Suppose that the series $\sum \limits_{n=1}^{\infty}a_n$ is absolutely convergent and let $I\subseteq \mathbb{N}$ such that $I=\bigsqcup\limits_{k=1}^{\infty}I_k$. Then show that $$\sum \limits_{n\in I}a_n=\sum \limits_{k=1}^{\infty}\sum \limits_{n\in I_k}a_n. \qquad (*)$$

I don't have any idea how to solve it.

I do know that in any absolute convergent series permutation of terms does not change the sum and I guess it should be used somehow in order to prove equality $(*)$.

Can anyone show the rigorous proof of equality $(*)$, please?

  • $\begingroup$ here is a proof. Here is another. $\endgroup$ – Masacroso May 16 at 20:14
  • $\begingroup$ What's the definition of summation here? $\endgroup$ – Hashem Ben Abdelbaki May 16 at 20:14
  • $\begingroup$ @Masacroso, link which you provided have not nothing in common with my question. I do know that in absolute convergent series any permutation does not change the sum. $\endgroup$ – ZFR May 16 at 20:22
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    $\begingroup$ @ZFR you had written "I do know that in any absolute convergent series permutation of terms does not change the sum. But I want to prove it rigorously and cannot do it." Hence my comment providing two formal proofs. Make clear your question please. $\endgroup$ – Masacroso May 16 at 20:23
  • $\begingroup$ @Masacroso, sorry about that. Done $\endgroup$ – ZFR May 16 at 20:27

First assume that $a_n \ge 0$ and define $\sum_{n \in I} a_n = \sup_{J \subset I, J \text{ finite}} \sum_{n \in J} a_n$. Note that it follows that if $I \subset I'$ then $\sum_{n \in I} a_n \le \sum_{n \in I'} a_n$.

From https://math.stackexchange.com/a/3680889/27978 we see that if $K = K_1 \cup \cdots \cup K_m$, a disjoint union, then $\sum_{n \in K} a_n = \sum_{n \in K_1} a_n + \cdots + \sum_{n \in K_m} a_n$.

Since $I'=I_1 \cup \cdots \cup I_m \subset I$ we see that $\sum_{n \in I} a_n \ge \sum_{n \in I'} a_n = \sum_{k=1}^m \sum_{n \in I_k} a_n$. It follows that $\sum_{n \in I} a_n \ge \sum_{k=1}^\infty \sum_{n \in I_k} a_n$. This is the 'easy' direction.

Let $\epsilon>0$, then there is some finite $J \subset I$ such that $\sum_{n\in J} a_n > \sum_{n \in I} a_n -\epsilon$. Since $J$ is finite and the $I_k$ are pairwise disjoint we have $J \subset I'=I_1 \cup \cdots \cup I_m$ for some $m$ and so $\sum_{k=1}^\infty \sum_{n \in I_k} a_n \ge \sum_{k=1}^m\sum_{n \in I_k} a_n \ge \sum_{k=1}^m\sum_{n \in J \cap I_k} a_n = \sum_{n\in J} a_n > \sum_{n \in I} a_n -\epsilon$.

(It is not relevant here, but a small proof tweak shows that the result holds true even if the $a_n$ do not have a finite sum.)

Now suppose we have $a_n \in \mathbb{R}$ and $\sum_{n \in I} |a_n| = \sum_{n=1}^\infty |a_n|$ is finite. We need to define what we mean by $\sum_{n \in I} a_n$. Note that $(a_n)_+=\max(0,a_n) \ge 0$ and $(a_n)_-=\max(0,-a_n) \ge 0$. Since $0 \le (a_n)_+ \le |a_n|$ and $0 \le (a_n)_- \le |a_n|$ we see that $\sum_{n \in I} (a_n)_+ = \sum_{k=1}^\infty \sum_{n \in I_k} (a_n)_+$ and similarly for $(a_n)_-$.

This suggests the definition (cf. Lebesgue integral) $\sum_{n \in I} a_n = \sum_{n \in I} (a_n)_+ - \sum_{n \in I} (a_n)_-$.

With this definition, all that remains to be proved is that $\sum_{k=1}^\infty \sum_{n \in I_k} a_n = \sum_{k=1}^\infty \sum_{n \in I_k} (a_n)_+ - \sum_{k=1}^\infty \sum_{n \in I_k} (a_n)_-$ and this follows from summability and the fact that for each $k$ we have $\sum_{n \in I_k} a_n = \sum_{n \in I_k} (a_n)_+ - \sum_{n \in I_k} (a_n)_-$.

Note: To elaborate the last sentence, recall that I defined $\sum_{n \in I_k} a_n$ to be $\sum_{n \in I_k} (a_n)_+ - \sum_{n \in I_k} (a_n)_-$, so all that is happening here is the definition is applied to $I_k$ rather than $I$. Then to finish, note that if $d_k,b_k,c_k$ are summable and satisfy $d_k=b_k-c_k$ then $\sum_{k=1}^\infty d_k= \sum_{k=1}^\infty b_k- \sum_{n=1}^\infty c_k$, where $d_k = \sum_{n \in I_k} a_n$, $b_k = \sum_{n \in I_k} (a_n)_+$ and $c_k = \sum_{n \in I_k} (a_n)_-$.

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  • $\begingroup$ Great answer! But probably you meant bijection $\sigma: \mathbb{N}\to I$, right? $\endgroup$ – ZFR May 18 at 18:42
  • $\begingroup$ @ZFR: Correct, I will fix. $\endgroup$ – copper.hat May 18 at 18:58
  • $\begingroup$ When you wrote that "It is straightforward to show that for any bijection $\sigma:\mathbb{N} \to I$ we have $\sum_{n \in I} a_n = \sum_{n=1}^\infty a_{\sigma(n)}$." Do you mean here that $a_n\geq 0$ or not? I am a bit confused $\endgroup$ – ZFR May 18 at 19:17
  • $\begingroup$ (+1) This is less fussy, more self-contained, more authoritative, and better connected to the literature than my answer, so it ought to replace it as the accepted answer. (I haven't read @Matematleta's answer yet, because it looks harder to understand than these two, so I reserve judgement on it, for now!) On this answer: am I right in thinking that one can also write $\sum_{n \in I}a_n = s$ iff for all $\epsilon > 0$ there exists finite $J \subseteq I$ such that $\left\lvert\sum_{n \in J}a_n- s\right\rvert < \epsilon$? $\endgroup$ – Calum Gilhooley May 18 at 19:18
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    $\begingroup$ You really helped me to understand the thing which i did not understand for a long time. Thank you so much! I really appreciate your permanent help! $\endgroup$ – ZFR May 18 at 20:50

Suppose for the moment that the result is known to be true for convergent series of non-negative terms.

If $\sum_{n=1}^\infty a_n$ is an absolutely convergent series of real numbers, define $a_n = b_n - c_n,$ for all $n \geqslant 1,$ where $c_n = 0$ when $a_n \geqslant 0$ and $b_n = 0$ when $a_n \leqslant 0.$ Then $|a_n| = b_n + c_n,$ therefore $\sum_{n=1}^\infty b_n$ and $\sum_{n=1}^\infty c_n$ are convergent series of non-negative terms, therefore: \begin{align*} \sum_{n \in I}a_n & = \sum_{n \in I}b_n - \sum_{n \in I}c_n \\ & = \sum_{k=1}^\infty\sum_{n \in I_k}b_n - \sum_{k=1}^\infty\sum_{n \in I_k}c_n \\ & = \sum_{k=1}^\infty\left( \sum_{n \in I_k}b_n - \sum_{n \in I_k}c_n\right) \\ & = \sum_{k=1}^\infty\sum_{n \in I_k}(b_n - c_n) \\ & = \sum_{k=1}^\infty\sum_{n \in I_k}a_n. \end{align*} So it is enough to prove the result on the assumption that $a_n \geqslant 0$ for all $n \geqslant 1.$

Given any set $K \subseteq \mathbb{N},$ I shall use the Iverson bracket notation: $$ [n \in K] = \begin{cases} 1 & \text{if } n \in K, \\ 0 & \text{if } n \notin K. \end{cases} $$ I shall assume that, however the notation $\sum_{n \in K}a_n$ has been defined, it satisfies the identity: $$ \sum_{n \in K}a_n = \sum_{n=1}^\infty a_n[n \in K]. $$ Let $J_k = I_1 \cup I_2 \cup \cdots \cup I_k$ ($k = 1, 2, \ldots$). Because the $I_k$ are disjoint, we have $$ [n \in J_k] = [n \in I_1] + [n \in I_2] + \cdots + [n \in I_k], $$ therefore $$ \sum_{n \in I_1}a_n + \sum_{n \in I_2}a_n + \cdots + \sum_{n \in I_k}a_n = \sum_{n \in J_k}a_n \leqslant \sum_{n \in I}a_n, $$ therefore $$ \sum_{k=1}^\infty\sum_{n \in I_k}a_n \leqslant \sum_{n \in I}a_n, $$ and the outer infinite sum on the left hand side exists, because its partial sums are bounded above by the sum on the right hand side. On the other hand, for all $m \geqslant 1,$ \begin{align*} \sum_{n=1}^ma_n[n \in I] & = \sum_{n=1}^ma_n[n \in I_1] + \sum_{n=1}^ma_n[n \in I_2] + \cdots + \sum_{n=1}^ma_n[n \in I_r] \\ & \leqslant \sum_{n \in I_1}a_n + \sum_{n \in I_2}a_n + \cdots + \sum_{n \in I_r}a_n \\ & \leqslant \sum_{k=1}^\infty\sum_{n \in I_k}a_n, \end{align*} where $$ r = \max\{k \colon n \leqslant m \text{ for some } n \in I_k\}, $$ therefore $$ \sum_{n \in I}a_n \leqslant \sum_{k=1}^\infty\sum_{n \in I_k}a_n, $$ and the two inequalities together prove (*).

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  • $\begingroup$ Thanks a lot for your answer! I read your answer and I have two questions: 1) I am a bit confused by the definition of $r$? Could you explain it in details, please? 2) Have you ever used that the series $\sum \limits_{n=1}^{\infty}a_n$ is absolutely convergent? $\endgroup$ – ZFR May 17 at 2:16
  • $\begingroup$ I would define $r$ in this way: for each $n\in I$ s.t. $1\leq n\leq m$ there is $p(n)$ s.t. $n\in I_{p(n)}$ and let's take $r=\max \{p(1),\dots,p(m)\}$ then $[n\in I]=[n\in I_1]+\dots+[n\in I_r]$. Is my reasoning correct? And still I would be happy if you can explain and point the moments where you have used that the series is absolutely convergent. $\endgroup$ – ZFR May 17 at 2:30
  • $\begingroup$ Your definition of $r$ looks right to me, and looks equivalent to mine. Certainly I had much the same idea in mind. I used the absolute convergence of $\sum_{n=1}^\infty a_n$ when I inferred that the two series of non-negative terms $\sum_{n=1}^\infty b_n$ and $\sum_{n=1}^\infty c_n$ are convergent. I'm going to try to get some more sleep now, but I'll have another look at the whole thing after lunch, and see if I can make it any clearer. (Assuming it isn't all just a load of dingo's kidneys, of course!) $\endgroup$ – Calum Gilhooley May 17 at 9:04
  • $\begingroup$ I deliberately chose not to write \begin{gather*} b_n = \frac{|a_n| + a_n}2 \geqslant 0, \\ c_n = \frac{|a_n| - a_n}2 \geqslant 0, \end{gather*} but perhaps it would have been clearer had I done so. $\endgroup$ – Calum Gilhooley May 17 at 15:00
  • $\begingroup$ Perhaps I should also have stated explicitly that the identity I took to be satisfied by $\sum_{n \in K}a_n$ is a possible definition of that notation; but it is not the only possible definition; and you didn't say what definition you were using; so I left it open. $\endgroup$ – Calum Gilhooley May 17 at 15:05

I think there is an elementary proof (one without measure theory), that we can adapt from a similar claim in Apostol's Analysis book. Without loss of generality, $I=\mathbb N$. For each $k\in \mathbb N,\ I_k$ may be regarded as a map from some subset $\{1,2,\cdots,\}\subseteq \mathbb N$, to $\{\sigma_k(1),\sigma_k(2),\cdots,\}$ which may or may not be infinite, so $\sigma_k$ is an injective map from the susbet of $\mathbb N$ of the same cardinality as $|I_k|,$ starting at $1$, to the $\textit{set}\ I_k.$ If $|I_k|=j$, extend $I_k$ to all of $\mathbb N$ by mapping $n\in \mathbb N\setminus \{1,2,\cdots, j\}$ to $\mathbb N\setminus \{\sigma_k(1),\sigma_k(2),\cdots, \sigma_k(j)\}$ injectively and defining $a'_n:=0$ for all $n\in \mathbb N\setminus \{\sigma_k(1),\sigma_k(2),\cdots, \sigma_k(j)\}$. This construction will not affect any of the sums, so without loss of generality, $I_k$ maps $\mathbb N$ to a subset of $\mathbb N$ such that

$\tag1 I_k\ \text{is injective on}\ \mathbb N $

$\tag2 \text{the range of each}\ I_k \ \text{is a subset of } \ \mathbb N, \text{say}\ P_k $

$\tag3 \text{the}\ P_k\ \text{are disjoint}$

Now put $\tag4 b_k(n)=a_{I_{k}(n)}\ \text{and}\ s_k=\sum^\infty_{n=0}b_k(n)$

which is well-defined by $(1)-(3).$ We have to prove that

$\tag5 \sum^\infty_{k=0}a_k=\sum^\infty_{k=0}s_k$

It's easy to show that the right hand side of this converges absolutely. To find the sum, set $\epsilon>0$ and choose $N$ large enough so that $\sum^\infty_{k=0}|a_k|-\sum^n_{k=0}|a_k|<\frac{\epsilon}{2}$ as soon as $n>N.$ This implies also that


Now choose $\{I_1,\cdots, I_r\}$ so that each element of $\{a_1,\dots ,a_N\}$ appears in the sum $\sum^\infty_{n=0}a_{I_{1(n)}}+\cdots +\sum^\infty_{n=0}a_{I_{r(n)}}=s_1+\cdots+ s_r.$ Then, if $n>r,N$ we have

$\tag 7\left|\sum^n_{k=0}s_k-\sum^n_{k=0}a_k\right|<\sum^\infty_{n=N+1}<\frac{\epsilon}{2}$

Now $(5)$ folllows from $(6)$ and $(7).$

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  • $\begingroup$ I am a bit confused with your definition of $I_k$? Could you clarify it, please? $\endgroup$ – ZFR May 16 at 23:26
  • $\begingroup$ $I_k$ is some subset of $\mathbb N$. So it may be regarded as a function on the subset of $\mathbb N$ of the same cardinality. For instance, if $I_k=\{2,5,7\}$ then $I_k$ the function would map $\{1,2,3\}$ to $\{2,5,7\}$. If $I_k$ is infinite, then it is countable, so there is an injective function from $\mathbb N$ to it. That would be our $I_k$ considered as a function on $\mathbb N.$ $\endgroup$ – Matematleta May 16 at 23:31
  • $\begingroup$ Hmm. I guess that i got you. Also could you show why the RHS of (5) converges absolutely, please? $\endgroup$ – ZFR May 16 at 23:37
  • $\begingroup$ Hint: show directly by comparison with $\sum |a_n|$ that $\{s_k\}$ has bounded partial sums. $\endgroup$ – Matematleta May 16 at 23:42

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