How to set up Zeno's paradox as a summation Let's suppose that a projectile travels 1 m/s and I want to find out how long it takes to travel 2m -- it should be two seconds, of course. But how could this be set up with a summation? For example, I would think that it would be something along the lines of:
$$
\sum_{t=0}^{?}
delta_t *rate = 2m
$$
Or, how could this be shown with a summation? I'm not very good at latex but perhaps something along the lines of (just halving the time each time until it gets to 2...but this of course assumes the answer is already 2...):
$$
\sum_{i=1}^{∞}
time * rate =
\sum_{i=1}^{∞}
1/2^{n-1} * rate = 2m
$$
I know this isn't a programming site, but here is the C code I used to emulate the last summation:
int n=1, max_n;
double rate=1.0, distance=2.0;

printf("Enter the number of iterations you want: \n");
if (scanf("%d", &max_n)  != 1) printf("Error! Items parsed != 1\n");

for (double t=0.0, distance=0.0, dt; n <=max_n ; n++) {
    dt =  1.0 / pow(2, n-1);
    t += dt;
    distance += dt * rate;
    printf("%2d. Time: %8.4f | Distance: %.4f \n", n, t, distance);
}


Enter the number of iterations you want: 12 
1. Time:   1.0000 | Distance: 1.0000 
2. Time:   1.5000 | Distance: 1.5000 
3. Time:   1.7500 | Distance: 1.7500 
... 
12. Time:   1.9995 | Distance: 1.9995

 A: Per OP's request:
First see my comment.  I derived the expression in my comment simply by noting that for $|x| < 1, ~\sum_{t=0}^{\infty} x^t = \frac{1}{1 - x}.$
Since I was looking for a value of $x$ so that $\frac{1}{1-x} = 2$, I set $x = \frac{1}{2},$ and then noted that $\left(\frac{1}{2}\right)^t = 2^{(-t)}.$
A: There are two initial conditions to Zeno's paradox:  The relative speed of the tortoise and the amount of head-start the tortoise is given.  Say the tortoise is given a 50cm head start and Achilles moves twice as fast.  Then Achille's movement becomes:
$$\frac{1}{2}+\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{2^2}...$$
So set your starting conditions Headstart=H and speed differential=$\frac{1}{s}$.
Your summation is then:
$$\sum_{n=0}^{\infty} H\frac{1}{s^n}$$
For your example, pick the initial distance of the projectile - say H=1, and the relative size you want of each term - say $\frac{1}{2}$.  That will give you 2m.
More generally, we know
$$\sum_{n=0}^{\infty} \frac{1}{x^n} = \frac {1}{1-\frac{1}{x}} = \frac{x}{x-1}$$  So if you want an answer of 2m, you need $H.\frac{s}{s-1}=2$ or $H=2\frac {s-1}{s}$.  Interestingly, note that this answer is completely independent of the speed of the projectile, since the important criterion is that speed relative to another object. Are you comparing an ICBM to a bullet or a bullet to an arrow?
