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?
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.