I am trying to make understand gravity on a flat 2-D torus. To help myself get my head around it, I simplify the problem into a 1-D torus. Let's have a 1-D space (or line). Instead of the left part of the line going infinitely to the left and the right part of the line going infinitely to the right, I am going to make them meet, like a torus. So we have circle instead, which radius is R. Limited space, but unbounded. Let's have a point mass O on the line. The acceleration any particle P will experience towards O (classical mechanic) is a=Gmo/r2 where G is the gravitational constant, m is the mass of O and r is the distance between O and P. (I know there are a lot of issue with using inverse square law in 1-D, but lets just ignore it for the moment). Since this is just for conceptual purposes, I would just assume that G and mo are equal to 1. The formula for acceleration will be reduced to: a=1/r2 However, in a non toroidal space, the gravitational field would touches particle P once and just extend infinitely, never touching P again. Yet, if we say that the particle P is located r distance to the left of O, it will not only experience a=1/r2 going to the right, but also a=1/(R-r)2 going to the left. The resulting acceleration would be a=1/r2 - 1/(R-r)2. And yet, it doesn't stop there, the gravitational field that to the left, pulling accelerating everything to the right, will meet point P at r, and then circle the whole space and meet point P again at r+R and circle the space again and meet point P again at r+2R and so on. The same also apply with the gravitational field going to the right. The total acceleration would be: sum_(a=1)^infinity(1/(a b+c)^2-1/(a b-c)^2) Now, Wolfram alpha cannot simplify that sum for me, neither can I myself. http://www.wolframalpha.com/input/?i=sum+1/(ab+c)^2-1/(ab-c)^2,a=1+to+a=infinity My questions are: 1. Did I made any mistake in my reasoning? 2. Can the sum be simplified? 3. Would the 2-D solution of the same problem be "nice"