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.

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.
