Is there an algebra of summable series?

Question

Let $D$ denote a divergent series and let $C$ denote a convergent series. Furthermore, let $s : \{ Series \} \to \{ numbers \} $ be a regular, linear divergent series operator, which is either one of these operators:

(the hyperlinks will direct you to the wiki page of the relevant summation method, not the person who invented/discovered it)

• Borel summation
• Abel summation
• Euler summation
• Césaro summation
• Lambert summation
• Ramanujan summation
• Summing the series by means of Analytic continutation
• Some Regularization method

I am wondering if there is any meaningful way to answer the following questions (Assuming $D_1 , D_2$ are summable with $s$):

1. What does $s(D_1 + D_2)$ equal? Is it always equal to $s(D_2 + D_1)$ ? How does it relate to $s(D_1)$ and $s(D_2)$ ?
2. What does $s(D_1 \cdot D_2) $ equal? Is it always equal to $s(D_2 \cdot D_1)$ ? How does it relate to $s(D_1)$ and $s(D_2)$ ?
3. What happens when we add convergent series into the mix? And what if we're summing linear combinations of $n$ convergent and $m$ divergent series?

Do the results differ for different summation methods, listed above?

Answer 1

Short answer : no. Convergent series have the addition property s(A+B) = s(A)+s(B)$ but no multiplication property.

if $A_n=1/n$, $s(A) =\infty$ but $A\cdot A$ is convergent.

I didn't look in details at your links but what I'm saying here is pretty general. Multiplying series feels weird.

Answer 2

For (1) and (3) I think all summation methods listed are linear summation methods. So $s(D_1+D_2)=s(D_1)+s(D_2)$, when $s(D_1)$ and $s(D_2)$ exist.

For (2), $D_1\cdot D_2=D_2\cdot D_1$. So, $s(D_1\cdot D_2)=s(D_2\cdot D_1)$, when they exist.

On the other hand, consider $D:=\{(-1)^n\}_{n=0}^{\infty}$, and $s$ to be Cesaro limit.

Then $s(D)=\lim_n\frac{\sum_{k=0}^{n}(-1)^k}{n}=0$, where there are $n$ ones in the numerator.

But $D^2=\{1\}_{n=0}^{\infty}$. Then the Cesaro limit is $s(D^2)=\lim_n\frac{\sum_{k=0}^{n}1}{n}=\lim_n\frac{n+1}{n}=1$.

So, even when $s(D_1)$ and $s(D_2)$ exist, $s(D_1\cdot D_2)\neq s(D_1)s(D_2)$.

So, for Cesaro summation, and therefore for any summation method extending it, the limit is not multiplicative.