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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)

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?

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    $\begingroup$ I think all the summation methods listed are linear. So questions (1) and (3) are not interesting. It remains question (2). $\endgroup$ – OR. Nov 14 '13 at 7:14
  • $\begingroup$ Dear Max: Cool question. What is the product of two divergent series? Does the order of summation matters? Also, why do you introduce $C$ if you never use it? $\endgroup$ – Bruno Joyal Nov 15 '13 at 23:18
  • $\begingroup$ I think that the interesting question is not about multiplying series, but rather multiplying limits of sequences, which for series would be their partial sums. So, the question is actually about the multiplicativity of the generalized limits. The usual limit is multiplicative in the sense that if $\lim f$ and $\lim g$ exist, then $(\lim f)(\lim g)=\lim (fg)$, and the latter exist too. $\endgroup$ – OR. Nov 18 '13 at 19:23
  • $\begingroup$ @BrunoJoyal I am so sorry, I forgot to answer your questions. I am not sure what you mean by "the order of summation". I should have used a $C$ indeed, it was meant to be included within the third question. $\endgroup$ – Max Muller Apr 10 '14 at 21:28
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    $\begingroup$ You should explain how you define multiplication $X=D_1 \cdot D_2$ of series $D_1$ and $D_2$: the termwise multiplication $x_k=d_{1,k} \cdot d_{2,k}$ ? The cauchy-product? Or the product of the partial evaluations up to some index $k$? And if the latter: of the original terms? Or of the terms after the transformation by the divergent summation-procedure? $\endgroup$ – Gottfried Helms Aug 27 '14 at 0:05
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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.

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  • $\begingroup$ ABC, please see my comment which I put in an answerbox because it is too long for the commentfunction $\endgroup$ – Gottfried Helms Aug 26 '14 at 23:54
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This is a comment to @ABC 's answer.
I seem to have difficulties to understand things correctly, maybe it's simply too late.

Here is a list of

  • a: the alternating series of units,
  • b: the partial sums (=using Cesaro-order 0),
  • c: the partial sums using Cesaro-order 1,
  • d: and the squares of $c$, (where it is focused that the cesaro-partial sum up to $n$ is used and then squared).
  • e: the terms of the cauchy-products $D_1 \cdot D_1$ up to index n :$ e_n=\sum_{k=0}^n d_k \cdot d_{n-k} $
  • f: the Cesaro(order 2)-partial-sums of e

We see, that $d$ has a clear limit (the expected one: for every $n$ the square of the partial sum so far in $c$). We needed to use $c$ and not $b$ because $b$ is not converging and thus has no limit which we could use for the multiplication. Similarly this is with the entries in f: the sequence of partial sums of the terms in e ( which are all Cauchy-products) when transformed by the Cesaro-summation (order 2) approaches the expected limit.

Here is the table:

$\qquad \qquad$ table

(The effect is even more drastical if we use Euler-summation instead of Cesaro-summation, because it is much more powerful with such type of series.

Here is the table:

$\qquad \qquad$ table

)

But perhaps I missed/misunderstood some idea in your answer, it was late evening...

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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.

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  • $\begingroup$ I think that the interesting question is not multiplying series like that, but multiplying the partial sums. So, the question is actually about the multiplicativity of the generalized limits. The usual limit is multiplicative in the sense that if $\lim f$ and $\lim g$ exist, then $(\lim f)(\lim g)=\lim (fg)$, and the latter exist too. $\endgroup$ – OR. Nov 18 '13 at 19:22
  • $\begingroup$ do you have any example ? $\endgroup$ – Thomas Nov 18 '13 at 20:04
  • $\begingroup$ I think there are. See the answer I added. $\endgroup$ – OR. Nov 18 '13 at 21:57
  • $\begingroup$ eh he... nice example but I was talking about things "generally" not with examples. You can also consider the series A0=a, An=0 for n>0, and similarly for B. And this is an algebra that has lots of properties too :) $\endgroup$ – Thomas Nov 19 '13 at 8:26

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