Is the regularization of an otherwise diverging two-sided sum always equal to zero?

As a first example, take the divergent series of all powers of two $1+2+4+8+...=\sum\limits_{k=0}^\infty 2^k$ which can be regularized by using the analytical continuation of the geometric series $\sum\limits_{k=0}^\infty q^k = \frac1{1-q}\Big|_{|p|<1}$ to obtain $1+2+4+8+...=-1$, while on the other hand, the sum $\frac12 + \frac14 + \frac18 + ... = 1$, such that $$\sum_{k=-\infty}^\infty 2^k = 0$$

As a second example, take $... -3-2-1+0+1+2+3+...$, which is clearly zero as well (while the half-sided sum requires (Riemann) zeta regularization to obtain $1+2+3+4+...=-\frac1{12}$).

But is this generally the case or did I just pick some exceptional examples?

As a third example - that I am not sure about - take $...+1+1+1+1+...$: $$\underbrace{...+1+1+1}_{=\zeta(0)=-\frac12} + \underbrace{1}_{\stackrel{\text{from}}{k=0}} + \underbrace{1+1+1+1+...}_{=\zeta(0)=-\frac12} = 0$$ - I am not sure here since I pretend that $\sum_{k=1}^\infty\frac1{(-k)^s}\Big|_{s=0}$ is also $\zeta(0)$ due to the expression's symmetry.

• Nice conjecture :-). I've arrived at the same some years ago and it looked quite good - but was faulty.I'd approached this with the concept of Carleman-matrices and made a bold conjecture: go.helms-net.de/math/tetdocs/Tetration_GS_short.pdf But it came out to be false in the generality and I had to learn about Ramanujan-summation and the additional integral, the "magic gravitation-center of a series" (Ramanujan) (The text I've linked to was one of my first tries to write some coherent thing in number theory, so please bear with my amateurish approach and little formal style there) – Gottfried Helms Oct 1 '13 at 7:07
• @GottfriedHelms Thanks - can you give me a reference on that magic gravitation-center? Google seems to fail there, suggesting computing centres instead... – Tobias Kienzler Oct 1 '13 at 7:24
• I've found that phrase in an article which I've downloaded a couple of years ago, perhaps you can find it using google, otherwise email me. The header in this article is:$\text{Ramanujan’s Summation} \\ \text{Eric Delabaere} \\ \text{Universit´e d’Angers (France)}\\ \text{December 3, 2001}\\ \text{Summary by Vincent Puyhaubert}\\$ – Gottfried Helms Oct 1 '13 at 7:38
• @GottfriedHelms I did: algo.inria.fr/seminars/sem01-02/delabaere2.pdf Thanks, that'll be my "Gutenachtlektüre" then :7 – Tobias Kienzler Oct 1 '13 at 7:41
• ...let the moon shine over your dreams... – Gottfried Helms Oct 1 '13 at 9:37

It's too much to ask that the regularization of any two-sided divergent series be equal to zero. Clearly there is an extra symmetry in the examples you picked, both sides being given by the same expression. Otherwise, one could define either side separately to be any arbitrary divergent series and get all kinds of answers.

It's clearly true for any geometric series $\cdots + q^{-2} + q^{-1} + 1 + q + q^2 + \cdots$ if you regularize the two sides separately. The sum $\sum_{k=1}^\infty q^k = \frac{q}{1-q}$ plus the sum $\sum_{k=1}^\infty q^{-k} = \frac{1/q}{1-1/q} = \frac{-1}{1-q}$ is $-1$ which cancels out with the $q^0$ term to give 0.

It works trivially for the odd $\zeta$-sums, as in $\sum_{n=-1}^{-\infty} \frac{1}{n^{2k+1}} = -\zeta(2k+1)$, but fails for the even ones, such as $$\cdots + \frac{1}{(-3)^2} + \frac{1}{(-2)^2}+\frac{1}{(-1)^2} = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \cdots = \frac{\pi^2}{6}$$

I played around with these divergent sums a while ago and found the same examples as you have, but no others. The third one was especially tantalizing, but I eventually convinced myself it's a coincidence (though of course there's no proof of that).

One subtle phenomenon is the the lack of "shift-invariance" when we assign limits to divergent sums. It's known that we can define, in a non-unique way, a linear function $\lim_{n\rightarrow \infty}$ on all sequences, such that it agrees with the normal limit on convergent ones, as long as we don't expect $\lim_{n \rightarrow \infty} a_{n} = \lim_{n \rightarrow \infty} a_{n+1}$.

This can already be seen in the sums you gave. For example take

$$a_n = \left\{ \begin{array}{c} 1\text{ if }n\text{ is odd}\\0\text{ otherwise}\end{array}\right.\ \ \ \ \ b_n = \left\{ \begin{array}{c} 1\text{ if }n\text{ is even}\\0\text{ otherwise}\end{array}\right.$$

Then $a_n + b_n$ is the constant 1 sequence, but $b_n = a_{n+1}$. If they have the same (non-zero) limit , they can't cancel out to 0.

• You're of course right, though an interesting question would be under what conditions to the series coefficients it would work... – Tobias Kienzler Oct 1 '13 at 7:24

I have a very nice example introducing what I call "alternating iteration series" ("AIS").

Some years ago I considered the iteration of the function $\small g(x) = \sqrt{x+1/2}$ and its inverse $\small f(x) = x^2-1/2$.

Preliminaries
Let's introduce the following notation for convenience: we assume an initial value $x_0$ . We denote an iterate $\small x_0 \to x_1 = g(x_0)$ and in general $\small x_{h+1} = g(x_h)$ . (I use the letter $h$ here because in general I use also the term "h-eight" for the number of iterations, which initially stems from the study of analysis of "power towers" and their visual represention as vertically stacked symbols).
Similarly $\small x_{h-1} = f(x_h)$.

There are two fixpoints $\small t_f = (1-\sqrt 3)/2$ and $t_g = (1+ \sqrt 3)/2$ so for all initial values $x_0$ in the range $t_f \lt x_0 \le t_g$ the iterates $x_{-\infty}$ and $x_{+\infty}$ converge to that constant values.

This allows to Cesaro-sum (indicated by the underset $\small \mathfrak C$) the alternating iteration series $$\begin{array}{rlll} A_{g,p}(x_0) &\underset{\mathfrak C}= x_0^p - x_1^p + x_2^p - ... + ... \\ A_{f,p}(x_0) &\underset{\mathfrak C}= x_0^p - x_{-1}^p + x_{-2}^p - ... + ... \\ \end{array}$$
Even more, once we have that possibility of assigning meaningful values, we can put them together and define $$\begin{array}{rlll} A_p(x_0) &= A_{g,p}(x_0) +A_{f,p}(x_0) -x_0^p \\ &= \cdots - x_{-3}^p +x_{-2}^p-x_{-1}^p + x_0^p - x_1^p + x_2^p - x_3^p + \cdots \end{array}$$

Heuristics
Heuristically I found that $$\begin{array}{rlll} A_4(x_0) &= 0 \\ &= \cdots - x_{-3}^4 +x_{-2}^4-x_{-1}^4 + x_0^4 - x_1^4 + x_2^4 - x_3^4 + \cdots \end{array}$$ for all initial values $\small x_0$ of the allowed range. Moreover, the two partial series can be regularized to simple polynomials in $x$ (which has just recently been proven here in MSE), which also explain that zero-results.

A generalization - perhaps that still fits your request - are that compositions of two two-way-infinite series $$\begin{array}{rlll} A_p(1) + A_p(1/2) &= 0 & \qquad&\text{ for all } p \\ A_1(x) + A_2(x) &= 0 &&\text{ for all x in the allowed range} \\ \end{array}$$ but for which I've no proof yet...