Generalizations of summation methods of divergence series If one looks at the "summation proofs" of divergent series such as Grandi's series, one might see a pattern that most of the computation rely on linearity and comparability with the shift operator of summation. These, of course, are not real proofs, since the series do not converge, but one might try to generalize the concept of summation to such series. $\DeclareMathOperator{\shft}{sh}$
Thus, inspired by Lebesgue measure on the reals, one might define a summability space as a pair $(\mathcal{S}, \sigma)$ such that:

*

*$\mathcal{S} \subset \mathbb{R}^\mathbb{N}$ is a vector subspace which contains the space $\mathcal{C}$, of real sequences whose sum converges, and is closed under $\shft$.

*$\sigma \colon \mathcal{S} \to \mathbb{R}$ is a linear operator.

*Regular: For every $(a_n)_n \in \mathcal{C}$ we have that
$\sigma ( (a_n)_n ) =\Sigma_n a_n=a_1+a_2+\cdots$.

*Translative: For every $(a_n)_n \in \mathcal{S}$ we have that $\sigma((a_n)_n)=a_1 +\sigma(\shft((a_n)_n))$.

Here $\shft$ is the shift operator, i.e., $\shft(a_1, a_2, \dots)=(a_2,a_3,\dots)$.
What is the largest possible summability space (which is nice is some way)? Has this idea already been studied? Can we strengthen the definition of a summability space to get a canonical largest summability space (in the same way that Lebesgue measure is the "largest nice" measure on $\mathbb{R}$)?
I know that there are many ways to sum divergent series, such as Cesaro summation, Abel summation, etc. But I am asking about the "best" summation method, which "unifies" all other summation method.  (Note that we can't simply apply the Hahn-Banach theorem or such a result since we would like to preserve translativity as well).
 A: $\newcommand{\R}{\mathbb R}\newcommand{\N}{\mathbb N}\newcommand{\si}{\sigma}\newcommand{\SSS}{\mathcal S}\newcommand{\CC}{\mathcal C}\newcommand{\sh}{\operatorname{sh}}$In the previous answer, it was shown that there are infinitely many maximal summability spaces, in the sense of the OP. That was done using a very fast growing sequence $b\notin\CC$.
Here it will be shown that we can instead use any sequence $b=(b_1,b_2,\dots)\in\R^\N\setminus\CC$ with
\begin{equation*}
    b_n\to0. \tag{0}\label{0}
\end{equation*}
Let $b$ be such a sequence indeed; for instance, $b_n=1/n$ will do.
Here we will keep the following notations from the previous answer:
$\CC$ will denote the set of all sequences in $\R^\N$ summable in the sense that the corresponding sequence of partial sums is convergent, $T:=\sh$, and $\N_0:=\{0\}\cup\N$.
Take any $t\in\R$. Let
\begin{equation*}
    \SSS_t:=\text{span}(\CC\cup\{T^k b\colon k\in\N_0\}).  
\end{equation*}
Take any $s\in\SSS_t$. Then
\begin{equation*}
    s=a+\sum_{j=0}^k c_j T^jb
\end{equation*}
for some $a=(a_1,a_2,\dots)\in\CC$, some $k\in\N_0$, and some real $c_0,\dots,c_k$.
Let then
\begin{equation*}
    \si_t(s):=\si_t(a)+\sum_{j=0}^k c_j \si_t(T^jb), \tag{2}\label{2}
\end{equation*}
where
\begin{equation*}
    \si_t(a):=\sum_{n=1}^\infty a_n \tag{3}\label{3}
\end{equation*}
and
\begin{equation*}
    \si_t(T^j b):=t-\sum_{n=1}^j b_n.  \tag{5}\label{5}
\end{equation*}
Then $\si_t$ is a well-defined linear functional on $\SSS_t$. To check this, it suffices to verify the implication (A)$\implies$(B), where (A) and (B) are the following conditions:
\begin{equation*}
    0=a+\sum_{j=0}^k c_j T^jb \tag{A}\label{A}
\end{equation*}
and
\begin{equation*}
    0=\si_t(a)+\sum_{j=0}^k c_j \si_t(T^jb), \tag{B}\label{B}
\end{equation*}
for any given $k\in\N_0$ and any given real $c_0,\dots,c_k$.
Assume indeed that (A) holds. Then
\begin{equation*}
    0=a_n+\sum_{j=0}^k c_j b_{n+j} 
\end{equation*}
for all $n\in\N$. Hence, for any $N\in\N$, letting
\begin{equation*}
    s_N:=\sum_{n=1}^N b_n,
\end{equation*}
we have
\begin{equation*}
\begin{aligned}
    0&=\sum_{n=1}^N a_n+\sum_{j=0}^k c_j \sum_{n=1}^N b_{n+j} \\ 
    &=\sum_{n=1}^N a_n+\sum_{j=0}^k c_j (s_{N+j}-s_j) \\ 
    &=\sum_{n=1}^N a_n+\Big(\sum_{j=0}^k c_j\Big)s_N+\sum_{j=0}^k c_j(s_{N+j}-s_N-s_j).  
\end{aligned}
\tag{10}\label{10}
\end{equation*}
To obtain a contradiction, suppose that  $\sum_{j=0}^k c_j\ne0$.
Letting now $N\to\infty$ and noting that, by \eqref{0}, $s_{N+j}-s_N\to0$ for each $j$,
we see that \eqref{10} implies
\begin{equation*}
    s_N\to-\frac1{\sum_{j=0}^k c_j}\,\Big(\sum_{n=1}^\infty a_n-\sum_{j=0}^k c_j s_j\Big)\in\R
\end{equation*}
which contradicts the assumption $b\notin\CC$.
So,  $\sum_{j=0}^k c_j=0$ and hence, by \eqref{10}, \eqref{3}, and \eqref{5},
\begin{equation*}
\begin{aligned}
    0&=\sum_{n=1}^\infty a_n+\sum_{j=0}^k c_j(-s_j) \\  
    &=\sum_{n=1}^\infty a_n+\sum_{j=0}^k c_j t+\sum_{j=0}^k c_j(-s_j) \\   
    &=\sum_{n=1}^\infty a_n+\sum_{j=0}^k c_j(t-s_j) \\   
    &=\si_t(a)+\sum_{j=0}^k c_j \si_t(T^jb).   
\end{aligned}
\end{equation*}
So, the implication (A)$\implies$(B) is verified, which shows that $\si_t$ is indeed a well-defined linear functional on $\SSS_t$.
Now it is straightforward to complete the checking that $(\SSS_t,\si_t)$ is a summability space.
Clearly, no summability space can contain two summability spaces of the form $(\SSS_t,\si_t)$ with two different values of $t\in\R$. On the other hand, by Zorn's lemma, each summability space of the form $(\SSS_t,\si_t)$ is contained in a maximal summability space $(\SSS^*_t,\si^*_t)$, and all these maximal summability spaces $(\SSS^*_t,\si^*_t)$ are distinct from one another.
Thus, as claimed, there are infinitely many maximal summability spaces and, therefore, as stated in the previous comment, there is no one largest summability space.
A: $\newcommand{\R}{\mathbb R}\newcommand{\N}{\mathbb N}\newcommand{\si}{\sigma}\newcommand{\SSS}{\mathcal S}\newcommand{\CC}{\mathcal C}\newcommand{\sh}{\operatorname{sh}}$First of all, as was noted in the previous comment, $\CC$ should be defined, not as the set of all convergent sequences in $\R^\N$, but as the set of all sequences in $\R^\N$ summable (say) in the sense that the corresponding sequence of partial sums is convergent.
Let now $T:=\sh$ and $\N_0:=\{0\}\cup\N$.
Partially ordering the summability spaces by inclusion and using Zorn's lemma, we see that there is a maximal summability space.
Actually, there are infinitely many maximal summability spaces.
Indeed, take any $t\in\R$. Let
\begin{equation*}
    \SSS_t:=\text{span}(\CC\cup\{T^k b\colon k\in\N_0\}), \tag{1}\label{1}
\end{equation*}
where
\begin{equation*}
    b:=(b_1,b_2,\dots)\quad\text{with}\quad b_n:=n!.    
\end{equation*}
Note that, for each nonzero sequence $a\in\CC$, the sequences $a,b,Tb,T^2b,\dots$ are linearly independent. To see why this is true, suppose that, to the contrary, $c_{-1} a+c_0 b+c_1 Tb+\dots+c_k T^kb=0$
for some $k\in\N_0$ and some real $c_{-1},\dots,c_k$ such that $c_k\ne0$. Then for all $n\in\N$ we have
\begin{equation}
c_{-1} a_n+c_0 n!+c_1(n+1)!+\dots+c_{k-1}(n+k-1)!+c_k(n+k)!=0.      
\end{equation}
Letting $n\to\infty$, we get
\begin{equation}
c_{-1} a_n+c_0 n!+c_1(n+1)!+\dots+c_{k-1}(n+k-1)!=o((n+k)!)     
\end{equation}
and hence $c_k=0$, a contradiction.
So, we have the summability space $(\SSS_t,\si_t)$ with $\SSS_t$ as in \eqref{1} and the linear functional $\si_t\colon\SSS_t\to\R$ defined by the conditions
\begin{equation*}
    \si_t(a):=\sum_{n=1}^\infty a_n
\end{equation*}
for $a=(a_1,a_2,\dots)\in\CC$ and
\begin{equation*}
    \si_t(T^k b):=t-\sum_{n=1}^k b_n 
\end{equation*}
for $k\in\N_0$ (so that $\si_t(b)=t$).
(Added detail: Indeed, in view of \eqref{1} and because the sequences $a,b,Tb,T^2b,\dots$ are linearly independent for any nonzero $a\in\CC$, each sequence $s\in\SSS_t$ can be uniquely represented by the formula $s=a+c_0 b+c_1 Tb+\dots+c_k T^kb$ for some $a\in\CC$, some $k\in\N_0$, and some real $c_0,\dots,c_k$; and then $\si_t(s)=\si_t(a)+c_0 \si_t(b)+c_1 \si_t(Tb)+\dots+c_k \si_t(T^kb)$ by the linearity of $\si_t$.)
Clearly then, no summability space can contain two summability spaces of the form $(\SSS_t,\si_t)$ with two different values of $t\in\R$. On the other hand, by Zorn's lemma, each summability space of the form $(\SSS_t,\si_t)$ is contained in a maximal summability space $(\SSS^*_t,\si^*_t)$, and all these maximal summability spaces $(\SSS^*_t,\si^*_t)$ are distinct from one another.
Thus, as claimed, there are infinitely many maximal summability spaces and, therefore, as stated in the previous comment, there is no one largest summability space.
(This is probably why apparently "nobody studied" such a nonexistent entity.)
