(The above photo is from Philippe Petit’s famous tightrope walk across the WTC)

This is a really nice question about limits from a uni assignment I was given. Say you are at the start of a stretchy tightrope, a kilometre long. You walk along it at 1 metre per second – but at the end of every second, the rope stretches another kilometre longer. Do you ever reach the end?

It’s easy to think not – after all, the rope stretches away a thousand times faster than your walking speed! How could you ever catch up? Well, to think that would be forgetting something – when the rope stretches, it doesn’t just increase the distance in front of you, it also stretches you away from the starting point, and the further you get from the start, the faster it stretches you away from the start. So how do you try and answer a problem like this?

How do you begin?

Figuring out a system for tracking the distance travelled across the rope might seem like a daunting task – and indeed, it is! In the first instance, it took me a while to figure it out fully. But then I realised something – instead of focusing on the distance travelled along the rope, it is much more helpful to look at what percentage of the rope you have crossed.

Say we define a function R[t], that tells you what portion of the rope you have covered at time t (in seconds). During the first second, you walk forward 1 metre, so you have covered a thousandth of the 1000 metre rope, so R[t] = 1/1000.

Key Idea! (Does R[t] get to 1)

When R[t] = 1, you have crossed the whole length of the rope. Why? well, R[t] is the ratio between the distance you have walked, and the length of the rope. If they are the same (so you are at the end), then R[t] is 1. Ultimately, this means we can rephrase the question “do you get to the end” as “does R[t] get to 1?”

What does stretching do to R[t]?

The answer is simple – nothing! Stretching the rope keeps you exactly the same portion of the way along the rope. How do you show this? Well – say you are currently N metres along the rope, which is 1000M metres long at the moment (because it always increases 1000 metres at a time, at any given moment it will be 1000 times some integer long). This means that R[t] = N/1000M.

The first thing to figure out – how much is the rope stretched? We know it’s stretched 1000 metres longer, so the length goes from 1000M to (1000M+1000); and therefore, the stretching ratio (the number you multiply it by to take it from the length before stretching to the length after) is (1000M+1000)/1000M, or (M+1)/M. So all parts of the rope are now (M+1)/M longer than beforehand; for convenience, we’ll let S=(M+1)/M. Now the N metres behind you is stretched by S to get N*S, and the total length of the rope is also stretched by S, to give 1000M*S.As such, R[t] is now S*N/S*1000M, which simplifies back down to N/1000M, which is exactly what it was before. This shows that stretching the rope doesn’t actually change how far you are along the rope, proportional to the length of the rope, as the rope stretches equally behind you, and in front.

How is this helpful?

Calm down, I was getting to that. Because we know that stretching doesn’t change R[t], we can totally ignore that part of the process in our calculation – all we have to do is think about how far you get each second walking along the rope. This is where it gets interesting – so you walk at 1 metre a second, so R[t] goes up by 1/1000 every second, right? No – because the rope increases in length, the portion of the rope you cover goes down every second, so you get  1/1000  during the 1st second,  1/2000 during the second (as the length is now 2000 m, after stretching 1000 m), y 1/3000 (as you are only covering 1 of 3000 metres), and so on. So R[t] is:

R[t] = 1/1000 + 1/2000 + 1/3000 +…+ 1/1000t1/1000 ( 1/1 + 1/2 + 1/3 +… + 1/t)

This series on the right ( 1/1 + 1/2 + 1/3 +… + 1/t) is very well known – it’s called the harmonic series (series are the sum of the terms of an infinite sequence). Here it helps to remember that the question “do you reach the end” is the same as “Does R[t], the ratio between distance walked and rope length, reach 1?” What we want to do is find out what happens to R[t] if t gets very large, and if it can ever get to 1.

A bit about series

Now the question is – what happens to  ( 1/1 + 1/2 + 1/3 +… + 1/t), the harmonic series, as t gets larger and larger? there are generally 2 kinds of series. Convergent ones have the property that adding the infinitely many terms just gets you closer to some specific number (and if you add all the infinite terms, it is exactly that number). For example, if you add 1,1/2,1/4,1/8, and so on, instead of heading off to infinity, it approaches 2. (You can try adding these together yourself to see this, it’s pretty neat!)

The other kind of series is a divergent series, which doesn’t approach a neatly defined value. A good example of this is just 1+2+3+4+5… which heads off to infinity. In general, a divergent series either heads off to infinity or negative infinity, or just oscillates – for example, 1-1+1-1+1-1+1-1… just dances around between 1 and 0, so it doesn’t have a nice limit.

Now, where does this relate to our stretchy rope problem? Well, if we can show that, for example, the harmonic series (remember, it’s  1/1 + 1/2 + 1/3 +… + 1/t), then we know that R[t] also heads off to infinity as t gets larger – which means that, at some point, R[t] must get larger than 1, and the entire rope will be crossed. The great thing is, if we find out the sum is infinite, it doesn’t matter exactly what it is, or how fast it goes to infinity – as long as it ends up infinite, we’re all good. The easiest way to show the that the harmonic series is infinite is with a really neat idea called the integral test.

The really neat idea called the integral test

Essentially, what we’re going to do here is compare the harmonic series to the function 1/n. If we draw them both on a graph (well, at least between x=0 and x=5), we get something like this (again, this is only a small portion):

Harmonic Series

The area of those rectangles adds up to the harmonic series, because they are all 1 wide, and all their heights are the values of the terms in the harmonic series. Because the rectangles will always touch the curve on the left hand side (because they have the same value whenever the x value of the function is an integer!) it is pretty clear that the area of the rectangles is always going to be larger than the area underneath the graph – this means that if we can show the area underneath the graph is infinite, the area of the rectangles is also infinite, because it is “larger” than the area under the graph. Or, in the language of maths (the most romantic of languages):

\sum_{n=1}^{k}\frac{1}{n} > \int_{1}^{k+1}\frac{1}{n} dn

What on earth is that?

Good question! There’s two key elements here – on the left hand side, that big \sum is a symbol for “sum” – this is the harmonic series part, and it essentially means that you want to add up the first k terms of the harmonic series. On the right is an integral, which is a really neat tool to find the area under a curve. \int is the integral symbol, while the little numbers at the bottom and top tell it you want to find the area between 1 and k+1, which is the area that that is covered up by the rectangles. This inequality says that the sum of the first k terms of the harmonic series is greater than the area under the curve 1/n between 1 and k+1,
which is what the graph shows us.

Evaluating the integral

Let’s figure this integral out:

\int_{1}^{k+1}\frac{1}{n} dn=[log(|n|)]_{1}^{k+1}=log(k+1)

If you don’t understand any of that, it doesn’t really matter – the important bit is that we can replace the integral with a log function, and update our inequality:

\sum_{n=1}^{k}\frac{1}{n} >log(k+1)

Logarithm functions head off to infinity as the stuff inside the brackets gets large. And the inequality shows us that the harmonic series is bigger than the corresponding logarithm – so this means that the harmonic series also heads off to infinity.

Back to the rope

What this tells us is that, as t gets large, R[t], which if you remember is a thousandth of the harmonic series, heads off towards infinity (because infinity divided by a thousand is still infinity.) This means that, at some point, R[t]=1, and the person has walked the entire length of the rope.

I’m sure you’d all love to know how long it takes to reach the end. The answer is that it takes a little while. Sorry, did I say a little while? I meant roughly 10434 seconds, or about 10416 times the age of the universe. (You can calculate this by solving R[t]=1 for t.)

A final word

Sequences and series are actually very important in mathematics – they are tied deeply into the fields of calculus and analysis, and can get you some very interesting results (google 1+2+3+4+5+…=-1/12, for example.) This problem is just a single interesting case, concerning the harmonic series, which is both one of the most well known series, and one of the less understood ones.