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

Question

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?

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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.

Answer 1

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...

Answer 2

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.

Answer 3

I thought I'd write something since I find the fact that this is true is quite nice.

First, let's look at the situation symbolically. Take $$f(x) = \sum_{n=0}^\infty (-1)^n p(n) x^n$$ Where is a polynomial $p(n) = \sum_{k=0} a_k n^k$. Interchanging sums, we obtain $$\sum_{n=0}^\infty \sum_{k=0} a_k n^k (-1)^n x^n = \sum_{k=0}a_k\sum_{n=0}^\infty (-1)^n x^n n^k$$ By properties of the polylogarithm, $\sum_{n=0}^\infty (-1)^n x^n n^k = -\sum_{n=1}^\infty (-1)^nx^{-n}(-n)^k$ (if your inclinded to prove this, it can be seen by verifying it is true for the geomtric series (i.e. k=0), and then proving the series stay equal after applying $x \frac{d}{dx}$ to both sides).

Using this fact, we obtain $$\sum_{k=0}a_k\sum_{n=0}^\infty (-1)^n x^n n^k=-\sum_{n=1}^\infty(-1)^nx^{-n}\sum_{k=0}a_k (-n)^k = -\sum_{n=1}^\infty (-1)^n p(-n)x^{-n} $$

That's pretty cool-- we can analytically continue a function just by flipping $n$ to be negative and slightly changing the summation. For instance, here's a tricky sum that's easy to analytically continue with this method $$\sum_{n=0}^{\infty}ne^{n}\cos\left(n-2\right)x^{n} \to -\sum_{n=1}^{\infty}\left(-n\right)e^{-n}\cos\left(-n-2\right)x^{-n}$$

So, as long as your divergent series regularizations coincides with analytical continuation, we should find that for a large class of functions (basically those where $a_n$ has a power series that represents an entire function, or something similar) your proposition is true, that is $$ \left(\sum_{n=0}^\infty a_n x^n\right) + \left(\sum_{n=1}^\infty a_{-n} x^{-n}\right)=0$$

I can write more details about why the proposition is true more broadly and specify where it fails-- though this is an old question so for now I will leave the answer as it is unless there is a demand to expand it.