# Properties of monotone sequences

## 1. Introduction

We have already seen the definition of montonic sequences and the fact that in any Archimedean ordered field, every number has a monotonic nondecreasing sequence of rationals converging to it. Monotonic sequences are particularly straightforward to work with and are the key to stating and understanding the completeness axiom for the reals. This page sets out some simple properties of monotonic sequences that will be useful later on. We shall continue to work with the set of real numbers $ℝ$, but all the results here are valid in any ordered field; they don't even require the Archimedean property.

## 2. A convergent monotonic sequence is bounded by its limit

The theorem of this section is paraphrased by the title above. There are two versions of the result, one for nonincreasing sequences and one for nondecreasing sequences.

Theorem.

Suppose a sequence $( a n )$ is monotonic nondecreasing and converges to a limit $l ∈ ℝ$. Then $a n ⩽ l$ for all $n ∈ ℕ$.

Proof.

We suppose otherwise, that $l < a K$ for some $K ∈ ℕ$.

Subproof.

Let $ε = a K - l > 0$ and let $N ∈ ℕ$ such that

Let $k = max ( N , K )$ so $k ⩾ N , K$. Then $| a k - l | < ε = a K - l$. It follows that $l - ε < a k < l + ε = a K$. But this is impossible as $k ⩾ K$ so $a k ⩾ a K$ since $( A n )$ is nondecreasing.

A similar result for nonincreasing sequences is proved in exactly the same way.

Theorem.

Suppose a sequence $( a n )$ is monotonic nonincreasing and converges to a limit $l ∈ ℝ$. Then $a n ⩾ l$ for all $n ∈ ℕ$.

Proof.

Exercise.

## 3. Comparing two monotonic sequences

We now look at when two monotonic sequences converge to the same limit. The next theorem provides information on this.

Theorem.

Suppose sequences $( a n )$ and $( b n )$ are convergent monotonic nondecreasing sequences, and suppose the following holds

i.e., each value of $( a n )$ is bounded above by some value of $( b n )$. Then $lim a n ⩽ lim b n$.

Proof.

Let $l = lim a n$ and $m = lim b n$ and suppose (to get a contradiction) that $l > m$.

Let $ε = l - m > 0$ and let $N ∈ ℕ$ such that

Then $l + ε > a n > l - ε = m$.

This is impossible as, by assumption, there is $b k$ with $a N ⩽ b k$. This means that $m < a N ⩽ b k$ which contradicts the theorem above saying that the terms of the nondecreasing sequence $( b k )$ are bounded above by its limit $m$ .

Suppose sequences $( a n )$ and $( b n )$ are convergent monotonic nondecreasing sequences, and suppose that both

and

Then $lim a n = lim b n$.

Proof.

Both $lim a n ⩽ lim b n$ and $lim a n ⩾ lim b n$ by the previous theorem.

There is a similar result for nonincreasing sequences.

Theorem.

Suppose sequences $( a n )$ and $( b n )$ are convergent monotonic nonincreasing sequences, and suppose

Then $lim a n ⩾ lim b n$. If additionally we have

then $lim a n = lim b n$.

Proof.

Exercise.

## 4. A criterion for convergence

If $( a n )$ is a monotonic nondecreasing sequence and is convergent, then this sequence is bounded, since all convergent sequences are bounded. The converse of this statement is in fact true for the reals, but cannot be proved from the axioms of Archimedean ordered fields. This converse is called the Monotone Convergence Theorem and is discussed in a later web-page.

The point of this short section is that by a theorem above, a monotonic nondecreasing sequence is bounded above by its limit $l$. Because of this, it is often quite a lot easier to show that the sequence actually converges to $l$, as we see now. You may use the following result to prove that a monotone sequence converges to some particular limit $l$ ; if you don't know what the limit should be you need to invoke the Monotone Convergence Theorem, in which case all it will tell you is there is a limit, and not what the limit actually is.

Theorem.

Suppose a sequence $( a n )$ is monotonic nondecreasing (nonincreasing) and let $l ∈ ℝ$ with if it is nondecreasing ( if it is nonincreasing). Suppose also that

Then $a n → l$ as $n → ∞$.

The point is that the condition is not strong enough to imply convergence for ordinary sequences, but for monotonic sequences it is.

Proof.

We are given that $( a n )$ is monotonic. Suppose it is nondecreasing. (The nonincreasing case is proved in an identical way, you should check this as an exercise.) So

We are also given that

and

The proof follows the usual pattern.

Subproof.

Let $ε > 0$ be arbitrary.

Subproof.

Let $N ∈ ℕ$ such that $| a N - l | < ε$.

Subproof.

Let $n ⩾ N$ be arbitrary.

Then $l - ε < a N ⩽ a n ⩽ l$ so $| a n - l | < ε$.

So .

So .

So .

## 5. Conclusion

The results on this web page are not particularly difficult. They are even results that you might have guessed anyway. It does save a small about of time to remember them and use them when you need them, rather than repeat the proofs or the ideas in the proofs, but the choice is yours.