The Monotone Convergence Theorem

Why Monotonicity and Boundedness Work Together

The previous tutorial used comparisons between sequences to transfer information about their size. Monotonicity provides another kind of structure: once a sequence is nondecreasing, later terms never fall below earlier ones; once it is nonincreasing, later terms never rise above earlier ones. But monotonicity alone does not guarantee a finite limit. The sequence \(a_n=n+1\) is nondecreasing and grows without bound. A bound in the direction of motion is what prevents this kind of escape.

The Monotone Convergence Theorem combines these two conditions. It is one of the central ways to prove that a sequence converges when no candidate limit is immediately apparent. Its proof uses the completeness of the real numbers: an appropriate set of sequence values has a least upper bound or a greatest lower bound. The next tutorial proves the theorem; here we state it, examine how its hypotheses work, and practice using its conclusion.

Theorem (Monotone Convergence Theorem): Every nondecreasing real sequence that is bounded above converges to a finite real number. Every nonincreasing real sequence that is bounded below converges to a finite real number.

For a nondecreasing sequence, a lower bound is already available: \(a_0\leq a_n\) for every \(n\), by the result from “Monotone Sequences.” Thus the additional condition is an upper bound. Similarly, a nonincreasing sequence already has an upper bound, and the condition that matters is a lower bound. In either case, the sequence is bounded, but the one-sided bound in the direction of motion is the essential hypothesis.

The conclusion is convergence to a real number, not convergence to positive or negative infinity. A nondecreasing sequence with no upper bound may instead tend to \(+\infty\); a nonincreasing sequence with no lower bound may tend to \(-\infty\). Such sequences do not satisfy the hypotheses of the theorem.

Consequences for the Terms and Their Limit

The theorem guarantees that a limit exists, but it does not by itself identify the limit. The order of the terms gives a useful constraint on any limit that does exist: for a nondecreasing convergent sequence, the limit is an upper bound for every term. The corresponding statement for a nonincreasing convergent sequence is that the limit is a lower bound.

Theorem (The Limit Bounds Every Term of a Monotone Sequence): If \((a_n)\) is nondecreasing and \(a_n\to L\), then \(a_m\leq L\) for every \(m\in\mathbb{N}_0\). If \((a_n)\) is nonincreasing and \(a_n\to L\), then \(L\leq a_m\) for every \(m\in\mathbb{N}_0\).

Proof. First suppose \((a_n)\) is nondecreasing and \(a_n\to L\). Fix an arbitrary \(m\in\mathbb{N}_0\). If \(a_m>L\), let \(\varepsilon=(a_m-L)/2\), which is positive. By the definition of convergence, there is an \(N\in\mathbb{N}_0\) such that \(n\geq N\) implies \(|a_n-L|<\varepsilon\), and hence \(a_n<L+\varepsilon=(L+a_m)/2<a_m\). Choose \(n\geq\max\{N,m\}\). Since the sequence is nondecreasing and \(m\leq n\), \(a_m\leq a_n\), contradicting \(a_n<a_m\). Therefore \(a_m\leq L\).

Now suppose \((a_n)\) is nonincreasing and \(a_n\to L\). Fix \(m\). If \(a_m<L\), set \(\varepsilon=(L-a_m)/2>0\). For all sufficiently large \(n\), convergence gives \(a_n>L-\varepsilon=(L+a_m)/2>a_m\). Choose such an \(n\geq m\). Nonincreasing behavior gives \(a_n\leq a_m\), a contradiction. Thus \(L\leq a_m\). Since \(m\) was arbitrary, both conclusions hold for every term. \(\square\)

This theorem is useful when checking a proposed limit: a claimed limit below any term of a nondecreasing convergent sequence, or above any term of a nonincreasing convergent sequence, must be wrong. It also explains why bounds can help identify a limit after the Monotone Convergence Theorem has established that a limit exists.

Worked Applications

Worked Example: A Rational Sequence Increasing to One

Let \(a_n=(n+1)/(n+2)\) for \(n\in\mathbb{N}_0\). To check monotonicity, calculate the consecutive difference:

a_{n+1}-a_n = \frac{n+2}{n+3}-\frac{n+1}{n+2} = \frac{(n+2)^2-(n+1)(n+3)}{(n+2)(n+3)} = \frac{1}{(n+2)(n+3)}>0.

The numerator identity follows from \((n+2)^2=n^2+4n+4\) and \((n+1)(n+3)=n^2+4n+3\). Thus \((a_n)\) is nondecreasing. Also \(a_n=1-1/(n+2)<1\), so it is bounded above by \(1\). The Monotone Convergence Theorem guarantees a finite limit. To find it, use \(1/(n+2)\to0\) and the limit law for differences: \(a_n=1-1/(n+2)\to1\).

Worked Example: Partial Sums of a Geometric Sequence

Define \(s_n=\sum_{k=0}^{n}(1/3)^k\). The next term added to each partial sum is positive, so \(s_{n+1}-s_n=(1/3)^{n+1}>0\). Hence \((s_n)\) is nondecreasing. The finite geometric-sum identity gives

s_n=\frac{1-(1/3)^{n+1}}{1-1/3} =\frac{3}{2}\left(1-\frac{1}{3^{n+1}}\right) \leq\frac{3}{2}.

The identity can be checked by multiplying the sum by \(1-1/3\): all intermediate powers cancel, leaving \(1-(1/3)^{n+1}\). Since \(3^{-(n+1)}\geq0\), the displayed formula verifies the upper bound \(s_n\leq3/2\). The Monotone Convergence Theorem therefore guarantees convergence. Moreover, \(3^{-(n+1)}\to0\) by the theorem on powers with a ratio between zero and one, so the formula shows that the limit is \(3/2\).

Worked Example: A Recursive Sequence

Let \(x_0=1\) and \(x_{n+1}=\sqrt{2+x_n}\). We first show \(1\leq x_n<2\) for every \(n\). This holds at \(n=0\). If it holds for \(n\), then \(3\leq2+x_n<4\), so \(1<\sqrt{2+x_n}<2\); in particular, \(1\leq x_{n+1}<2\). Induction proves the bounds.

Next, \(x_1=\sqrt3>1=x_0\). If \(x_n\geq x_{n-1}\), then \(2+x_n\geq2+x_{n-1}\), and the square-root function preserves order on nonnegative numbers. Therefore \(x_{n+1}\geq x_n\). Induction shows that \((x_n)\) is nondecreasing. It is bounded above by \(2\), so the Monotone Convergence Theorem guarantees a limit \(L\).

The shifted sequence \((x_{n+1})\) has the same limit \(L\), since it is a subsequence of the convergent sequence \((x_n)\). Taking limits in the recurrence, using the limit law for sums and the limit theorem for square roots, gives \(L=\sqrt{2+L}\). Because \(x_n\geq1\) for all \(n\), the limit satisfies \(L\geq1\). Squaring the equation gives \(L^2=L+2\), or \((L-2)(L+1)=0\). The roots are \(2\) and \(-1\); \(L\geq1\) rules out \(-1\), so \(L=2\).

Monotonicity on a Tail Is Enough

A sequence need not be monotone from its first term onward for the Monotone Convergence Theorem to be useful. If it becomes monotone after some index, apply the theorem to that tail. Convergence is unaffected by changing finitely many terms, so convergence of the tail gives convergence of the entire sequence.

Theorem (Convergence of an Eventually Monotone Bounded Sequence): Every eventually monotone real sequence that is bounded converges to a finite real number.

Proof. Suppose \((a_n)\) is bounded and eventually nondecreasing. Then there is an \(N\in\mathbb{N}_0\) such that \(a_m\leq a_n\) whenever \(N\leq m\leq n\). Define the tail sequence \(b_k=a_{N+k}\) for \(k\in\mathbb{N}_0\). It is nondecreasing and bounded above because \((a_n)\) is bounded. By the Monotone Convergence Theorem, \(b_k\) converges to some real number \(L\). For every \(n\geq N\), \(a_n=b_{n-N}\), so \(a_n\to L\) as \(n\to\infty\). The finitely many terms before \(N\) do not affect convergence.

If instead \((a_n)\) is eventually nonincreasing, its tail is nonincreasing and bounded below. Applying the nonincreasing case of the Monotone Convergence Theorem to that tail again gives convergence of the tail and hence of the full sequence. These are the two possible directions of eventual monotonicity, so the result follows. \(\square\)

Worked Example: A Finite Initial Irregularity

Define \(c_0=5\) and \(c_n=2+1/n\) for \(n\geq1\). The first term does not continue the pattern: \(c_1=3<5=c_0\). But for every \(n\geq1\),

c_{n+1}-c_n = \left(2+\frac{1}{n+1}\right)-\left(2+\frac{1}{n}\right) = -\frac{1}{n(n+1)}<0.

Thus the sequence is eventually nonincreasing. Its tail is bounded below by \(2\), and the sequence is bounded (the initial term is \(5\), and all later terms lie between \(2\) and \(3\)). The theorem for eventually monotone bounded sequences guarantees convergence. Since the tail formula gives \(2+1/n\to2\), the full sequence converges to \(2\); the exceptional first term does not change that limit.

Checking the Hypotheses Carefully

The Monotone Convergence Theorem is a sufficient condition for convergence, not a claim that all convergent sequences are monotone. For example, \(u_n=(-1)^n/(n+1)\) changes direction as \(n\) varies, even though \(u_n\to0\). Nor does monotonicity by itself suffice: \(v_n=n+1\) is nondecreasing but has no finite upper bound, and it does not converge to a real number.

A reliable application therefore separates the work into three questions:

Is the sequence nondecreasing or nonincreasing, at least from some index onward?
Is it bounded in the direction required by that monotonicity?
After existence is established, is there a formula, a limit law, or an equation that identifies the limit?

The first two questions establish convergence. The third is a separate task, as the examples illustrate: monotonicity and boundedness guarantee a limit, while a formula or the recurrence may be needed to calculate it. Keeping existence and identification distinct prevents a common gap in sequence arguments.

Check Your Understanding

Use the theorem and examples in this tutorial to answer the following questions.

Which one-sided bound is required for a nondecreasing sequence, and which is required for a nonincreasing sequence?
Why does the sequence \(a_n=n+1\) not satisfy the hypotheses of the Monotone Convergence Theorem?
If a nondecreasing sequence converges to \(L\), why must every term be at most \(L\)?
What theorem can be applied if a sequence is bounded and becomes nonincreasing after some index?
For the recurrence \(x_{n+1}=\sqrt{2+x_n}\) with \(x_0=1\), which root of \(L=\sqrt{2+L}\) is consistent with the bounds on the sequence?