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 $( a n )$ converges to $0$ if

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

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, $[ x ]$. Now consider the sequence with terms $a n = [ 100 / n ]$.

Proposition.

$( a n )$ verconges to zero.

Proof.

We must prove that holds.

Remark.

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

Subproof.

Let $N ∈ ℕ$ be $101$.

Subproof.

Let $ε > 0$ in $ℝ$ be arbitrary.

Subproof.

Let $n ∈ ℕ$ be arbitrary.

Subproof.

Assume $n ⩾ N$.

Then $a n = [ 100 / n ] = 0$ since $N = 101$ so $n > 100$.

So $| a n | < ε$ as $ε > 0$.

Hence $( n ⩾ N ⇒ | a n | < ε )$.

Hence .

Hence .

Hence .

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 $( b n )$ with terms $b n = 1 / n$.

Proposition.

$( b n )$ does not verconge to zero.

Proof.

Remark.

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

As the first quantifier is a $∀$ we must start by taking $N$ arbitrary. Then we have to give a positive value of $ε$ and some $n ⩾ N$. It makes sense to take $n = N$ to make $| b n |$ as large as possible and it seems therefore that $ε = 1 / N$ would be a good choice for $ε$. (Be sure you realise why we are allowed to let $ε$ depend on the value of $N$. 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 $N ∈ ℕ$ be arbitrary.

Subproof.

Let $ε = 1 / N$ and $n = N$ so $ε$ is a positive real number and $n ⩾ N$.

So $( n ⩾ N ∧ | b n | ⩾ ε )$.

So .

So .

So , as $N ∈ ℕ$ 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 $( a n )$ verconges to $0$ then it also converges to $0$. This is in fact correct.

Exercise.

Prove that if a sequence $a n$ 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.,

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 $( a n )$ verconges to $0$ if and only if it is eventually zero.

Proof.

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

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

Subproof.

Let $N ∈ ℕ$ be such that .

Subproof.

Let $ε > 0$ in $ℝ$ be arbitrary.

Subproof.

Let $n ∈ ℕ$ be arbitrary.

Subproof.

Assume $n ⩾ N$.

Then $a n = 0$, so $| a n | < ε$ as $ε > 0$.

Hence $( n ⩾ N ⇒ | a n | < ε )$.

Hence .

Hence .

Hence .

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

Subproof.

Let $N ∈ ℕ$ be such that .

Subproof.

Let $n ⩾ N$ be arbitary. We must show that $a n = 0$.

Subproof.

Suppose $a n ≠ 0$.

Let $ε = | a n | / 2$, which is clearly positive, $ε > 0$.

Then $ε > 0$ and $n ⩾ N$ so by our asumption on $N$ we have $| a n | < ε$.

But this is absurd as it shows $| a n | < ε = | a n | / 2 < | a n |$.

So $a n = 0$ since assuming $a n ≠ 0$ lead to a contradiction.

So as $n ⩾ N$ was arbitary.

So , as required.

Exercise.

Define what it means for a sequence $( a n )$ to be is eventually constant with value $l$ . Prove that $( a n )$ to be is eventually constant with value $l$ if and only if it verconges to $l$ , i.e.,