Question on suitability of zeta-regularization for certain divergent series

Question

In another question here in MSE I arrived at the idea to decompose a divergent series, which was not Abel-summable, termwise into combination of alternating and non-alternating zeta series-terms, and then use the accordingly regularized zeta()/eta()-values in the found composition for the regularization of the original series.
This looked very convincing, however I've got a serious (and I think respectable) comment which was sceptical about the applicability if this method:

• _{"But I still believe that the fact that the arguments are shifted from s to s−1 and s−2 and you are combining several of them, even though the original sum had a uniform argument s, if you get my point, is illegitimate in the zeta regularization. There may be a set of rules in which your would be an allowed value - but maybe those rules would be allow any value. I sort of feel why your calculation would produce physically wrong results in physics." - comment at my own answer}

Thus my question here explicitely:

Is the regularization of a divergent series, which can term-by-term be decomposed in a weighted sum of $\zeta()$ and $\eta()$ (finitely many), equivalent to the same composition of the evaluated/regularized $\zeta()$ and $\eta()$ values? And: if not - what are the logical or formal obstacles?

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Example from my earlier question

I've defined the sequence $$A = \left \lbrace a_k \right \rbrace_{k=0}^\infty = (-1)^k \binom{2+k}{k} \tag 1$$ looking like $[1,-3,6,-10,..., (-1)^k\binom{2+k}{2},...]$
Then I defined $B$ as the sequence of partial sums of the $a_k$ giving $$B = \lbrace b_k \rbrace _{k=0}^\infty =[1,-2,4,-6,9,\cdots ] \tag 2$$ This sequence has the generating function $g(x)_B= 1/(1+x)^3/(1-x) $ and because there is the $(1-x)$-expression in the denominator it is not Abel-summable (or Cesaro- or Euler-summable).

Heuristically I found that the sequence $B$ can be composed termwise by combining the terms of the $\zeta()$ and $\eta()$ -series according to $$ \begin{array} {rr} 8 b_k &=& 1\cdot& 1 \cdot (1+k)^0 \\ &&+ 1\cdot&(-1)^k \cdot (1+k)^0 \\ &&+ 4\cdot&(-1)^k \cdot (1+k)^1 \\ &&+ 2\cdot&(-1)^k \cdot (1+k)^2 \end{array} \tag 3 $$

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Update to include the "regularization" part more explicite (copied from my own older answer to satisfy request from the comments):
Denote the original series by $T$, then let $$ f(s) = {1\over 1^s }-{2\over 2^s}+{4\over 3^s}-{6\over 4^s}+{9\over 5^s}- \cdots + \cdots \tag 1$$ of course attempting to justify $ T = \lim_{s \to 0} f(s)$.

This is convergent for $s \gt 3$ . For this cases we can decompose $$ \begin{array}{rcrll} 8 f(s) &=& & {8\over 1^s }-{16\over 2^s}+{32\over 3^s}-{48\over 4^s}+{72\over 5^s}- \cdots + \cdots \\ &=& 1(&{1\over 1^s }+{1\over 2^s}+{1\over 3^s}+{1\over 4^s}+{1\over 5^s}+ \cdots + \cdots &)\\ & & + 1(&{1\over 1^s }-{1\over 2^s}+{1\over 3^s}-{1\over 4^s}+{1\over 5^s}- \cdots + \cdots&) \\ & & + 4(&{1\over 1^s }-{2\over 2^s}+{3\over 3^s}-{4\over 4^s}+{5\over 5^s}- \cdots + \cdots & ) \\ & & + 2(&{1\over 1^s }-{4\over 2^s}+{9\over 3^s}-{16\over 4^s}+{25\over 5^s}- \cdots + \cdots & ) \\ &\underset{s \gt 3}=& &1 \zeta(s)+1\eta(s)+4\eta(s-1)+2\eta(s-2) \end{array} \\ \phantom{dummy } \\ f(s) \underset{s \gt 3} = {\zeta(s)+\eta(s)\over 8} + {\eta(s-1)\over 2} + {\eta(s-2)\over 4} \qquad \qquad \qquad \tag 2 $$ and from this, assuming it is regularizable for setting $s=0$ $$ T = f(0) \underset{\mathcal Z}{=} {\zeta(0)+\eta(0)\over 8} + {\eta(-1)\over 2} + {\eta(-2)\over 4} = 0 + {1\over4}\cdot{1\over 2} + 0 = {1\over 8} \tag 3$$ $\qquad \qquad $ where "$\mathcal Z$" means zeta-regularization

end update

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Thus I assumed that it is possible to regularize the sum of the sequence $B$ by the accordingly composed regularized zeta/eta() values: $$ \sum_{k=0}^\infty b_k \underset{\mathcal Z}{=} {1 \zeta(0) + 1 \eta(0) + 4 \eta(-1) + 2 \eta(-2) \over 8} = \frac18$$ $\qquad \qquad \qquad $where $\underset{\mathcal Z}=$ means "equals by zeta-regularization"

Additional remark: this decomposition I found manually, but it is easy to uncover such compositions for many sequences (where this is applicable at all) using multiple regression with the sequence $B$ as $y$-vector.
after having added the update I should enhance on the the focus of my question:

So this question does not only concern the given example but concerns the applicability of this as a method at all.

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Some random examples, by varying the generating function:

seq   = [1, -2, 4, -6, 9, -12, 16, -20, 25, -30, 36, -42, ...]
compos= +1/8*zeta(0)+1/8*eta(0)+1/2*eta(-1)+1/4*eta(-2)
value = 1/8  (by regularization)

seq   = [1, -1, 3, -3, 6, -6, 10, -10, 15, -15, 21, -21, ...]
compos= +3/16*zeta(0)+1/8*zeta(-1)+3/16*eta(0)+3/8*eta(-1)+1/8*eta(-2)
value = 1/12  (by regularization)

seq   = [1, 1, 4, 0, 9, -3, 16, -8, 25, -15, 36, -24, ...]
compos= +1/8*zeta(0)+3/4*zeta(-1)+1/8*eta(0)+-1/4*eta(-1)+1/4*eta(-2)
value = -1/8  (by regularization)

Answer 1

Yes, it is a consequence of the uniqueness of the analytic continuation of an absolutely convergent Dirichlet series. By the way, this is not the zeta regularization method. And Abelian means give in general distinct sums to a series, for example, the sum of the Grandi series $1 - 1 + 1 - 1 + \cdots$ ($a_1 = 1$) is $0$ if $\lambda_{2k -1}= \lambda_{2k}$ or 1 if $\lambda_{2k} = \lambda_{2k + 1}$.

Answer 2

I think things are clearer if you write in summations.

$\sum_{n=1}^{n=m}(-1)^{n+1}*(n+1)*(n)/2=1/8(-2*(-1)^m*m^2-4*(-1)^m*m-(-1)^m+1)$

You wanted the regularised value of this: $$\sum_{m=1}^{\infty} 1/8(-2*(-1)^m*m^2-4*(-1)^m*m-(-1)^{m}+1)$$

This can be done by taking the regulated value seprated.

$$\sum_{n=1}^{\infty} -1/4((-1)^m*m^2) =\eta(-2)/4=0$$ $$\sum_{n=1}^{\infty} -1/2((-1)^m*m =\eta(-1)/2=1/8$$ $$\sum_{n=1}^{\infty} -1/8((-1)^m) =\eta(0)/8=1/16$$ $$\sum_{n=1}^{\infty} 1/8(1)=\zeta(0)/8=-1/16$$

So the answer is indeed 1/8 because addition applies to regularization. You can't just randomly reorder the numbers of a series, see multiplication bellow. If you reorder them you have to keep the original order of the numbers intact (see also my take on Cesàro summation).

There's a trick for the orignal sum to see the answer "directly" if you don't know the eta function, which is taking the function equation of the summed series, which yields the solution aswell. Notice it's negative n*(n+1)/2 on the even values and positive on the uneven values of the summed series (1,-1,3,-3,6,-6, ... ). $$-(n/2*(n/2+1)/2)*(1+(-1)^n)/2$$ gives the even integers. $$((n+1)/2*((n+1)/2+1)/2)*(1-(-1)^n)/2$$ give the uneven integers.

You only need to use non-alternating parts, and the constant value. The constant value is 3/16, delta the "real" growing part, n/8, and regularise it (e.g. zeta(0)) which ends up at -1/16, which makes the regularised value of the orginal sum 1/8.

somewhat the same way is to make Cesàro sumable by removing the constant value of the original function (the +1/8 or 1/8*n from the summed value, knowing you have to add -1/16 to the finale value). But that's more like actually finding the value's of the eta function (You can find the value really quickly "numberically" by just taking the averge 3 times of the sum's result).

If there aren't shortcuts to Cesàro sum it, from old notes haven't checked, this was my take on Cesàro summable, but that's here unnecessary complex as there are lots of easier "shotcuts" here.

For an integer d>1; m will get canceld out and is to help; and k is whatever integer you want, besides a multiple of d ofcouse. $$ \sum_{n=1}^{\infty} f(n) e^{\frac{2*i *\pi*k}{d}}=\frac{1}{d}\sum_{s=0}^{d-1}\sum_{j=0}^{d-1}\sum_{n=1}^{\frac{m-s+j}{d}}f(dn-j) e^{\frac{2*i *\pi(-j)*k}{d}}$$

Addition applies to regularization but know multiplication doesn't apply the way you are used to but you can work around it if you apply a delta. And this is tricky if you use infinity as same notation for regularization because sometimes you want a regularization of a multiplication, and some you want the multiplication of a regularization.

$$(\sum_{n=1}^{\infty} f(n))^2)=\sum_{m=1}^{\infty} \Delta_{m} (\sum_{n=1}^{m}f(n))^2$$

for example $$(\sum_{m=1}^{\infty} 1^2)=\sum_{m=1}^{\infty} \Delta_{m} m^2=\sum_{m=1}^{\infty} 2m-1$$

So what would be wrong is for example summing/reorder diagonally to $\sum_{m=1}^{\infty}\sum_{c=1}^{m} 1$, or also incorrectly is to assume $\zeta(0)^2$