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Ok so as far as I know the infinite sum for a geometric progression with common ratio $r$ can be derived as follows:

$S_{n} = a(1 + r + r^{2} + ...+r^{n-1}) $

$rS_{n} = a(r + r^{2} +...+r^{n})$

$S_{n}(r-1) = a(r^{n}-1)$

$S_{n} = \frac{a(-1+r^{n})}{r-1} =$ $\frac{a(1-r^{n})}{1-r}$

Using this we can say that for $|r|<1$, $S_{\infty} = \frac{a}{1-r}$

In the case of $r = \frac{1}{2}, a = 1$, this gives $S_{\infty} = 2$

But now consider the following argument:

A runner has $2$metres left till he reaches the finishing line. In order to reach it, he must first cross halfway i.e. $1m$. Then he must cross half of that i.e. $\frac{1}{2}$metres, taking his distance travelled to $1 + \frac{1}{2}$ metres etc... Given that he eventually does cross the finishing line can we conclude that the $S_{\infty} =2$ from this argument alone? Or perhaps that $S_{\infty} \geq 2$. Or is my assumption that he does cross the finishing line not allowed?

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    This is obviously related to [Zeno's Achilles and the tortoise paradox](https://en.wikipedia.org/wiki/Zeno's_paradoxes#Dichotomy_paradox) for a stationary tortoise, but with this twist: `Given that he eventually does cross the finishing line`. Could you elaborate on what precisely you mean by that i.e. which is the assumption you make there. Is the runner's speed constant (as Achille's was), or is the series $\sum t_n$ assumed to converge where $t_n$ is the time taken for the $n^{th}$ step, or ...? – 2017-03-05
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    @dxiv I'm not sure if I fully understand the intricacies of each of those assumptions - what I meant by "given he eventually does cross the finishing line" is that the distance he must have covered must be greater than or equal to his initial distance from the finishing line. That is to say, he was originally 2 metres from the finishing line, and hence his distance covered must be greater than or equal to 2. I don't really understand why the time taken or speed is necessarily relevant to this. Perhaps I'm missing something. – 2017-03-05
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    Compare your wording to Zeno's from the previous link. You *state* that `he *eventually* does cross the finishing line`, and that hides *some* assumption about time and speed. For an extreme example, if the runner takes constant time for each of the smaller steps of length $1/2^n$ then he'll never actually cross the finish line (in finite time). What I asked for was clarification on what precisely your assumption is, since the answer *may* depend on that. – 2017-03-05
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    Well from what I understand, the assumption would be that he has a constant speed – 2017-03-05

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A tiring runner that, at $2$ metres from the finish line, takes $1$ step of $1$ metre, then $1$ step of $1/2$ a metre, then $1$ step of $1/4$ of a metre, and so on (each step covering half the ground as the previous step), will never cross, or even reach, the endline. On the other, no matter how close a point to the endline you set, he will eventually cross that point.

The easiest way to see it, is to look at the distance the runner needs to cover to reach the endline, and verify that each step covers exactly half the distance left, leaving the runner at a new distance equal to the step just taken. At $2$ metres, a $1$ metre step leaves the runner at $1$ metre from the endline. At $1$ metre, a $1/2$ metre step leaves the runner at $1/2$ a metre from the endline. And so on. Thus, no step will ever cover the entire distance left, but for any $\epsilon>0$ steps eventually will grow smaller than $\epsilon$ and so will the runner's distance from the endline.

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    I got slightly confused reading your argument. So let me ask this: a runner which has already crossed the finishing line, will have had to have come from 2 metres before the finishing line, and then crossed it. So given the fact that this event occurred, we surely know that the runner does cross the finishing line, and so the sum of the gp must be 2 or more? – 2017-02-27
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    @mrnovice It's my turn to be confused. If his steps are in a geometric progression, and their sum *exceeds* the $2$ metres separating him from the endline, then obviously their sum exceeds $2$ metres... In fact, you can prove that in such a case a *finite* number of steps will carry the runner beyond the endline. – 2017-02-27
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    Well I don't know that it must necessarily be greater, since any distance covered beyond the finish line would not be in the same GP. I think it only means that the limit must be at least 2(assuming all my logic thus far is correct) – 2017-02-27