Robot Adding Paper to An Infinite Stack A friend posed a question to me that she heard in a math logic class. If a robot is in a room, lays down two papers, then removes two, then adds four, then removes four, then adds eight, then removes eight, then adds sixteen and so on, for two minutes, every half of the remaining time interval, will the stack be infinitely tall or empty at the end of two minutes? 
This seems to be the same as Thompson's Lamp, where if you claim it is infinite, the robot could have removed all the papers in the next half interval, and the same goes if it 'ends' with an empty pile.
What's the solution?
 A: The solution, much like that of Thompson's Lamp, is that the mathematical situation described here does not converge and does not correspond to any experiment that is physically executable. It is a thought experiment with a weird result, but it's a thought experiment that stretches what we are willing to accept as physically possible to the point that everything just collapses as a heap of nonsense. 
A: Let's write this out as a sequence. Let $S_n$ be the number of papers in the stack after n halving intervals. Then $S_0 = 2$, $S_1 = 0$, $S_2 = 4$, etc. We quickly see the sequence is formally defined as $S_n = 2^{n+1}$ for odd n, $0$ for even n. As $n \to \infty$, $S_n \to$... what? $0$? The next odd term is gonna be a positive integer. $\infty$? The term right after is zero. The sequence does not converge.
Okay, but we know that sometimes functions don't have limits but do have values. So let's look at this at a different way and work backwards. Maybe there ends up being $0$ papers on the stack after two minutes. This would mean that at the previous time, $2-\epsilon$, there are what we formally call "many" papers on the stack. But between $2$ and $2-\epsilon$ there is the halfway point $2-\epsilon/2$, and the stack would have zero papers then. Which means that "2 minutes" happens after a zero paper time, so it would have the ultra-rigorous number "2 many" papers on the stack. Again, we have a contradiction.
What I'm trying to say here is that the behavior at the two minute mark is ill-defined, and the answer is clearly $0/0$.
