In a previous post, I tried to build a model of a ball bouncing off the floor. I found that, mathematically speaking, there was no contradiction in having the ball bounce an infinite number of times in a finite amount of time and I teased the idea that this fact was related to Zeno's paradox. Since the writing of that post, I have realised that there is much more to Zeno's paradox — philosophically speaking — than I had previously thought and that it is not simply an old-fashioned problem that once puzzled the ancients prior to eventually being solved with the tools of modern mathematics.

Zeno's paradox raises a number of profound questions regarding the existence and the possibility of motion, the principle of non-contradiction, etc. All of which are all extremely interesting and beyond my expertise. However I would encourage anyone interested to have a look at this book by the famous logician Graham Priest.

In the present post, I will merely use one of the common statements of the paradox to make a few points about the mathematical concepts of limit and series.

Suppose that an archer shoots its arrow at a target. In order for the arrow to reach its destination, it must first travel half the distance to the target. Once it has done so, it must proceed to travel half of the remaining distance. And then half of the distance that remains after that, and so on, and so forth. It would therefore seem like the arrow, having to travel and infinite number of places, should never reach its target. This is contrary to common sense, hence the paradox.

Modern mathematics offers a solution to this puzzle via the concept of limits.

Although the arrow does number of places, which we will call intervals, along its trajectory, it also travels smaller intervals quicker. Supposing that the arrow has a constant velocity, which is very close to being the case for short distances, then by definition the arrow takes twice as much time to go through a distance twice as long. From this reasoning, we understand how the arrow goes through intervals with an infinitesimal length in an infinitesimal amount of time.

We can write all of this using the language of series.

Working in units in which the numerical value of the distance between the target and the starting position is equal to one the sequence of intervals described above is

$$
\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},...
$$

each following term being half of its predecessor. The whole sequence can be summarized in the following form.

$$
\left(\frac{1}{2^n}\right)_{n\in\mathbb{N}_0}~.\label{eq:halves}
$$

In order to recover the complete distance, one sums over all the terms in the sequence. This series is a simple example of a geometrical progression which are series defined as the limit of sums of the following form

$$
\sum_{n}^N\alpha^n=\frac{\alpha^{n_0}-\alpha^{N+1}}{1-\alpha}~,
$$

where $n_0$ denotes the first rank in the sum. When $-1<\alpha<1$, the limit converges when $N\rightarrow\infty$ and we have

$$
\begin{equation}
\sum_{n=0}^{\infty}\alpha^n=\frac{1}{(1-\alpha)}~.\label{eq:seriesalphan}
\end{equation}
$$

We will use this result again at the end of this note. From the above we can see that the series of terms in the sequence Eq. \eqref{eq:halves} converges to

$$
\sum_{n=1}^{\infty}\frac{1}{2^n}=1~.
$$

All is well, as we have just proved that the sequence covers the whole distance between the archer and the target. This result, however, should not depend on the particular splitting of the original interval. For example, if we wanted to divide the same interval in successive "thirds of the distance", we would expect to arrive to the same conclusion.

One evening, we were thinking about this problem with friends and we realized that it required to be a bit more cautious than we had originally thought.

Take the example of "dividing in thirds". It is really tempting to guess that the sequence of intervals to consider should be successive power of $1/3$. Summing over the terms of such a sequence, however, gives

$$
\sum_{n=1}^{\infty}\frac{1}{3^n}=\frac{1}{2}~.
$$

Something is wrong. The error comes from the fact that above actually represents the fact of travelling successive powers of a third of the original distance. Whereas, what we really want to have is an arrow that travels a third of the distance left to be travelled at each step.

With that in mind, we can get an idea of the correct series by computing its first few terms explicitly

$$
\begin{align}
1^\text{st} \text{term}:&&&\frac{1}{3},\\
2^\text{nd} \text{term}:&&&\frac{1}{3}(1-\frac{1}{3})=\frac{2}{9},\\
3^\text{rd} \text{term}:&&&\frac{1}{3}(1-\frac{1}{3}-\frac{2}{9})=\frac{4}{27},\\
4^\text{th} \text{term}:&&&\frac{1}{3}(1-\frac{1}{3}-\frac{2}{9}-\frac{4}{27})=\frac{8}{81}~.\\
&&&\vdots\nonumber
\end{align}
$$

There is a pattern emerging here. In fact, the terms describe the following sequence

$$
\left(\frac{2^{n-1}}{3^n}\right)_{n\in\mathbb{N}_0}
$$

In order to convince ourselves that this is no accident, let us go through the same process with quarters. We get,

$$
\begin{align}
1^\text{st} \text{term}:&&&\frac{1}{4},\\
2^\text{nd} \text{term}:&&&\frac{1}{4}(1-\frac{1}{4})=\frac{3}{16},\\
3^\text{rd} \text{term}:&&&\frac{1}{4}(1-\frac{1}{4}-\frac{3}{16})=\frac{9}{64},\\
4^\text{th} \text{term}:&&&\frac{1}{4}(1-\frac{1}{4}-\frac{3}{16}-\frac{9}{64})=\frac{27}{256}.\\
&&&\vdots\nonumber
\end{align}
$$

The pattern is repeating with the sequence now being

$$
\left(\frac{3^{n-1}}{4^n}\right)_{n\in\mathbb{N}_0}~.
$$

This kind of pattern is no accident. In fact, it can be written as

$$
\left((1-\alpha)^{n-1}\alpha\right)_{n\in\mathbb{N}_0}~,
$$

or equivalently, since $0<\alpha<1$ ,

$$
\left((1-\alpha) \alpha^{n-1}\right)_{n\in\mathbb{N}_0}~.
$$

Summing on all the terms in the sequence, we can use Eq. (2) to obtain

$$
\sum_{n=0}^{\infty}(1-\alpha)\alpha^{n}=1~.
$$

This identity is valid for all $0<\alpha<1$ . Personally, I find it very elegant that the formula giving the limit of a geometric progression is related so closely to one of the most ancient problems in mathematics.