Whenever people agree with me I always feel I must be wrong. (Hans-Peter Geerdes)

This is the second post of my series on infinity, be sure to read the first one.

We start this chapter with a nice story, "Achilles and the Tortoise", one of Zeno's Paradoxes, though we slightly adapt it. So, Achilles races against a tortoise. He gives the tortoise a 1 mile head start. Assuming Achilles is four times as fast as the tortoise, when he arrives 1 mile, the tortoise is at 1.25 miles. When he arrives 1.25, the tortoise is at 1.3125 miles. When he arrives at 1.3125, the tortoise is at 1.328125. And so on. Whenever Achilles reaches the point the tortoise was when he started, the tortoise already went further. This way, Achilles would have to go through infinitely many points, before reaching the tortoise.

Of course, this is not really a paradoxon. The problem lies in the fact that we have an infinitary process in here. As mathematicians do, we write the problem down more formally. Let a_i be the i-th point where Achilles stands, and t_i the point where the tortoise stands. We have a_0 = 0 and t_0=1, respectively. By definition, we have a_{n+1} = t_n, because Achilles always reaches the former point of the turtle. As the turtle has a 1 mile head start, and has \frac{1}{4} times of Achilles' speed, we set t_{n+1}=1+\frac{a_{n+1}}{4}, which is t_{n+1}=1+\frac{t_{n}}{4}. It is easy to see, that t_n = 1+\frac{1}{4}+\frac{1}{16}+\ldots+\frac{1}{4^n}, that is, t_n=\sum\limits_{i=0}^n \frac{1}{4^i}, which is a shorter mathematical notation for such sums.

Using the axioms we saw in Part 1, we can prove this formally (if you are already a little confused, just skip this proof, it is not necessary for the further understanding): Assume there is an n such that t_n\neq\sum\limits_{i=0}^n\frac{1}{4^n}, then there is a smallest such n. Clearly, t_0=1=\frac{1}{4^0}=\sum\limits_{i=0}^0\frac{1}{4^i}, so n\neq 0. Therefore, it has a predecessor n-1, for which we must have t_{n-1}=\sum\limits_{i=0}^{n-1}\frac{1}{4^i}, since n was minimal. But then, t_{n} = 1+\frac{t_{n-1}}{4} = 1+\frac{1}{4}\cdot\sum\limits_{i=0}^{n-1}\frac{1}{4^i}=1+\sum\limits_{i=1}^{n}\frac{1}{4^i}=\sum\limits_{i=0}^{n}\frac{1}{4^i}. Contradiction.

Obviously, Achilles will pass the tortoise at some point, and obviously, this point is greater than all t_n. We want to find out more about this point where he passes the tortoise.



The point we are looking for, is \frac{4}{3}. In fact, this is provable. But it is a bit harder than the above induction.  What we have here, is a so-called geometric series. As we have t_{n}=1+\frac{1}{4}+\frac{1}{16}+\ldots+\frac{1}{4^n}, we have \frac{1}{4}t_{n}=\frac{1}{4}+\frac{1}{16}+\ldots+\frac{1}{4^{n+1}}, that is, \frac{1}{4}t_{n} and t_{n} differ in the terms 1 and \frac{1}{4^{n+1}}, and thus we have

t_{n}-\frac{1}{4}t_{n}=1-\frac{1}{4^{n+1}}
t_{n}(1-\frac{1}{4})=1-\frac{1}{4^{n+1}}
(\frac{3}{4})t_n=1-\frac{1}{4^{n+1}}
t_n = \frac{1-\frac{1}{4^{n+1}}}{(\frac{3}{4})}
t_n = \frac{4}{3}(1-\frac{1}{4^{n+1}})

From this, we see that, the larger n gets, the smaller \frac{1}{4^{n+1}} gets, and therefore, for very large n, t_n = \frac{4}{3}(1-\frac{1}{4^{n+1}}) approaches \frac{4}{3}(1-0)=\frac{4}{3}. In fact, if we did a little more preliminary work, this would be a valid mathematical proof, and mathematicians would write \lim\limits_{n\rightarrow\infty} t_n = \frac{4}{3}, where \infty is the symbol for "infinity", and call \frac{4}{3} the limit of this series. As we wrote \sum\limits_{i=0}^{n}\frac{1}{4^i} for the finite sum, we can also write \sum\limits_{i=0}^{\infty}\frac{1}{4^i} for its limit. The following graphic gives a certain geometric intuition of this fact:


(source)

Of course, it is not always that easy, not every infinite process has a finite outcome. For example, obviously, the infinite sum \sum\limits_{i=1}^\infty i=1+2+3+\ldots does not. However, we know that its finite parts get arbitrarily large, so we might say that \infty=\sum\limits_{i=1}^\infty i. This series diverges, while the above geometric series converges.

But even worse, consider the series \sum\limits_{i=0}^\infty (-1)^i=1-1+1-1+1-1+\ldots. Its finite parts are either 1 or 0. But you cannot find any "tendency" on what happens during infinty. This series neither converges nor diverges. \sum\limits_{i=0}^\infty (-1)^i is not well-defined.

The above example with the tortoise was one special geometric series, the general (infinite) geometric series is \sum\limits_{i=0}^\infty a^n, which converges for 0\le a<1, and then has the limit  \sum\limits_{i=0}^\infty a^n=\frac{1}{1-a} - in the case, we had a=\frac{1}{4}, for a general proof, consider the Wikipedia-article, or any good introduction to calculus.

Even more general, we have \sum\limits_{i=0}^\infty ba^n = \frac{b}{1-a}. With this formula, we can prove a fact quite a lot of people are not willing to understand: 0.\overline{9}=1, where 0.\overline{9} means the zero with infinitely many nine-digits after the decimal point. But in fact, we have 0.\overline{9}=\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\ldots=\sum\limits_{i=0}^\infty \frac{9}{10^{n+1}}, which is \sum\limits_{i=0}^\infty \frac{9}{10}\cdot\frac{1}{10^{n}}, which is, according to our formula, \frac{9}{10}\cdot\frac{1}{1-\frac{1}{10}}=\frac{9}{10}\cdot\frac{10}{9}=1.

There is a general law about recurring decimal numbers you may know from school, namely, 0.\overline{a_1a_2\ldots a_n} = \frac{a_1a_2\ldots a_n}{99\ldots 9}. For example, from this follows the above, 0.\overline{9}=\frac{9}{9}=1. More precisely written, we have \sum\limits_{i=1}^\infty \frac{x}{10^{k\cdot i}} = \frac{x}{10^k-1}, which follows by our above formula by \sum\limits_{i=1}^\infty \frac{x}{10^{k\cdot i}}=\frac{x}{10^k}\sum\limits_{i=0}^\infty \frac{1}{10^{k\cdot i}}=\frac{x}{10^k}\cdot\frac{1}{1-\frac{1}{10^k}}=\frac{x}{10^k}\cdot\frac{1}{\frac{10^k-1}{10^k}}=\frac{x}{10^k-1}.

As you see, studying this kind of objects can get very complicated, but produces a lot of interesting outcomes. If you did not quite get the last part, do not worry, it is rather sophisticated for a non-mathematician.