$\newcommand{\R}{\mathbb R}\newcommand{\N}{\mathbb N}\newcommand{\si}{\sigma}\newcommand{\CC}{\mathcal C}$[This previous question](https://mathoverflow.net/q/432344/36721) 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](https://mathoverflow.net/a/432535/36721).
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](https://mathoverflow.net/a/432721/36721), $b$ is very good.


It was suggested (in some other terms) in [the comment](https://mathoverflow.net/questions/432344/generalizations-of-summation-methods-of-divergence-series/432721#comment1114106_432721) 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 $b\_n\to0$.



On summation methods of divergent series
$\renewcommand{\R}{\mathbb R}\renewcommand{\N}{\mathbb N}\renewcommand{\si}{\sigma}\newcommand\ep\varepsilon$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 $l\ne0$.


