A classic puzzle* goes like this:

Imagine a one-metre length of rubber band, with an ant at one end. The ant starts walking towards the other end of the band, at a speed of 1 cm/s. At the end of each second the band is instantly and uniformly stretched by 1 m. Will the ant ever reach the other end of the band?

At first glance the answer seems obvious. The ant can’t possibly reach the end – the band is stretching by a metre for every centimetre the ant walks. Impossible! But think about it for a moment and you’ll see that, maybe, there’s a thin slice of possibility that the ant will make it to the other end.

It all rests on realising that the band is stretched evenly along its length, so the length of band *behind* the ant will increase, as well as the length in front. In fact, when the length of the band changes, it doesn’t change the proportion of the band that the ant has covered. Don’t believe me? Well, let’s work through some numbers.

After 1 second, the ant has travelled 1 cm. It has covered $\frac{1}{100}$ of the total length of the band. The band is then stretched by 1 m, so its total length is now 200 cm. Crucially, the stretch occurs evenly across the full length of the band. The 99 cm in front of the ant is doubled to 198 cm, but the 1 cm behind the ant is also doubled, to 2 cm. So the ant has covered $\frac{2}{200} = \frac{1}{100}$ of the total length of the band. The proportion covered hasn’t changed.

This works in general. After $n$ seconds, the length of the band will increase from $100n$ cm to $100(n+1)$ cm, an increase by a factor of $\frac{\left(n+1\right)}{n}$. The distance behind the ant will increase by the same factor, so the stretching of the band doesn’t affect the proportion the ant has covered. And because the ant continues to walk along the band, this proportion continues to increase. If it ever reaches 1 then the ant will have made it to the other end of the band!

So the crucial question becomes, is the proportion covered increasing in a way that means it will eventually get to 1?

One way to work this out is to express the proportion covered as a series.

Let $a_n$ be the distance the ant has covered after $n$ seconds. As we’ve already seen, this will be $\frac{n+1}{n}\left(a_{n-1}+1\right)$. We can use this to work out the first few terms in the series.

$a_0=0$

$a_1=\frac{2}{1}\left(a_0 + 1\right)= 2$

$a_2=\frac{3}{2}\left(a_1 + 1\right)=\frac{3}{2}\left(2+1\right)=3+\frac{3}{2}=3\left(1+\frac{1}{2}\right)$

$a_3=\frac{4}{3}\left(a_2 + 1\right)=\frac{4}{3}\left(3+\frac{3}{2}+1\right)=4+\frac{4}{2}+\frac{4}{3}=4\left(1+\frac{1}{2}+\frac{1}{3}\right)$

$a_4=\frac{5}{4}\left(a_3 + 1\right)=\frac{5}{4}\left(4+\frac{4}{2}+\frac{4}{3}+1\right)=5+\frac{5}{2}+\frac{5}{3}+\frac{5}{4}=5\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)$

We can begin to see a pattern here:

$a_n=\left(n+1\right)\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots +\frac{1}{n}\right)$

Now we can write the proportion of the band covered after $n$ seconds, call it $p_n$, by dividing the distance by the total length after $n$ seconds, which is $100\left(n+1\right)$:

$\begin{aligned} p_n&=\cfrac{\left(n+1\right)\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots +\frac{1}{n}\right)}{100\left(n+1\right)} \\ &=\cfrac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots +\frac{1}{n}}{100} \end{aligned}$

So if there is a value of $n$ for which $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots +\frac{1}{n}$ is equal to 100 then the proportion covered will be equal to 1 and the ant will reach the end of the rubber band.

Happily for the ant, the series $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots +\frac{1}{n}$ is a well-known series, called the harmonic series. It is known to diverge, which means that it gets bigger and bigger with no limit; whatever positive integer value you choose, there will be a value of $n$ for which $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots +\frac{1}{n}$ is larger than that value. If you’re interested, you can find a couple of different proofs of this on Wikipedia. For the ant, this means that there is a value of $n$ for which $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots +\frac{1}{n}>100$, which means that the proportion of the band the ant has covered will eventually be greater than 1 and the ant will escape. Hooray!

Unfortunately for the ant, the harmonic series grows extremely slowly. According to Wolfram MathWorld, the minimum value of $n$ required for the partial sum of the harmonic series to exceed 100 is 15,092,688,622,113,788,323,693,563,264,538,101,449,859,497. Since the value of $n$ corresponds to the number of seconds for which the ant has been walking, this is approximately $5 \times 10^{35}$ years. Sadly, I suspect that this is longer than the lifespan of an ant.

To see what the ant might think of this puzzle, check out The Ant’s Lament.

*The earliest reference I’ve found to this puzzle is by Denys Wilquin, published in December 1972 in Science et Vie. The version here is slightly adapted.

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