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If $\{a_n\}_{n\in\mathbb N}$ is any sequence real numbers and the series $\sum_{n=1}^\infty a_n$ converges, then is the equation $$ \sum_{n=1}^\infty a_n = \sum_{k=1}^\infty a_{2k-1}+\sum_{k=1}^\infty a_{2k} $$ true? I believe the answer should be no, but if the series were absolutely convergent then the answer would be yes. However, I seem to have a proof and I'm not sure where it goes wrong for arbitrary series.

Let $\{s_n\}_{n\in\mathbb N}$ be the sequence of partial sums

$$ s_n = \sum_{k=1}^n a_k $$

Of course $\sum_{n=1}^\infty a_n = \lim_{n\to\infty}s_n$. Moreover $s_n=\sum_{k=1}^{n/2} a_{2k-1}+\sum_{k=1}^{n/2}a_{2k}$ if $n$ is even and similar if $n$ is odd. By distributing the limit $\sum_{n=1}^\infty a_n = \sum_{k=1}^\infty a_{2k-1}+\sum_{k=1}^\infty a_{2k}$.

Although general rearrangements are only generally valid for absolutely convergent series, could it be that partitioning the terms of the series is always valid even for arbitrary series (perhaps so long as the terms in the partition preserve their ordering)?

I suspect it's more likely that I'm carelessly ignoring something in this argument but I'm not sure. And I also suspect it's where I hand-waved the two cases of $n$ being even and odd. I tried proving that step more rigorously with an $\varepsilon,N$ argument but realized at a certain point that I wasn't able to do it unless the terms were non-negative. But this realization isn't quite a proof of anything, so I wanted to see if someone could just sort of point out what is true and what is false here.


The question: If a series converges can you always partition the terms, find each series of terms in each cell of the partition, and then sum the series? If not, is absolute convergence a sufficient condition for this?

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    $\begingroup$ It is true that for a convergent series $$\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty \left(a_{2n-1}+a_{2n}\right)$$But the example $a_n=\frac{(-1)^n}{n}$ shows that the splitting of even and odd terms as separate series is not generally legitimate. $\endgroup$
    – Mark Viola
    Commented Jun 12, 2021 at 18:48
  • $\begingroup$ Intuitively, it should work with absolute convergence., since any rearrangement gets to the same sum. $\endgroup$
    – Alan
    Commented Jun 12, 2021 at 18:56

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It is true that for a convergent series

$$\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty \left(a_{2n-1}+a_{2n}\right)\tag1$$

But the example $a_n=\frac{(-1)^n}{n}$ shows that the splitting of even and odd terms as separate series is not generally legitimate.

This result is not a surprise since we know that it is not true in general that $\lim (a+b)=\lim a+\lim b$. However, if both individual limits exist, then so does the limit of the sum. Note that if both $a\ge0$ and $b\ge0$, then $\lim (a+b)=\lim a+\lim b$.


To see this clearly, we can write

$$\sum_{n=1}^{2N}a_n=\sum_{n=1}^N (a_{2n-1}+a_{2n})\tag2$$

and

$$\sum_{n=1}^{2N+1}a_n=\sum_{n=1}^N (a_{2n-1}+a_{2n})+a_{2N+1}\tag3$$

As the series on the right-hand sides of $(2)$ and $(3)$ are convergent, we can let $N\to \infty$ and arrive at $(1)$.


However, for the example $a_n=\frac{(-1)^n}{n}$, we see that

$$\begin{align} \sum_{n=1}^{2N} \frac{(-1)^n}{n}&=\sum_{n=1}^N \frac{(-1)^{2n-1}}{2n-1}+\sum_{n=1}^N \frac{(-1)^{2n}}{2n}\\\\ &=\sum_{n=1}^N \frac{1}{2n}-\sum_{n=1}^N \frac{1}{2n-1}\tag4\\\\ &=-\sum_{n=1}^N \frac1{2n(2n-1)}\tag5 \end{align}$$

So, while the series on the right-hand side of $(5)$ converges (in fact, absolutely), neither of the series on the right-hand side of $(4)$ by itself converges.


To summaries, $(1)$ is correct. But it is not true in general that $\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty a_{2n-1}+\sum_{n=1}^\infty a_{2n}$.

Obviously, if both the series of even terms and the series of odd terms converge, then the series of the sum of even and odd terms converges. This condition will be satisfied for an absolutely convergent series.

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  • $\begingroup$ Are you aware of a general treatment of this question? For instance, is there any resource which discussed an arbitrary partition of the terms of an absolutely convergent series? $\endgroup$
    – Addem
    Commented Jun 12, 2021 at 20:01
  • $\begingroup$ Hi Addem. I'm sure that most books on Real Analysis or on Advanced Calculus would discuss this. $\endgroup$
    – Mark Viola
    Commented Jun 12, 2021 at 21:14
  • $\begingroup$ Page 76 of Rudin discusses rearrangements but not partitions. And I don't see any of the problems at the end of that chapter which seem relevant to the topic. $\endgroup$
    – Addem
    Commented Jun 12, 2021 at 21:20
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Ok, after spending a day on this, and scanning a bunch of Analysis textbooks, I think the most comprehensive treatment that I've found is in Abbott's Understanding Analysis. Starting on page 80 he begins a sequence of exercises and theorems which, on page 82, concludes with the fact that the double-sum can be computed by computing the diagonals of the rectangular array, indexed by just a single index. It will take me a long time to process all of the exercises which are found here, and the proofs probably consume a few pages, so I won't try to produce all of that myself here. So this answer will necessarily be incomplete and anyone else is free to complete it. (Note: This textbook has a solution manual and in it the proof for this fact takes up about 3 1/2 pages. I think that's a bit of an indication that this particular problem is not quick and easy.)

But once you have proved that the double-sum can be computed by computing the diagonals with a single index, you can then use the fact that any sum of a single index, composed of non-negative terms, can be rearranged in any manner you want. Hence any enumeration of the terms will be equivalent to the one which proceeds by diagonals.

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