**Nanfan Yi**

Our last discussion ended with Georg Cantor’s famous Diagonal Argument to show that the real numbers cannot be counted. Here, instead of analysing the number of points on a line (i.e. “how many” situations), we will look at how infinity is used to describe the magnitude of something as immeasurable. Again, we will start with the basics – dots and lines.

Imagine you’re walking 100m back home;

- You passed ½ of the distance – 50m (yay!)

- You passed ½ of the remaining distance – 25m (amazing!)

- You passed another ½ of the remaining distance – 12.5m (we are getting close!)

…and so on.

We can find the next point and then repeat the above procedure infinitely many times, because there are infinitely many points between any line segments (a nice result from part I)! And so it seems that you need to complete an *infinite *number of “tasks” before you arrive home; this might baffle you (and me) when you know that Usain Bolt could sprint through this distance in less than 10 seconds. Take a moment and think about it: do you go home, or do you go wild?

In fact, if you try adding up all the distance you’ve travelled in each task,

50+25+12.5+6.25+3.125…

This sum, which has *infinitely* many terms, is called an *infinite series*. As you may have guessed, its sum turns out to be exactly 100! And yes, you will eventually get home. In particular, this series is called *geometric*, because each term differs from the previous term by a factor of ½. Using this property, we could show, mathematically, that this sum is equal to 100.

Start by giving this sum a name, say X.

X = 50+25+12.5+6.25+3.125…

What about 1/2 X?

1/2 X = 25+12.5+6.25+3.125+…

Since adding a 0 to 1/2 X would not change its value, we have:

1/2 X = 0+25+12.5+6.25+3.125+…

Now X and 1/2 X are both infinite geometric series, and if we subtract 1/2X from X by subtracting each of the terms and grouping them together as follows:

X-1/2 X = (50-0) + (25-25) + (12.5-12.5) + (6.25-6.25) + (3.125-3.125) + …

we have

1/2X= 50 + 0 + 0 + 0 + 0 +… = 50,

as all the terms after the first one cancel out. So X = 100, by multiplying both sides by 2.

Note that 100 is a *finite *number, despite the fact we are doing *infinitely* many additions. Maybe it is not too surprising that the total is 100; after all, each term is decreasing in value. Okay, then how about this *infinite series*:

1 + 1/2 + 1/3 + 1/4 +…+ 1/n +…

Obviously, this sum also has numbers decreasing in value, as you may (or may not yet) have learned, 1/n will decrease to a value near 0 as n increases. What is your guess – is the sum going to be a finite number? Well, it actually blows up to infinity! Let me give you an informal proof if you are somewhat interested to see how this differs from the last one.

Consider if we group the terms like this:

1 > 1/2

1/2 + 1/3 > 1/4 + 1/4

1/4 + 1/5 + 1/6 + 1/7 + 1/8 > 1/8 + 1/8 + 1/8 + 1/8

1/8 + 1/9 + … +1/15 > 1/16 + 1/16 + … + 1/16

…

On the left-hand side of the inequality, the terms add up to our series. On the right-hand side, we have terms adding up to 1/2 for each inequality. We have infinitely many such inequalities and thus we need to add up infinitely many 1/2s. Hence, the sum involving 1/2s is going to infinity, yet our series is even larger! Things have gone wild!

Some infinite series might add up to a finite value, while the same thing might not be true for others. There are different methods to see whether an infinite series will sum to a finite value or not, some of which mathematicians identify as convergence tests. How to actually determine the value of the sum of the infinite series is another story, requiring more advanced mathematics. However, this following series is interesting in that we can use a “lazy” way to find its sum:

Hmm. First hint: what is 1/n – 1/(n+1)? Combining the fractions we see it is exactly 1/(n(n+1))! Let’s try to write down all the terms in this series using this “difference” expression:

What do you notice? Some of the terms cancel out!

Everything but 1/1 will cancel out since for every -1/(n+1), there is a +1/(n+1) from expressing 1/[(n+1)(n+2)] into 1/(n+1) – 1/(n+2). So, the total of the sum is 1!

The more mathematically-minded amongst you may notice that 1/n(n+1) has a similar form to 1/n^2. So, what will happen when we sum up 1/(1^2) + 1/(2^2) + 1/(3^2) + …+ 1/(n^2) + …? Is the sum finite or not?

A natural thing to do might be to compare this sum with the last one.

Adding up the left-hand side of the inequality, we get 1/(2^2) + 1/(3^2) + … + 1/(n^2), and for the right-hand side we have the same sum as before which we know sums to 1.

Adding 1 to both sides of the inequality:

And since 1 = 1/(1^2), we will get:

This above inequality immediately tells you the 1/(n^2) series sums to a finite value that is less than 2! In fact, we could go even further and show the sum is (pi^2)/6. For safety issues, I will not include the proof here, but recommend you take a look at this video.

If this is your first time seeing the sum of infinite series, you probably will be shocked – we’ve had an example of infinite non-negative terms adding up to a finite value, another one going wild; one involving infinitely many rational numbers adding up to an integer, and another one to an irrational number. Bizarre!

The series concludes with the final article here.

*******Not-A-Mind-Blowing Complement**********

Warning: all the proofs in this article __are not__ mathematically rigorous.

When we want to see whether an infinite series sums up to a finite value or not, we need to consider the sequence of its partial sums (given a series, the partial sum of the first n terms is the sum from the first term up to and including the nth term), and try to find out if the sequence has a limit or not. What we’ve done here does not involve much technicalities, whereas a formal mathematical proof will need us to build everything from scratch. I will leave the explanation and the rest of the job to real analysis, and recommend you to read introductory analysis books.