Tomaz Kristan said:
The right pointing component is always smaller. It's also always a finite sum of a finite number of finite right pointed forces.
The resulting force is always negative, for each ball.
Yes and the resulting sum of all such forces is undefined.
Consider the force on each ball simply from the one to the left of it.
The mass of the right ball is 2^{-N}. The mass of the ball to its left is 2^{-N-1}. The distance between the two balls is 10^{-N}-10^{-N-1}.
For the left pulling (negative) force on the Nth ball due to only the ball to its immediate left is
F_N = \frac{G 2^{-N} 2^{-N-1}}{\left(10^{-N}-10^{-N-1}\right)^2}= \frac{G 2^{-2N-1} }{10^{-2N}\left(1-10^{-1}\right)^2}
F_N = G\frac{5^{2N}}{2 \left(\frac{9}{10}\right)^2}=\frac{100 G}{162} 5^{2N}
So the sum of all these leftmost forces(a sum less than the total leftmost forces is:
S = \sum_{N=0}^{\infty} F_N = \sum_{N=0}^{\infty} \frac{100 G}{162} 5^{2N}
S = \frac{100 G}{162} \sum_{N=0}^{\infty} 5^{2N}
5^{2N} is obviously, no immediately obviously a divergent sum. In case you don't believe me, here's the first terms 1, 25 , 625 , 15625, 390625, 9765625, 244140625. This isn't going to diverge.
What more do you want? The force on the center of mass is not defined. It doesn't have a mathematically defined motion.
).