Prove the Monotone Convergence Theorem

Monotonicity Turns Bounds into Limits

The Squeeze Theorem proves convergence when a sequence is trapped between two sequences approaching the same limit. Another route is available when the terms move in only one direction. If a sequence is nondecreasing and bounded above, its terms may keep increasing, but they cannot pass their least upper bound. Completeness of the real numbers ensures that this bound exists, and the sequence must get arbitrarily close to it.

This result is called the Monotone Convergence Theorem. Its proof is a direct application of the supremum property of \(\mathbb R\), rather than a search for two explicit bounding sequences. We first prove the increasing case and then obtain the decreasing case by using the infimum.

“Nondecreasing” allows equal consecutive terms; strict increase is not required. If a sequence is nondecreasing, repeated use of its defining inequality gives \(a_n\leq a_m\) whenever \(n\leq m\). For a nonincreasing sequence, the corresponding inequality is \(a_m\leq a_n\).

The Increasing Case: Convergence to a Supremum

Proof. Set \(A=\{a_n:n\geq1\}\). This set is nonempty and bounded above, so its supremum \(s\) exists by completeness of the real numbers. We use the Epsilon Characterization of the Supremum: \(s\) is an upper bound for \(A\), and for every \(\varepsilon>0\), there is an element of \(A\) greater than \(s-\varepsilon\).

Let \(\varepsilon>0\). Since \(s=\sup A\), choose a positive integer \(N\) such that \(s-\varepsilon<a_N\). Because \(s\) is an upper bound, \(a_n\leq s\) for every \(n\). And because the sequence is nondecreasing, \(a_N\leq a_n\) whenever \(n\geq N\). Thus, for every \(n\geq N\),

It follows that \(|a_n-s|<\varepsilon\) for every \(n\geq N\). Since \(\varepsilon>0\) was arbitrary, the definition of convergence gives \(a_n\to s\). \(\square\)

The key point is that one term lies within \(\varepsilon\) of the supremum from below, and every later term lies at least as high as that term. No later term can exceed the supremum. Together these facts keep the whole tail inside an interval of width \(\varepsilon\) below \(s\).

The Decreasing Case: Convergence to an Infimum

Proof. The set \(A=\{a_n:n\geq1\}\) is nonempty and bounded below, so its infimum \(\ell\) exists. Let \(\varepsilon>0\). By the Epsilon Characterization of the Infimum, there is a positive integer \(N\) such that \(a_N<\ell+\varepsilon\). Since \(\ell\) is a lower bound, \(\ell\leq a_n\) for every \(n\). Since the sequence is nonincreasing, \(a_n\leq a_N\) whenever \(n\geq N\). Therefore, for every \(n\geq N\),

Consequently \(|a_n-\ell|<\varepsilon\) for every \(n\geq N\), which proves \(a_n\to\ell\). \(\square\)

Together, the two cases give the standard form of the Monotone Convergence Theorem: every bounded monotone real sequence converges. More precisely, a nondecreasing sequence converges to the supremum of its terms, while a nonincreasing sequence converges to their infimum. The proof also explains why the appropriate bound matters: an increasing sequence needs an upper bound, and a decreasing sequence needs a lower bound.

Worked Applications

Boundedness Is Necessary for a Monotone Sequence to Converge

The theorem supplies a sufficient condition for convergence. The next result gives a useful converse: a convergent real sequence must be bounded, whether or not it is monotone. In particular, for a monotone sequence, boundedness is exactly what distinguishes convergence from unbounded growth in its direction of motion.

Proof. Suppose \(a_n\to L\). By the definition of convergence, there is a positive integer \(N\) such that \(|a_n-L|<1\) whenever \(n\geq N\). The triangle inequality then gives \(|a_n|\leq |L|+1\) for every \(n\geq N\). The terms before \(N\) form a finite set, so their absolute values have a finite maximum if there are any such terms. Choose a real number \(B\) at least \(|L|+1\) and at least every \(|a_n|\) with \(n<N\). If \(N=1\), there are no terms before \(N\), and \(B=|L|+1\) suffices. In either case, \(|a_n|\leq B\) for all \(n\), so the sequence is bounded.

Now suppose \((a_n)\) is monotone and bounded. If it is nondecreasing, boundedness supplies an upper bound, so the increasing case of the Monotone Convergence Theorem proves convergence. If it is nonincreasing, boundedness supplies a lower bound, so the decreasing case proves convergence. Conversely, if a monotone sequence converges, the first part of this proof shows that it is bounded. This proves the claimed equivalence. \(\square\)

What the Theorem Does—and Does Not—Require

An increasing sequence does not need to contain its limit. In the first worked example every term is strictly less than \(4\), yet the supremum is \(4\) and is the limit. The theorem says the sequence approaches its supremum; it does not say that some term equals the supremum. Similarly, a decreasing sequence may approach an infimum that is never attained.

The one-sided bound in each case is essential. For instance, \(a_n=n\) is increasing but has no upper bound, and it does not converge to a real number. The sequence \(b_n=-n\) is decreasing but has no lower bound, and it does not converge to a real number. By contrast, a nondecreasing sequence automatically has a lower bound, namely \(a_1\); what the theorem needs in addition is an upper bound. A nonincreasing sequence automatically has an upper bound \(a_1\); it needs a lower bound.

A common proof error is to verify boundedness without verifying monotonicity, or to verify monotonicity without the appropriate bound. Neither fact alone guarantees convergence: a bounded sequence can oscillate, while an unbounded monotone sequence can grow without limit. Check both hypotheses, in the correct directions, before applying the theorem.