[ELI5] how do we prove that something goes to infinity or – infinity since we can’t observe it?

Question

For instance how do we know that something weird doesn’t happen at some insanely large number?

In: 3

Answer 1

First of all, we need to have a good definition of “going to infinity”. It’s usually the following :

A sequence of numbers goes to infinity if for any threshold, I can find a rank (depending on that threshold) such that the sequence is larger than that threshold after that rank.

So usually, the recipe to prove that something goes to infinity is to give an explicit way to find a rank that satisfy the property when you are given an arbitrary threshold.

Let’s give two examples :

Say you have the sequence 1, 3/2 , 2, 5/2, 3, 7/2, … where you add 1/2 to the previous number at each step. It should be clear that this sequence is going to infinity, but let’s write the formal proof.

Fix an arbitrary threshold M (for convenience, let’s say that M is an integer). All you have to tell me is that there is a rank after which all the values of the sequence are larger than M. (note that it does not have to be the best rank, just one that works). For example, you can convince yourself pretty easily that after rank 2M, the values are all larger than M. That’s your proof. Nothing weird happens because you know the rules

Now, if you want a less trivial example, look at the sequence 1, 1+1/2 , 1+1/2+1/3 , 1+1/2+1/3+1/4 , … where you add 1/n to the previous number at each step where n is the rank in the sequence. It’s far less clear that this goes to infinity (and can be quite counter-intuitive as even after a billion terms, the sequence is still around 21 and is increasing veeeeery slowly).

But let’s write a proof anyway ! Fix a threshold M. Can you find a rank such that all values after that rank are bigger than M ? Not easy, but let’s convince ourselves that we can using the previous easy sequence and looking at the first terms :

The sequence starts like the previous one 1 , 3/2, so that’s a good start, but then it grows more slowly. But is there a term in our new sequence that is larger than 2 (which is the next term in our first easy sequence) ? Yes, because 1/3+1/4 > 1/4+1/4 = 1/2. So the 4th term is larger than 2.

Let’s get to the next level : is there a term in our new sequence that is larger than 5/2 ? Yes, because 1/5+1/6+1/7+1/8> 1/8+1/8+1/8+1/8 = 1/2. So the 8th term of the sequence is larger than 5/2.

Let’s get to the next level : is there a term in the sequence larger than 3 ? Yes because 1/9+…+1/16> 1/16+…+1/16=1/2. So the 16th term of the sequence is larger than 3.

So for any threshold M, you can convince yourself with that same trick that the value at rank 2*2^M of the sequence will be larger than M. Bingo, that’s a proof.