{N \to \infty}\sum_{j \leq N} \sum_{n \in A_{j}}a_{n}$ seems to be the same regardless of what $A_{i}$'s are, as long as each $\sum_{n \in A_{j}}a_{n}$ converges absolutely and this way of summing up is concerned.

And I would like to think about this question in the context of convergence of the product \[ \prod_{p \leq N} (1 + a(p)p^{-s}) = \sum_{n \in A_{1} \cup \cdots \cup A_{N}}\frac{a(n)}{n^{s}} \] with $|a(n)| = 1$, $s$ some number $0 < s < 1$, and $p$ primes. Here, the right series converges absolutely for $s > 1$. If the relation \[ \lim_{N \to \infty}\prod_{p \leq N} (1 + a(p)p^{-s}) = \lim_{N \to \infty}\sum_{n \in A_{1} \cup \cdots \cup A_{N}} \frac{a(n)}{n^{s}} \] is shown to be valid for $a < \text{Re}(s) \leq 1$ for some $0 < a < 1$, then we may have more restrictions on rearrangement of the sum on the right (since by analytic continuation, rearrangements are restricted to those which assign the series the same original value).

The next question is, then, can we see the sum \[ \lim_{N \to \infty}\sum_{n \in A_{1} \cup \cdots \cup A_{N}} \frac{a(n)}{n^{s}} \] as a Dirichlet series? To put it in formulas, is it true that \[ \lim_{N \to \infty}\sum_{n \in A_{1} \cup \cdots \cup A_{N}} \frac{a(n)}{n^{s}} = \sum_{ n \geq 1} \frac{a(n)}{n^{s}} \quad ? \]

Any objection, comments are welcome.

  • $\begingroup$ Your sets $A_i$ might as well all contain a single natural number, being the $i$th number that you want to have in a specific re-ordering that you consider. So then $\sum_{j\geq 1} \sum_{n \in A_j} a_n$ is just a way of writing the summation over the different ordering you want. So $\sum_{n \in A_j} a_n$ trivially "converges absolutely" for each $j$, since the sum contains a single term. So if your claim about "converges absolutely" were true, it would mean that all orderings give the same sum (which is false). The Mundron answer shows how to get a different sum from a different order. $\endgroup$ – Michael Jun 11 '18 at 1:50

The limit of its partial sums $\lim_{N \to \infty}\sum_{j \leq N} \sum_{n \in A_{j}}a_{n}$ seems to be the same regardless of what $A_{i}$'s are, as long as each $\sum_{n \in A_{j}}a_{n}$ converges absolutely and this way of summing up is concerned.

That is not true. Consider a series $\sum a_n$ converges but not $\sum |a_n|$ and an arbitrary $M\in\mathbb R$. Since $\sum |a_n|$ diverges, the series over the positive $a_n$ diverges as the series over the negative $a_n$. Hence, you can pick positive $a_n$ until there sum exceeds $M$ and define the set of indices $A_1$. Next, you pick negative $a_n$ until you become less than $M$ and define the set of indices $A_2$ and so on. In the end, you will get $$ \lim_{N\to\infty}\sum_{j=1}^N\sum_{n\in A_j}a_n = M. $$

For further details, you should study the proof in https://en.wikipedia.org/wiki/Riemann_series_theorem .

Since the theorem states the existence of a rearrangement, you can't directly deduce conditions on $A_1,\ldots$ such that the limits matches.

  • $\begingroup$ Yes. Actually, I kind of recall this proof. Any idea on conditions for the sets A's for which the series take the same value as the original one? $\endgroup$ – Grown pains Jun 11 '18 at 1:58
  • $\begingroup$ The construction gives sets $A_i$ which are finite and obviously the limits can differ if $A_i$ are not finite. Hence, I don't think that you can give some nice condition for $A_i$ for an arbitrary series $\sum a_n$. There might be special conditions for an explicit series, but why do you need it? I don't see a use for it. $\endgroup$ – Mundron Schmidt Jun 11 '18 at 7:37
  • $\begingroup$ "I don't see a use for it." As I edited my original question, under the stated conditions, I want to regard the conditionally convergent series $\lim_{N \to \infty}\sum_{n \in A_{1} \cup \cdots \cup A_{N}} a(n) n^{-s}$ as a Dirichlet series. For this purpose, we need some argument. $\endgroup$ – Grown pains Jun 13 '18 at 0:10
  • $\begingroup$ This is also covered in the old classic "Infinite Sequences And Series" by Bromwich, available in cheap re-print from Dover Publications, $\endgroup$ – DanielWainfleet Jun 13 '18 at 3:56
  • $\begingroup$ @DanielWainfleet "This is also covered in the old classic..." By this, would you mean that it covers what I wrote about regarding $\lim_{N \to \infty}\sum_{A_{1} \cup \cdots \cup A_{N}}a(n)n^{-s}$ as a Dirichlet series? If you are still tuned, please let me know. $\endgroup$ – Grown pains Jun 24 '18 at 23:54

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