# Sequences that are eventually constant

## 1. Introduction

We have seen how to write down statements using quantifiers and some of the rules for manipulating quantifier in proofs. We have also see the definition of convergence of a sequence, and noted that it was rather complicated. I said that these complications were in fact necessary, and nothing simpler will do. I'm going to partially justify this claim now by looking at a modification of the definition of convergence that at first sight looks reasonable, but which turns out to be very wrong.

Recall that a sequence ( ) converges to 0 if

>0 ( < )

We are going to see the effect of changing round the first two quantifiers. Say that a sequence ( ) verconges to 0 if

>0 ( < )

At first sight, this looks a reasonable idea, and might also capture the right notion of convergence. So our question is, what does it mean to say a sequence verconges to 0, and is it the same thing as converging to 0? To answer this, we first look at some examples.

## 2. A sequence that has the property

Recall the integer-part function, . Now consider the sequence with terms =100/ .

Proposition.

( ) verconges to zero.

Proof.

We must prove that >0 ( < ) holds.

Remark.

This statement starts with a so we must start by giving a value of . For reasons that will become clear we shall take =101. Then the next two quantifiers are quantifiers, so we must take arbitrary values for and . After that, the proof should be easy enough.

Subproof.

Let be 101.

Subproof.

Let >0 in be arbitrary.

Subproof.

Let be arbitrary.

Subproof.

Assume .

Then =100/ =0 since =101 so >100.

So < as >0.

Hence ( <) .

Hence ( <) .

Hence >0 ( <) .

Hence >0 ( <) .

That's good. At least there is some sequence that we know about now that converges to 0 and also verconges to 0. But sadly, there are examples of sequences that converge to 0 but don't verconge to 0.

## 3. A sequence that doesn't have the property

Now consider the sequence ( ) with terms =1/ .

Proposition.

( ) does not verconge to zero.

Proof.

Remark.

The first step is to push the not inside the quantifiers in not ( ) verconges to 0. When we do this carefully we realise that we must prove that >0 ( ) holds.

As the first quantifier is a we must start by taking arbitrary. Then we have to give a positive value of and some . It makes sense to take = to make as large as possible and it seems therefore that =1/ would be a good choice for . (Be sure you realise why we are allowed to let depend on the value of . If it is not obvious to you right now, it should be clear when you have read the following proof.) Here is the proof.

Subproof.

Let be arbitrary.

Subproof.

Let =1/ and = so is a positive real number and .

So ( ) .

So ( ).

So >0 ( ) .

So >0 ( ) , as was arbitrary.

## 4. So what is this new notion?

Of course these two examples don't prove anything, but they do give us ideas.

One idea is that it might be the case that if a sequence ( ) verconges to 0 then it also converges to 0. This is in fact correct.

Exercise.

Prove that if a sequence verconges to 0 then it also converges to 0.

The other idea comes from the feature of our first example that it is eventually zero, i.e.,

( =0 ) .

Perhaps all such sequences that are eventually zero verconge to zero? and Perhaps all sequences verconging to zero are eventually zero? It may not be 100% obvious, but this is in fact the case.

Proposition.

A sequence ( ) verconges to 0 if and only if it is eventually zero.

Proof.

We must prove both directions. First, assume that is eventually zero. That is, assume

( =0 )

We must show that >0 ( < ) holds. We must start by giving an , and the only that seems sensible to choose is some that is given by eventually zero such that ( =0 ). After we realise this, the rest of the proof goes like the first example above.

Subproof.

Let be such that ( =0 ).

Subproof.

Let >0 in be arbitrary.

Subproof.

Let be arbitrary.

Subproof.

Assume .

Then =0 , so < as >0.

Hence ( <) .

Hence ( <) .

Hence >0 ( <) .

Hence >0 ( <) .

For the other direction, assume that ( ) verconges to zero. We must show that ( =0 ) . So we must start by giving an . The only sensible choice is such that >0 ( < ) . (This is just taken from the definition of verconge.) Once we realise this the proof is straightforward.

Subproof.

Let be such that >0 ( < ) .

Subproof.

Let be arbitary. We must show that =0 .

Subproof.

Suppose 0 .

Let = /2 , which is clearly positive, >0.

Then >0 and so by our asumption on we have < .

But this is absurd as it shows <= /2< .