Madelung's constant by neutral spheres Madelung's Constant is informally defined as:
$$M = \sum_{i,j,k=-\infty}^{+\infty} \frac{(-1)^{i + j + k}}{\sqrt{i^2 + j^2 + k^2}}$$
where the sum does not include the $i = j = k = 0$ term. The sum is conditionally convergent, and the wiki article includes references to proofs of the fact that:

*

*Summing over expanding spheres will cause the sum to diverge

*Summing over expanding cubes around the origin will cause the sum to converge to a finite constant, this is the actual definition of $M$.

It seems to me the essential problem of using expanding spheres is that the spherical shells are striated into shells containing only even lattice points, and shells containing only odd lattice points, causing the partial sums to oscillate, whereas the cubical shells are all `electrically neutral'. I'm wondering if using 'electrically neutral' shells guarentees convergence to $M$.
Formally, suppose that $S_1 \subseteq S_2 \subseteq \cdots$ is a sequence of convex, closed, compact subsets of $\mathbb{R}^3$, such that $\varinjlim S_i = \mathbb{R}^3$, such that the number of even lattice sites contained in $S_i$ is equal to the number of odd lattice sites contained in $S_i$. Is it true that the partial sums over $S_i$ are guarenteed to converge to $M$?
 A: It’s a good idea, but I’m afraid it doesn’t work.
Consider the octahedron $\{(x,y,z)\in \mathbb Z^3\mid|x|+|y|+|z|\le n\}$ for $n\in\mathbb N$. Its faces are all of one charge and carry a total charge of about $4n^2$, so as $n$ increases, the total charge of the octahedron oscillates by about $(-1)^n2n^2$ as $n$ increases.
To make it electrically neutral, you can remove roughly one half of the faces. There are different ways of doing that while keeping the remainder convex. In particular, you can either symmetrically remove the most remote half of the faces, or you can asymmetrically remove the half with, say, negative $z$.
Your suggested theorem would imply that the partial sums for both sequences of modified octahedra, the symmetrical and the asymmetrical one, would converge to $M$. But that implies that the sum for their difference converges to $0$. The difference is a dipole with one charge more remote from the origin than the other. The magnitude of the charge increases with $n^2$, but its distance from the origin only increases with $n$, and the ratio of the potentials caused by the two charges at different average distances from the origin is roughly constant. Thus, there’s an excess potential that increases with $n$ and oscillates according to the parity of $n$.
Since the sum for the difference diverges, the sum for at least one of the sequences of modified octahedra must diverge.
