Monotone Convergence Theorem

In real analysis, the monotone convergence theorem states that if a sequence increases and is bounded above by a supremum, it will converge to the supremum; similarly, if a sequence decreases and is bounded below by an infimum, it will converge to the infimum. Let us learn about the monotone convergence theorem and its proof, as well as its two cases, in this post.

Introduction to Monotone Convergence Theorem

If a sequence of real numbers (a_n) is either increasing or decreasing, it is said to be monotone. In addition, if ∀n∈N, a_n≤a_n+1, a sequence (a_n) increases, and if ∀n∈N, a_n≥a_n+1, a sequence (a_n) decreases.

We’ll now look at a vital theorem that states that bounded monotone sequences will converge.

Theorem

If (a_n) is said to be a monotone sequence of real numbers, (a_n) is convergent if (a_n) is bounded.

Proof of Monotone Convergence Theorem

Let’s say that (a_n) is a monotone sequence.

Let’s pretend (a_n) is convergent. Then (a_n) is considered to be bounded using the Boundedness of Convergent Sequences Theorem.

There are two scenarios to think about. When (a_n) is an increasing sequence, the first case is true; when (a_n) is a decreasing sequence, the second case is true.

Case 1: a_n is an Increasing Sequence

Let’s assume that (a_n) is a bounded increasing sequence. We have that a_n≤M since (a_n) is bounded ∃M∈R such that ∀n∈N. Take a look at the set {a_n:n∈N}. This set is bounded because sequence (a_n) is bounded. This set has a supremum in R, called L=sup{a_n:n∈N}, according to the completeness property of real numbers.

Give the value ϵ>0. L−ϵ is not an upper bound to {a_n:n∈N} because it is the supremum of {a_n:n∈N}, therefore∃aN from this sequence such that L−ϵ<aN. As (a_n) is an increasing sequence, we have a_N≤a_n for all n≥N, as follows:

L−ϵ<a_N≤a_n≤L<L+ϵ …(1)

We can see that if we leave out the unnecessary parts of the inequality, we get L−ϵ<a_n<L+ϵ for n≥N, and hence ∣a_n−L∣<ϵ. Because ϵ>0 is arbitrary, we can deduce that lim_n→∞a_n=L, implying that (a_n) is convergent to L.

Case 2: a_n is a Decreasing Sequence

Let’s suppose that (a_n) is a bounded decreasing sequence. We have that m≤a_n, because (a_n) is bounded ∃m∈R such that ∀n∈N. Take a closer look at the set {a_n:n∈N}. This collection is bounded because sequence (a_n) is bounded. In R, this set should have an infimum, which we’ll name L=inf{a_n:n∈N}.

Give the value ϵ>0. L+ϵ is not a lower bound to {a_n:n∈N} because L is the infimum of {a_n:n∈N}, therefore ∃a_N from this sequence such that a_N<L+ϵ. Because (a_n) is a decreasing sequence, we have a_N≥a_n, for all n≥N, as follows:

L−ϵ<L≤a_n≤a_N<L+ϵ … (2)

We can see that if we remove the unnecessary portions of the inequality, we get L−ϵ<a_n<L+ϵ and hence ∣a_n−L∣<ϵ for n≥N. Because ϵ>0 is arbitrary, we can deduce that lim_n→∞a_n=L, implying that (a_n) is convergent to L.

Here, (a_n) was convergent in both circumstances.

It’s worth noting that the Monotone Convergence Theorem only holds if the sequence (a_n) is bounded and eventually monotone (i.e., increasing or decreasing). As a result, (a_n) is convergent if the sequence (a_n) is bounded and ∃N∈N is such that ∀n≥N is either a_n≤a_n+1 or a_n≥a_n+1.

Corollary

If (a_n) is a growing and bounded sequence, then we have lim_n→∞a_n=sup{a_n:n∈N}; otherwise, lim_n→∞a_n=inf{a_n:n∈N}.

In both cases 1 and 2, there is no proof to the corollary when L is either the supremum (for increasing sequences) or the infimum (for decreasing sequences), as we already demonstrated in the monotone convergence theorem. We can see that a sequence that converges has a unique limit, which we have already discovered, according to the uniqueness of limits of a sequence theorem.

Solved Example

Example:

For any n ∈ N, let x_n = (-1)ⁿ. Prove that the sequence (x_n) is non-convergent.

Solution:

Assume that x_n → x₀ for certain x₀.

Then, according to the definition, there exists N such that |x_n − x₀| <¼ for all n ≥ N, and for ϵ=¼. Then |x_n − x_m| ≤ |x_n − x₀| + |x_m − x₀| ≤ (¼)+(¼) = 1/2 , for every m, n ≥ N, and it is not true because |x_n − x_n+1| = 2, for almost any n.