On summation methods of divergent series

Question

$\newcommand{\R}{\mathbb R}\newcommand{\N}{\mathbb N}\newcommand{\si}{\sigma}\newcommand{\CC}{\mathcal C}$This previous question introduced the following notion of a summability space.

Let $\N:=\{1,2,\dots\}$. Let $T$ be the shift operator on $\R^\N$ defined by the formula $Ts:=(s_2,s_3,\dots)$ for $s\in\R^\N$. Here and in what follows, $s_j$ will denote the $j$th coordinate of a sequence $s=(s_1,s_2,\dots)\in\R^\N$.

Let $C$ denote the set of all sequences in $\R^\N$ summable in the sense that the corresponding sequence of partial sums is convergent.

Say that a pair $(S,\si)$ is a summability space if the following conditions hold:

(i) $S$ is a vector subspace of $\R^\N$ containing $C$ and closed with respect to the shift $T$;

(ii) $\si\colon S\to\R$ is a linear functional;

(iii) for every $a\in C$ we have
$\si(a) =\sum_{n=1}^\infty a_n$;

(iv) for every $s\in S$ we have
$\si(s) =s_1+\si(Ts)$.

The set of all summability spaces is naturally ordered by inclusion. Clearly, $(C,\si)$ with $\si$ defined by condition (iii) above is a summability space, actually the smallest one.

Let us say that a sequence $g\in\R^\N$ is good if $g\in S$ for some summability space $(S,\si)$. Let us say that a sequence $v\in\R^\N$ is very good if for each real $t$ there is a summability space $(S_t,\si_t)$ such that $v\in S_t$ and $\si(v)=t$. Obviously, every very good sequence is good.

By Zorn's lemma, given any very good sequence $v$, for every each real $t$ there will exist a maximal summability space $(S_t^*,\si_t^*)$ such that $v\in S_t^*$ and $\si_t^*(v)=t$, so that these maximal summability spaces $(S_t^*,\si_t^*)$ will be distinct for distinct values of $t\in\R$.

For an illustration, consider the geometric sequence $b$ with $b_n=Cr^n$ for some nonzero real $C$, some nonzero real $r$, and all $n\in\N$. If $|r|<1$, then $b$ is in $C$ and therefore good. If $r=1$, then $Tb=b$ and therefore $b$ is not good, in view of condition (iv). If $r\notin(-1,1]$, then $Tb=rb$ and hence, if $b\in S$ for some summability space $(S,\si)$, then, by conditions (iv) and (ii),
we necessarily have $\si(b)=\frac{b_1}{1-r}=\frac{Cr}{1-r}$, so that in this case $b$ is good but not very good.

If a sequence $b\in\R^\N$ grows faster than geometrically, so that $b_{n+1}/b_n\to\infty$ (as $n\to\infty$), then $b$ is very good -- cf. this previous answer. On the other hand, if a sequence $b\in\R^\N$ is not in $C$ and $b_n\to0$, then, as shown in the other previous answer, $b$ is very good.

It was suggested (in some other terms) in the comment to the latter answer that a sequence $b\in\R^\N$ is not good if \begin{equation} b_n\to l \end{equation} for some nonzero real $l$.

This suggestion will be confirmed in the answer below. So, it will follow that a convergent sequence $b\in\R^\N\setminus C$ is very good iff $b_n\to0$.

Answer 1

$\renewcommand{\R}{\mathbb R}\renewcommand{\N}{\mathbb N}\renewcommand{\si}{\sigma}\newcommand\ep\varepsilon$Take indeed any sequence $b\in\R^\N$ such that \begin{equation*} b_n\to l \tag{0}\label{0} \end{equation*} for some real $l\ne0$.

To obtain a contradiction, suppose that $b$ is good, so that $b\in S$ for some summability space $(S,\si)$.

Let \begin{equation*} a:=Tb-b. \tag{1}\label{1} \end{equation*} Then, for any $N\in\N$, \begin{equation*} \sum_{n=1}^N a_n=\sum_{n=1}^N (b_{n+1}-b_n)=b_{N+1}-b_1. \end{equation*} Letting now $N\to\infty$ and using condition \eqref{0}, we see that $a\in C$. Moreover, by condition (iii) in the definition of a summability space, \begin{equation*} \si(a)=\sum_{n=1}^\infty a_n=l-b_1. \tag{2}\label{2} \end{equation*} On the other hand, by \eqref{1} and conditions (ii) and (iv) in the definition of a summability space, \begin{equation} \si(a)=\si(Tb)-\si(b)=-b_1. \tag{3}\label{3} \end{equation} Comparing \eqref{2} and \eqref{3}, we get a contradiction with the condition $l\ne0$.