Following this question and more particularly this question where the proof of the equivalence of two notions was left as an exercise, here is my attempt to solve that exercise:


  1. Is the definition below the correct particular case of the general one linked?

  2. Is my proof of the proposition below correct? If not, what mistakes did I make?

  3. In case my proof is incorrect: what would be a valid one?

  4. In case my proof is correct (statistically improbable ^^), is there a "neater" one? Or one which takes Riemann rearrangement theorem for granted but not Steinitz's theorem?

In the following $a_k\in\mathbb{C}$ ($k\in\mathbb{Z})$.

Definition: $\displaystyle\sum_{k\in\Bbb{Z}} a_k$ is the (unique) limit of the net $ \left\{\displaystyle\sum_{k \in F}a_k:F\subset\Bbb{Z}\text{ and $F$ is finite}\right\}$ as defined in here. That is, $\displaystyle\sum_{k\in\mathbb{Z}}a_k=L$ if, and only if, $$ \forall\epsilon>0,\exists F_{\epsilon}\subset\Bbb{Z}\text{ finite},\forall \Bbb{Z}\supset F\supseteq F_{\epsilon}\text{ finite}:\left|\sum_{k\in F}a_k-L\right|<\epsilon\tag{1} $$ Proposition: $$ \sum_{k\in\mathbb{Z}}a_k=L\iff\sum_{k=-\infty}^{\infty}a_k\text{ converges absolutely and }\sum_{k=-\infty}^{\infty}a_k=L $$ Proof: $[\Longrightarrow]$ (i) Let $\epsilon>0$ and $F_{\epsilon}$ be as in $(1)$. Let $N:=\max\{|a|:a\in F_{\epsilon}\}$. Then for all $m,n>N$, the subset $F_{m,n}:=[-m,n]$ contains $F_{\epsilon}$ hence $$ \left|\sum_{k=-m}^na_k-L\right|=\left|\sum_{k\in F_{m,n}}a_k-L\right|<\epsilon $$ So, $\displaystyle\sum_{k=-\infty}^{\infty}a_k=L$.

(ii) Suppose that absolute convergence doesn't hold. Then, WLOG, $\displaystyle\sum_{k=0}^{\infty}a_k$ converges conditionally. By Steinitz's theorem, there exists $z_1,z_2$, $z_1\neq z_2$ and permutations $\sigma_1,\sigma_2$ on $\mathbb{Z}_{\geq0}$ such that $$ \sum_{k=0}^{\infty}a_{\sigma_1(k)}=z_1\text{ and }\sum_{k=0}^{\infty}a_{\sigma_2(k)}=z_2 $$ By (i), $\displaystyle\sum_{k=1}^{\infty}a_{-k}$ converges, say to $\alpha$.

Given $\epsilon>0$, a standard argument then shows that there exists a $N$ such that $$ \left|\left(\sum_{k=1}^ma_{-k}+\sum_{k=0}^na_{\sigma_1(k)}\right)-(\alpha+z_1)\right|<\epsilon $$ whenever $m,n>N$.

Now, for any $\epsilon>0$, it is (clearly) possible to choose $F$ in $(1)$ such that $$ \sum_{k=1}^ma_{-k}+\sum_{k=0}^na_{\sigma_1(k)}=\sum_{k\in F}a_k $$ and $$ \left|\left(\sum_{k=1}^ma_{-k}+\sum_{k=0}^na_{\sigma_1(k)}\right)-(\alpha+z_1)\right|<\epsilon\\ \left|\sum_{k\in F}a_k-L\right|<\epsilon $$ simultaneously. This implies $L=\alpha+z_1$. Similarly, we must have $L=\alpha+z_2$. This is a contradiction because $z_1\neq z_2$.

Hence $\displaystyle\sum_{k=-\infty}^{\infty}a_k$ converges absolutely.

$[\Longleftarrow]$ Let $\epsilon>0$. Since $\displaystyle\sum_{k=-\infty}^{\infty}a_k$ converges absolutely, both $\displaystyle\sum_{k=0}^{\infty}a_k$ and $\displaystyle\sum_{k=1}^{\infty}a_{-k}$ converge absolutely. Consequently, we may take $N$ sufficiently large so that $$ \sum_{k=N+1}^{\infty}|a_k|<\frac{\epsilon}{2}\text{ and }\sum_{k=N+1}^{\infty}|a_{-k}|<\frac{\epsilon}{2} $$ Taking $F_{\epsilon}:=[-N,N]$, if $F\supseteq F_{\epsilon}$ then \begin{align} \left|\sum_{k\in F}a_k-L\right|&=\left|\sum_{k\in F}a_k-\left(\sum_{k=0}^{\infty}a_k+\sum_{k=1}^{\infty}a_{-k}\right)\right|\\ &=\left|\sum_{\substack{k\geq N+1\\k\not\in F}}a_{k}+\sum_{\substack{k\geq N+1\\k\not\in F}}a_{-k}\right|\\ &\leq\sum_{k=N+1}^{\infty}|a_{k}|+\sum_{k=N+1}^{\infty}|a_{-k}|\\ &<\epsilon \end{align} This shows that $\displaystyle\sum_{k\in\Bbb{Z}}a_k=L$. $\square$


I'll just answer your question 4. I am not saying the proof is neater, but it does not use anything deep. I will prove that if your limit (1) holds, then there is a constant $c>0$ such that $$ \sum_{k\in F}\left\vert a_{k}\right\vert \leq c $$ for every finite set $F\subseteq\mathbb{Z}$. I will use real numbers and not complex (you can just consider the real and immaginary sums separately).

Step 1: Let $F\subseteq\mathbb{Z}$ be a finite set. We claim that there exists a subset $G\subseteq F$ such that $$ \sum_{k\in F}\left\vert a_{k}\right\vert \leq2\left\vert \sum_{k\in G} a_{k}\right\vert . $$ Let $s:=\sum_{k\in F}\left\vert a_{k}\right\vert $. If $s=0$, take $G:=\emptyset$. If $s>0$, then writing $$ s:=\sum_{k\in F}\left\vert a_{k}\right\vert =\sum_{k\in F}a_{k}^{+}+\sum_{k\in F}a_{k}^{-}, $$ where $x^{+}:=\max\left\{ x,0\right\} $ and $x^{-}:=\max\left\{ -x,0\right\} $, we have that $\sum_{k\in F}a_{k}^{+}\geq\frac{s}{2}$ or $\sum_{k\in F}a_{k}^{-}\geq\frac{s}{2}$. Assume that $\sum_{k\in F}a_{k} ^{+}\geq\frac{s}{2}$ (the other case is similar). Let $G=\{k\in F:$ $a_{k}>0\}$. Then $$ \left\vert \sum_{k\in G}a_{k}\right\vert =\sum_{k\in G}a_{k}=\sum_{k\in F}a_{k}^{+}\geq\frac{s}{2}=\frac{1}{2}\sum_{k\in F}\left\vert a_{k}\right\vert . $$ Step 2: We claim that there exists $C>0$ such that $$ \left\vert \sum_{k\in F}a_{k}\right\vert \leq C $$ for every finite set $F\subseteq\mathbb{Z}$. Take $\varepsilon=1$ and let $E_1$ be the set given in (1). If $F\subseteq\mathbb{Z}$ then \begin{align} \left\vert\sum_{k\in F} a_{k}\right\vert= \left\vert\sum_{k\in (F\cup E_1)} a_{k}-\sum_{k\in (E_1\setminus F )} a_{k}\right\vert\le \left\vert\sum_{k\in (F\cup E_1)} a_{k}\right\vert+\sum_{k\in E_1} |a_{k}| \\ \le 1+|L|+ +\sum_{k\in E_1} |a_{k}|=:C \end{align}

Step 3: We claim that $$ \sum_{k\in F}\left\vert a_{k}\right\vert \leq2C $$ for every finite set $F\subseteq\mathbb{Z}$. This follows from Step 2 by applying Step 1.

  • $\begingroup$ This is very nice (+1)! It completely avoids the use of any rearrangement theorem. Thank you very much for sharing this. $\endgroup$ – NeedForHelp Feb 5 '17 at 5:43

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