The Monotone Convergence Theorem

Does the sequence $a_n = \left(1 + \frac{1}{n}\right)^n$ converge? Every term is a mess of binomial coefficients. No algebraic manipulation reveals a clean limit. Yet a single structural observation is enough to guarantee convergence: this sequence increases with $n$ but never exceeds $3$ . This holds without ever computing the limit itself. This is the power of the Monotone Convergence Theorem. It trades the labor of finding a limit for an easier task. That task is bounding a sequence and checking that it moves in one direction. The theorem sits at the foundation of real analysis. It is one of the few tools that produces existence of a limit from qualitative information alone. Most convergence tests — the Order Limit Theorem, the Squeeze Theorem — require you to already suspect what the limit is. This lets you compare the sequence to something known. The Monotone Convergence Theorem needs no candidate limit whatsoever. It converts the completeness of $\mathbb{R}$ , encoded in the existence of suprema and infima, directly into a convergence statement about sequences. This is not a minor technical convenience. The real numbers are distinguished from the rationals precisely by completeness. The sequence of rational truncations of $\sqrt{2}$ is monotone and bounded in $\mathbb{Q}$ . Yet it has no limit in $\mathbb{Q}$ . The Monotone Convergence Theorem is the sequential face of the same completeness axiom. That axiom guarantees suprema exist. Understanding this theorem is understanding why $\mathbb{R}$ is the right setting for calculus. This article gives the precise definitions of monotone and bounded sequences. It states and proves the theorem rigorously from the Completeness Axiom (existence of suprema). It works through two genuinely nontrivial examples, including the sequence defining Euler's number $e$ . It shows exactly which hypothesis is indispensable. It closes by connecting the theorem forward to Subsequences and to the convergence theory of Infinite Series and Partial Sums.

Key takeaways

A sequence is monotone if it is nondecreasing or nonincreasing for all $n$ ; it is bounded if its range is a bounded set.
The theorem: every bounded monotone sequence of real numbers converges.
The proof uses only the Completeness Axiom — the supremum (or infimum) of the range is the limit.
No formula for the limit is needed; monotonicity plus boundedness alone forces convergence.
The hypothesis "in $\mathbb{R}$ " is essential: the same sequence viewed in $\mathbb{Q}$ can be bounded and monotone yet fail to converge there.
Key idea: completeness turns "bounded and moving one way" into "must settle down."

Where this lives today

The Monotone Convergence Theorem is the engine behind convergence of infinite series with nonnegative terms: the sequence of partial sums $s_n = \sum_{k=1}^n b_k$ with $b_k \ge 0$ is automatically nondecreasing, so it converges exactly when it is bounded above — this is the comparison test in embryo, and it is exactly the bridge to Infinite Series and Partial Sums. It also underlies the construction of the Riemann integral, where upper and lower Darboux sums form monotone sequences (or nets) as partitions refine, guaranteeing the existence of the integral for well-behaved functions. In probability theory, the monotone convergence theorem for expectations (a measure-theoretic generalization of this result) is one of the three foundational convergence theorems (alongside Fatou's Lemma and Dominated Convergence), used constantly to justify interchanging limits and integrals/expectations for nondecreasing sequences of nonnegative random variables. In numerical analysis, iterative root-finding schemes like Newton's method (as in Example 2) are proved convergent precisely by exhibiting monotonicity and a bound, exactly as done here. Finally, the theorem is the standard tool for proving convergence of recursively defined sequences appearing throughout combinatorics and dynamical systems, and it foreshadows the Bolzano–Weierstrass theorem on Subsequences, which removes the monotonicity hypothesis entirely at the cost of only guaranteeing a convergent subsequence rather than convergence of the whole sequence.

Ask the tutor: Can you show me how the Monotone Convergence Theorem is used to prove the comparison test for infinite series?

(a_n)_{n=1}^{\infty}

Intuition

Picture a nondecreasing sequence as a point crawling rightward along the number line, one step at a time, never allowed to backtrack. If nothing stops it, it marches off to infinity. But if a wall is placed at some finite value $M$ that the point can never cross, something has to give: the point cannot keep making progress forever while staying under $M$ . Intuitively, it must crowd ever closer to some ceiling value — not necessarily $M$ itself, but the least ceiling consistent with all the positions the point has already visited. That least ceiling is exactly $\sup\{a_n : n \in \mathbb{N}\}$ . The genius of the argument is that this supremum is not just an upper bound floating above the sequence — because it is the least upper bound, the sequence must eventually enter any window just below it (otherwise a smaller upper bound would work, contradicting leastness). Combined with monotonicity, which prevents the sequence from ever leaving that window once inside near the top, this forces convergence to the supremum.

Figure

A nondecreasing sequence climbing toward its supremum

L=\sup a_n

The mirror picture holds for a nonincreasing sequence bounded below: it settles down onto its infimum from above. In both cases, completeness — the mere existence of the supremum or infimum as a real number — is what makes the crawl stop.

(a_n)_{n=1}^{\infty}

a_n

Worked examples

Example 1: The sequence defining $e$ . Define $a_n = \left(1 + \dfrac{1}{n}\right)^n$ for $n \ge 1$ . We show $(a_n)$ is nondecreasing and bounded above, hence converges (its limit is defined to be $e$ ). Boundedness above. By the Binomial Theorem,

a_n = \sum_{k=0}^{n} \binom{n}{k} \frac{1}{n^k} = \sum_{k=0}^n \frac{1}{k!}\cdot \frac{n(n-1)\cdots(n-k+1)}{n^k} = \sum_{k=0}^n \frac{1}{k!}\prod_{j=0}^{k-1}\left(1 - \frac{j}{n}\right).

Each factor $\left(1 - \frac{j}{n}\right) \in (0,1]$ , so every term is bounded above by $\frac{1}{k!}$ :

a_n \le \sum_{k=0}^n \frac{1}{k!} \le 1 + 1 + \sum_{k=2}^{n} \frac{1}{2^{k-1}} = 2 + \left(1 - \frac{1}{2^{n-1}}\right) < 3,

using $k! \ge 2^{k-1}$ for $k \ge 1$ (proved by induction: true at $k=1$ ; if $k! \ge 2^{k-1}$ then $(k+1)! = (k+1)k! \ge 2 \cdot 2^{k-1} = 2^k$ since $k+1\ge 2$ ) and summing the resulting geometric series. So $a_n < 3$ for all $n$ : $(a_n)$ is bounded above by $3$ . Monotonicity. Using the same expansion, write

a_n = \sum_{k=0}^n \frac{1}{k!}\prod_{j=0}^{k-1}\left(1 - \frac{j}{n}\right), \qquad a_{n+1} = \sum_{k=0}^{n+1} \frac{1}{k!}\prod_{j=0}^{k-1}\left(1 - \frac{j}{n+1}\right).

Compare term by term for $0 \le k \le n$ : since $\frac{j}{n+1} < \frac{j}{n}$ for $j \ge 1$ , we get $1 - \frac{j}{n+1} > 1 - \frac{j}{n}$ , and since all factors are nonnegative (for $n \ge k$ ),

\prod_{j=0}^{k-1}\left(1-\frac{j}{n+1}\right) \ge \prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right).

So each of the first $n+1$ terms of $a_{n+1}$ is at least the corresponding term of $a_n$ , and $a_{n+1}$ has one extra nonnegative term ( $k=n+1$ ) that $a_n$ lacks. Hence $a_{n+1} \ge a_n$ : the sequence is nondecreasing. By the Monotone Convergence Theorem, $(a_n)$ converges. Its limit is denoted $e$ , and the bound above shows $2 < e \le 3$ (in fact $e \approx 2.71828$ , and a finer estimate than the crude one above pins this down, but existence of the limit required none of that numerical precision — only monotonicity and some finite upper bound).

Example 2: A recursively defined sequence converging to $\sqrt{2}$ . Define $a_1 = 2$ and $a_{n+1} = \dfrac{1}{2}\left(a_n + \dfrac{2}{a_n}\right)$ for $n \ge 1$ (Newton's method for $\sqrt{2}$ , started asymmetrically at $2$ rather than at a "nice" value). We show $(a_n)$ is nonincreasing and bounded below by $\sqrt{2}$ , using no prior knowledge of the limit's exact numerical value beyond the defining property $(\sqrt 2)^2 = 2$ . Bounded below by $\sqrt 2$ , for $n \ge 2$ . For any $x > 0$ , by AM–GM applied to $x$ and $2/x$ (equivalently, directly from $(x - \sqrt2)^2 \ge 0$ expanded as $x^2 - 2\sqrt2 x + 2 \ge 0$ , i.e. $x + \frac2x \ge 2\sqrt2$ ):

\frac{1}{2}\left(x + \frac{2}{x}\right) \ge \sqrt{2}, \qquad \text{for all } x > 0.

Since $a_1 = 2 > 0$ , and inductively if $a_n > 0$ then clearly $a_{n+1} = \frac12(a_n + 2/a_n) > 0$ too (sum of positives), all terms are positive, and applying the displayed inequality with $x = a_n$ gives

a_{n+1} \ge \sqrt{2} \qquad \text{for every } n \ge 1.

So $a_n \ge \sqrt2$ for all $n \ge 2$ (and $a_1 = 2 > \sqrt2$ as well, so in fact for all $n\ge1$ ). Nonincreasing, for $n \ge 1$ . Compute, using $a_n \ge \sqrt2 > 0$ :

a_n - a_{n+1} = a_n - \frac{1}{2}\left(a_n + \frac{2}{a_n}\right) = \frac{1}{2}\left(a_n - \frac{2}{a_n}\right) = \frac{a_n^2 - 2}{2a_n}.

Since $a_n \ge \sqrt2$ , we have $a_n^2 \ge 2$ , so $a_n^2 - 2 \ge 0$ ; and $2a_n > 0$ . Hence $a_n - a_{n+1} \ge 0$ , i.e. $a_{n+1} \le a_n$ : the sequence is nonincreasing. By the Monotone Convergence Theorem (part (b), nonincreasing and bounded below), $(a_n)$ converges to some limit $L = \inf\{a_n\} \ge \sqrt2$ . To identify $L$ : since $a_n \to L$ and $a_n \ne 0$ for all $n$ with $a_n \to L \ge \sqrt2 > 0$ , the Order Limit Theorem (algebraic limit laws for quotients, valid since $L \ne 0$ ) gives $a_{n+1} = \frac12(a_n + 2/a_n) \to \frac12(L + 2/L)$ . But also $a_{n+1} \to L$ , since it is the same sequence shifted by one index (a shifted sequence has the same limit as the original, by the definition of convergence applied with $N' = N+1$ ). By uniqueness of limits,

L = \frac{1}{2}\left(L + \frac{2}{L}\right) \implies 2L = L + \frac2L \implies L = \frac2L \implies L^2 = 2 \implies L = \sqrt2

(taking the positive root since $L \ge \sqrt2 > 0$ ). So $a_n \to \sqrt2$ , with $a_1 = 2$ overshooting and the sequence decreasing monotonically down to $\sqrt2 \approx 1.41421$ .

\mathbb{R}

Scope, edge cases & common traps

Situation	What happens
Nondecreasing, bounded above	Converges to $\sup a_n$ (Theorem, part a)
Nonincreasing, bounded below	Converges to $\inf a_n$ (Theorem, part b)
Nondecreasing, not bounded above	Diverges to $+\infty$ (need not converge; e.g. $a_n = n$ )
Monotone only eventually (for $n \ge N_0$ )	Still converges — tails alone control convergence; the theorem applies to $(a_n)_{n \ge N_0}$
Not monotone, merely bounded	May diverge by oscillation (e.g. $a_n = (-1)^n$ ); theorem does not apply
Ambient space is $\mathbb{Q}$ instead of $\mathbb{R}$	Theorem fails, as shown above — completeness is essential
Sequence monotone and bounded, but you want the numerical value of the limit	Theorem only proves existence; identifying $L$ explicitly (as in Example 2) needs extra algebraic input

Common trapThe classic error: assuming the limit before proving it exists

A common mistake, especially with recursively defined sequences $a_{n+1} = f(a_n)$ , is to "solve for the limit" by setting $L = f(L)$ before establishing that $(a_n)$ actually converges. This silently assumes existence of the limit, which is exactly what must be proved first. If the sequence in fact diverges (say it oscillates or grows without bound), the equation $L = f(L)$ may still have a perfectly good solution — a phantom limit that the sequence never approaches. The correct order, followed in Example 2 above, is: (1) prove monotonicity, (2) prove boundedness, (3) invoke the Monotone Convergence Theorem to get existence of $L$ , and only then (4) use algebraic limit laws to pin down the value of $L$ .

The mathematician behind it

Origins · 1 minRichard Dedekind and how this was discovered

Richard Dedekind

1831–1916 · Rigorous foundations of the real numbers, Göttingen and Braunschweig

Dedekind, working alongside Cantor and in the tradition of Weierstrass and Bolzano, gave the first fully rigorous construction of the real numbers via what are now called Dedekind cuts — partitions of $\mathbb{Q}$ designed precisely so that every bounded-above set of cuts has a least upper bound. This construction proves the Completeness Axiom rather than merely postulating it, closing the logical gap that had let mathematicians manipulate limits, continuity, and convergence for over a century without a fully secure foundation. The Monotone Convergence Theorem, and indeed essentially all of elementary real analysis, is downstream of exactly this construction: once suprema are guaranteed to exist, sequential completeness follows by the argument given above.

History

- 1821 — Cauchy's Cours d'Analyse gives the first systematic

\varepsilon

-style treatment of limits and convergence, though without a rigorous construction of

\mathbb{R}

itself.

- 1858 — Dedekind, dissatisfied with relying on geometric intuition to justify that a bounded monotone sequence converges, begins developing the cut construction of the reals while lecturing on calculus in Zürich.

- 1872 — Dedekind publishes Stetigkeit und irrationale Zahlen, rigorously constructing

\mathbb{R}

from

\mathbb{Q}

via cuts and proving completeness as a theorem.

- 1872 — Cantor, independently and the same year, constructs

\mathbb{R}

via equivalence classes of Cauchy sequences of rationals, giving an alternative route to the same completeness property.

- 1976 (and after) — The Monotone Convergence Theorem becomes a standard early theorem in real analysis textbooks (Rudin, Apostol), typically the first place students see completeness converted into a concrete convergence guarantee.

Does the Monotone Convergence Theorem tell me what the limit equals?

Yes, in principle: the limit is exactly $\sup\{a_n\}$ (nondecreasing case) or $\inf\{a_n\}$ (nonincreasing case). But knowing this abstractly does not automatically hand you a closed-form numerical value — computing $\sup\{a_n\}$ explicitly can itself be hard, as with $e = \sup_n (1+1/n)^n$ . In practice, the theorem is used first to guarantee that a limit exists, and then separate algebraic techniques (like substituting $L = f(L)$ for recursively defined sequences, once convergence is already known) are used to identify the value.

Is boundedness above enough on its own, without monotonicity?

No. A bounded sequence that is not monotone need not converge at all — the classic example is $a_n = (-1)^n$ , which is bounded between $-1$ and $1$ but oscillates forever without approaching any single limit. Monotonicity is what rules out oscillation: it forces the sequence to move persistently toward its supremum (or infimum) rather than bouncing around near it. Boundedness alone only guarantees, via the Bolzano–Weierstrass theorem, that some subsequence converges — not the whole sequence.

Can the theorem be used to prove a sequence diverges?

Indirectly, yes, in the following sense: if you show a sequence is monotone and unbounded, then it cannot converge (a convergent sequence is always bounded, an earlier fact about boundedness of convergent sequences), so it must diverge to $+\infty$ or $-\infty$ . The Monotone Convergence Theorem itself, however, is a pure existence statement for the bounded case; the divergence conclusion for the unbounded case is really the contrapositive of "convergent implies bounded," not an extra clause added to this theorem.

Why does the proof only need monotonicity from some point onward, not from

n=1

Convergence is a statement about the tail behavior of a sequence: it asks whether $a_n$ eventually stays close to $L$ , and "eventually" already discards any finite initial segment. If $(a_n)$ is monotone and bounded only for $n \ge N_0$ , apply the theorem to the shifted sequence $(a_{N_0+k})_{k \ge 0}$ : it converges to the sup or inf of that tail, and since altering finitely many initial terms never changes whether or to what a sequence converges, $(a_n)$ itself converges to the same limit. This flexibility is used constantly, since many natural sequences are monotone only after some early "settling in" period.

How does this theorem relate to the existence of the number e?

The number $e$ is most rigorously defined as the limit of $(1+1/n)^n$ , and the only way to assert this limit exists — before any other characterization of $e$ (as a power series, as the base of the natural logarithm, etc.) is available — is via the Monotone Convergence Theorem. Example 1 above gives this proof in full: monotonicity and the bound $a_n < 3$ are both established purely combinatorially from the Binomial Theorem, with no circular appeal to properties of $e$ itself.

Does an analogous theorem hold for monotone sequences of functions, or for series?

Yes, in spirit. For series with nonnegative terms, the partial sums form a nondecreasing sequence, so the Monotone Convergence Theorem immediately implies: a series of nonnegative terms converges if and only if its partial sums are bounded above — this is the seed of the comparison test for Infinite Series and Partial Sums. In measure theory, the Monotone Convergence Theorem for integrals (Lebesgue's theorem) is a distinct, more powerful result about interchanging limits and integrals for monotone sequences of functions; it shares the name and the underlying philosophy but requires substantially more machinery to prove.

What is the relationship between this theorem and the Bolzano–Weierstrass theorem?

Both are completeness-driven convergence theorems, but they trade hypotheses differently. The Monotone Convergence Theorem requires monotonicity (a strong global structural condition) and rewards you with convergence of the entire sequence. The Bolzano–Weierstrass theorem, covered under Subsequences, requires only boundedness (no monotonicity at all) but only guarantees a convergent subsequence, not the whole sequence. In fact, one common proof of Bolzano–Weierstrass extracts a monotone subsequence from an arbitrary bounded sequence and then applies the Monotone Convergence Theorem to that subsequence — making this theorem a genuine stepping stone toward the more general result.