# Sum of infinite geometric progression (Paradox of Zeno)

In I.M Gelfand's Algebra book, there is a question which as follows: -

Achilles and a turtle have a race and turtle is given a head start. Achilles runs ten times faster than the turtle.

When the race starts Achilles comes to the place where the turtle initially was, but during that time even turtle moved $$\frac{1}{10}th$$ of the initial distance.

When Achilles covers that distance, then turtle moves $$\frac{1}{100}th$$ of the initial distance and so on.

So naturally we will get a Geometric Progression (G.P.) which as follows: -

$$1, \frac{1}{10}, \frac{1}{100}, \frac{1}{1000} ...$$

Where the common ration $$(q)$$ would then be equal to $$\frac{1}{10}$$

$$q = \frac{1}{10}$$

The sum of G.P. is given by the formula: -

$$1 + q + q^2 + q^3 + ..... + q^{(n-1)} = \frac{q^n - 1}{q - 1}$$

In our case since $$q = \frac{1}{10}$$ and $$n$$ will be infinitely bigger, we can write our summation formula as:-

$$\frac{1}{1 - q}$$

Where after subsitution of the value $$q = \frac{1}{10}$$, we shall get $$S = \frac{10}{9}$$

Till here I understood the logic of what Gelfand is trying to say,(that Achilles shall get ahead of the turtle after $$\frac{10}{9}$$ meters) but after that he continues the question and says: -

Imagine now that Achilles is running ten times more slowly than the turtle. When he comes to the place where the turtle initially was, it is at the distance ten times than the initial one, and so on.

So we shall get G.P. now as: -

$$1, 10, 100, 1000 ...$$

Where in this case the $$q = 10$$.

After that Gelfand asks us to put $$q = 10$$ in the formula: -

$$\frac{1}{1 - q}$$

After which we shall get an absurd answer of $$S = \frac{-1}{9}$$

My question is how can we put $$q = 10$$ in the above formula? Shouldn't we put it in the formula $$\frac{q^n - 1}{q - 1} ?$$

We transformed $$\frac{q^n - 1}{q - 1} to \frac{1}{1 - q}$$ only because of the fact that $$\displaystyle{\lim_{n \to \infty}} q^n = (\frac{1}{10})^n = 0$$.

But the second case is totally different. In second case $$q = 10$$.

Gelfand further goes on and says: -

Is it possible to give a reasonable interpretation of the (absurd) statement "Achilles will meet the turtle after running $$- \frac{1}{9}$$ meters"?

Hint. Yes, it is.

So can someone please explain how can we interpret the "absurd" statement? Is it because we used the wrong formula (my guess) or am I missing something here?

Edit: - I found the online solution for my question here (problem number 222). Howsoever I am not able to understand its connotation and denotation.

• As for interpreting the "absurd statement" of after running $-\frac{1}{9}$ meters they would have met... If we think about what happens if the "start of the race" was merely when we had started observing the race but that the race had already been going on for some time... we can conclude that there was some distance ago (specifically $\frac{1}{9}$ meters before the point where Achilles is at the time where we began observing him) where the turtle passes Achilles. Feb 4, 2020 at 14:02

It is often the case that if negative numbers appear in answer to the question "When does an event occur?", the correct interpretation is that the event occurred at a moment in the past. Such an interpretation makes sense in the context of this problem: a solution of $$-1/9$$ means that Achilles and the turtle were in the same place at some moment in the past. Specifically, they were in the same place when they were $$1/9$$ of a meter behind the starting line.
It might be more clear if we think along the following lines: suppose that we set up a camera which shows some patch of ground. Achilles and the turtle race past the camera, with the turtle running 10 times faster than Achilles. As they pass the camera, we take a picture which shows the turtle 1 meter ahead of Achilles. We can now reasonably ask "Where do Achilles and the turtle meet?" In this framework, where the race is thought of as ongoing, it makes sense to say that the turtle and Achilles already met, and that they met at a point which is $$1/9$$ of a meter behind Achilles' position when the photograph was taken.
Note that we can also tackle this algebraically. To fix some units, suppose that Achilles runs at a speed of $$1$$ meter per second, and that the turtle runs at a speed of $$10$$ meters per second. Then their locations at any given moment in time are given by \begin{align} A(t) &= t + x_0, &&\text{(Achilles' position at time t)} \\ T(t) &= 10t + (x_0+1), &&\text{(the turtle's position at time t)} \end{align} The constant $$x_0$$ represents Achilles' position at the moment we start paying attention. In the initial framework of the question, $$t=0$$ is the moment the race starts andd $$x_0$$ is the starting line; while in the framework of the racers passing a fixed camera, $$t=0$$ is the moment the photograph is taken and $$x_0$$ is Achilles' position at that moment. In either case, $$x_0$$ is just some place in space, and is largely irrelevant.
The question is "When is $$A(t) = T(t)$$?" This can be solved algebraically: $$A(t) = T(t) \implies t+x_0 = 10t + (x_0 + 1) \implies -9t = 1 \implies t = -1/9.$$ Hence the turtle and Achilles' are in the same location when $$t=-1/9$$; that is, they are in the same location $$1/9$$-th of a second in the past. This location is $$A(t) = x_0 - \frac{1}{9},$$ i.e. $$1/9$$-th of a meter behind Achilles' initial location.
• Wow so basically, when we put $q = 10$ what we actually did was that the speed of turtle suddenly became 10 times more than the speed of Achilles; and when they were 1 meter apart, (The time at which we started observing the race) the speed of Achilles then became 10 times the speed of the turtle (returning to the first G.P. mentioned by me on the above question). Feb 4, 2020 at 16:37