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Does the sum $$\sum_{z \in \mathbb{Z}^3\setminus \{(0,0,0)\}} \left( \frac{1}{|{\bf x} - {\bf z}|^2} - \frac{1}{|{\bf z}|^2} \right)$$
converge pointwise or even uniformly for $\varepsilon < |{\bf x}| < 1-\varepsilon$?

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Don't think it converges. At a distance $R\gg 1$ there are $\propto R^2$ points contributing, each with $1/(R-x)^2-1/R^2 \approx 2x/R^3$ such that the sum is logarithmically diverging. –  Fabian Mar 1 '11 at 19:59
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1 Answer 1

It seems to converge conditionally, with convergence if you sum over a sphere centered at the origin and then let the radius of the sphere increase without bound. Let ${\bf x}=(0,0,d)$ and ${\bf z}=(x,y,z)$ (sorry for reusing $x$). Then $$\begin{align}\sum_{z \in \mathbb{Z}^3\setminus \{(0,0,0)\}} \left( \frac{1}{|{\bf x} - {\bf z}|^2} - \frac{1}{|{\bf z}|^2} \right)&=\sum_{z \in \mathbb{Z}^3\setminus \{(0,0,0)\}} \left(\frac{1}{x^2+y^2+(z-d)^2}-\frac{1}{x^2+y^2+z^2}\right) \\ &=\sum_{z \in \mathbb{Z}^3\setminus \{(0,0,0)\}} \frac{(z-d)^2-z^2}{(x^2+y^2+(z-d)^2)(x^2+y^2+z^2)}\\ &=\sum_{z \in \mathbb{Z}^3\setminus \{(0,0,0)\}} \frac{-2dz+d^2}{(x^2+y^2+(z-d)^2)(x^2+y^2+z^2)}\end{align}$$

Now we add together the contributions from the points $(x,y,z)$ and $(x,y,-z)$ so the sum is over the upper half space $$\begin{align}&=\sum \frac{-2dz+d^2}{(x^2+y^2+(z-d)^2)(x^2+y^2+z^2)}+\frac{2dz+d^2}{(x^2+y^2+(z+d)^2)(x^2+y^2+z^2)} \\ & =\sum \frac{-8d^2z^2+d^2(x^2+y^2)}{(x^2+y^2+(z-d)^2)(x^2+y^2+(z+d)^2)(x^2+y^2+z^2)}\end{align}$$

Far from the origin, where convergence is determined, this goes down at least as fast as $\frac{1}{r^4}$ and there are $r^2$ points in a spherical shell at a given radius, so we have a sum of $\frac{1}{r^2}$, which converges.

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It looks like it depends in which order you add the terms up... I agree that the sum would converge if the original post would state that he only sums over the upper half space and if he would combine the terms with $z$ and $-z$. As it stand the sum does not converge. –  Fabian Mar 1 '11 at 20:01
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@Fabian: I was thinking about adding up a sphere around the origin, then letting the radius go to infinity. I think you are right that the convergence is conditional, but this is a natural way to do it. –  Ross Millikan Mar 1 '11 at 21:16
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