The monotone convergence theorem

Statement

Let \( (a_n)_{n \in \mathbb{N}} \) be a sequence of real numbers. If \( (a_n) \) is monotone increasing (\( a_n \le a_{n+1} \) for all \( n \in \mathbb{N} \)) and bounded above (\( \exists M \in \mathbb{R} \) with \( a_n \le M \) for all \( n \)), then \( (a_n) \) converges, and \[ \lim_{n \to \infty} a_n \;=\; \sup\{ a_n : n \in \mathbb{N} \}. \] Dually, if \( (a_n) \) is monotone decreasing and bounded below, then \( (a_n) \) converges to \( \inf\{ a_n : n \in \mathbb{N} \} \). In summary: every bounded monotone sequence of real numbers converges. The result is stated over \( \mathbb{R} \) and uses the completeness axiom of \( \mathbb{R} \) essentially.

Why it matters

This is the first theorem in analysis that lets you prove a limit exists without knowing its value in advance. The definition of convergence requires a candidate limit \( L \); the monotone convergence theorem removes that obstacle entirely for monotone sequences, replacing an \( \varepsilon \)–\( N \) verification by two structural checks (monotonicity and boundedness) that are usually provable by induction. It is the engine behind the construction of \( e \), the convergence theory of positive series, iterative root-finding schemes, and the nested interval theorem.

Conceptually, it is one face of the completeness of \( \mathbb{R} \): over an incomplete ordered field such as \( \mathbb{Q} \) the statement is false, and in fact the theorem is logically equivalent to the least upper bound axiom over the ordered field axioms. It is the sequence-level shadow of "the real line has no holes".

Hypotheses

Proof

We prove the increasing case in full; the decreasing case follows formally at the end. Assume \( (a_n) \) is increasing and bounded above.

Both cases are established, so the proof is complete. \( \blacksquare \)

Result

Corollaries & converses

• Series with non-negative terms. If \( c_k \ge 0 \) for all \( k \), the partial sums \( s_n = \sum_{k=1}^n c_k \) are increasing, so \( \sum c_k \) converges \( \iff \) \( (s_n) \) is bounded above. This single fact underwrites the comparison, condensation, ratio and integral tests.
• Nested interval theorem. If \( I_n = [x_n, y_n] \) with \( I_{n+1} \subseteq I_n \), then \( (x_n) \) increases, \( (y_n) \) decreases, both are bounded, and \( \bigcap_n I_n \supseteq [\lim x_n, \lim y_n] \neq \emptyset \).
• Existence of \( e \). \( \left(1 + \frac{1}{n}\right)^n \) is increasing and bounded above by \( 3 \), so its limit exists; we name it \( e \) (see Problem 3).
• Bolzano–Weierstrass. Every sequence has a monotone subsequence (the "peak point" lemma); combined with this theorem, every bounded real sequence has a convergent subsequence (see Problem 5).
• Convergence of monotone-driven recursions. For \( a_{n+1} = f(a_n) \), once monotonicity and a bound are proved by induction, the limit \( L \) exists and, for continuous \( f \), satisfies \( L = f(L) \).
• Converse fails. Convergent does not imply monotone: \( a_n = \frac{(-1)^n}{n} \to 0 \) but alternates in sign, so it is neither increasing nor decreasing. What is true of every convergent sequence is boundedness — so within the class of monotone sequences, convergence and boundedness are exactly equivalent.
• Partial converse (monotone case). A monotone sequence converges \( \iff \) it is bounded: the forward direction is this theorem, the reverse is the standard fact that convergent sequences are bounded.

Fails without

Common errors

• Solving the fixed-point equation before proving convergence. For \( a_{n+1} = f(a_n) \), passing to the limit in the recursion is only legal after convergence is established. Classic trap: \( a_{n+1} = 2a_n \), \( a_1 = 1 \) "gives" \( L = 2L \Rightarrow L = 0 \), which is nonsense — the sequence diverges.
• Claiming the limit is the supremum for a decreasing sequence (or the infimum for an increasing one). The limit of a bounded decreasing sequence is \( \inf_n a_n \); its supremum is just \( a_1 \).
• Believing the sequence must reach its supremum. \( a_n = 1 - \frac{1}{n} \) converges to \( \sup_n a_n = 1 \) but no term equals \( 1 \); \( \sup \) need not be \( \max \).
• Verifying monotonicity only for the first few terms. \( a_n \le a_{n+1} \) must hold for all \( n \) (or all \( n \ge N_0 \)), normally via induction; three computed terms prove nothing.
• Circular induction in recursions. Proving "increasing" often needs the bound, and "bounded" often needs monotonicity of \( f \); the clean pattern is to prove the bound by induction first, then monotonicity using the bound (see Worked example 1).
• Strict inequalities in the limit. From \( a_n \lt 2 \) for all \( n \) one may conclude only \( \lim a_n \le 2 \), not \( \lim a_n \lt 2 \); limits do not preserve strictness.

Discussion

The theorem is best understood as the sequential form of completeness. The proof used the least upper bound axiom exactly once (Step 2), and that use is unavoidable: over the ordered field axioms alone, "every bounded monotone sequence converges" implies the least upper bound property back again (given a non-empty set \( S \) bounded above, one builds an increasing sequence of elements of \( S \) and a decreasing sequence of upper bounds by repeated bisection; their common limit is \( \sup S \)). Thus it sits in the tight circle of equivalences: supremum axiom \( \iff \) monotone convergence \( \iff \) nested intervals + Archimedean property \( \iff \) Bolzano–Weierstrass \( \iff \) Cauchy completeness + Archimedean property.

Historically the statement was taken as self-evident throughout the eighteenth century — Euler manipulated \( \left(1+\frac{1}{n}\right)^n \) freely without asking why its limit exists. Bolzano (1817) and Cauchy (1821) began demanding proofs of such existence claims, and Weierstrass's Berlin lectures of the 1860s made the theorem a cornerstone of the "arithmetisation of analysis". A genuine proof only became possible once Dedekind (1858, published 1872) and Cantor constructed \( \mathbb{R} \) and isolated completeness as the property that separates \( \mathbb{R} \) from \( \mathbb{Q} \).

The result travels far beyond first-year sequences. In measure theory it reappears as the monotone convergence theorem of Beppo Levi — \( 0 \le f_n \uparrow f \) pointwise implies \( \int f_n \to \int f \) — where the same one-directional structure defeats the pathologies that break naive limit-integral interchange. In the theory of series it is the reason absolute convergence is so much better behaved than conditional convergence: non-negative partial sums are monotone, so rearrangement cannot change a supremum. Iterative numerics (Babylonian square roots, Newton's method on convex functions, fixed-point schemes) routinely deliver monotone bounded iterates, so the theorem certifies convergence before any error estimate is attempted.

In order-theoretic language, the theorem says \( \mathbb{R} \) is a conditionally complete lattice in which order convergence of monotone nets agrees with topological convergence: an increasing net bounded above order-converges to its supremum. This viewpoint generalises to monotone nets in \( \mathbb{R} \) (the sequential proof transfers verbatim), to Dini's theorem (monotone pointwise convergence of continuous functions on a compact space to a continuous limit upgrades itself to uniform convergence), and to monotone operator sequences in functional analysis (an increasing sequence of self-adjoint operators bounded above converges in the strong operator topology). What does not generalise is choice-free constructivity: constructively, the theorem fails for sequences whose monotonicity is not accompanied by a modulus, since \( \sup A \) need not be computable — a reminder that Step 2 is a genuinely non-constructive appeal to completeness.

Common misconceptions. The theorem does not say monotone sequences converge (boundedness is essential), nor that bounded sequences converge (monotonicity is essential), nor that the limit is a term of the sequence. It also does not compute the limit for you beyond identifying it as a \( \sup \) or \( \inf \); extracting a closed form usually needs extra structure such as a recursion or a series identity.

Worked examples

Example 1 (recursive sequence). Define \( a_1 = \sqrt{2} \) and \( a_{n+1} = \sqrt{2 + a_n} \). Show \( (a_n) \) converges and find its limit.

Example 2 (positive series). Show that \( \sum_{k=1}^{\infty} \frac{1}{k^2} \) converges.

Problems

1. Show that \( a_n = \dfrac{n}{n+1} \) is increasing and bounded above, deduce that it converges, and identify the limit as a supremum.
   Solution

   Monotonicity: \( a_{n+1} - a_n = \frac{n+1}{n+2} - \frac{n}{n+1} = \frac{(n+1)^2 - n(n+2)}{(n+1)(n+2)} = \frac{1}{(n+1)(n+2)} \gt 0 \), so \( (a_n) \) is strictly increasing. Boundedness: \( a_n = \frac{n}{n+1} \lt 1 \) for all \( n \), since \( n \lt n+1 \). By the monotone convergence theorem \( (a_n) \) converges to \( L = \sup_n a_n \le 1 \).

   To see \( L = 1 \): given \( \varepsilon \gt 0 \), choose \( n \gt \frac{1}{\varepsilon} \) (Archimedean property); then \( a_n = 1 - \frac{1}{n+1} \gt 1 - \varepsilon \), so no number below \( 1 \) is an upper bound. Hence \( \sup_n a_n = 1 \) and \( \lim_{n\to\infty} \frac{n}{n+1} = 1 \). Note the supremum is not attained: \( a_n \lt 1 \) for every \( n \).

2. (Babylonian square roots.) Let \( a_1 = 2 \) and \( a_{n+1} = \dfrac{1}{2}\left( a_n + \dfrac{2}{a_n} \right) \). Prove that \( (a_n) \) converges and that its limit is \( \sqrt{2} \).
   Solution

   Positivity. By induction \( a_n \gt 0 \): \( a_1 = 2 \gt 0 \), and if \( a_n \gt 0 \) then \( a_{n+1} \) is half the sum of two positive numbers.

   Lower bound \( a_n \ge \sqrt{2} \) for all \( n \). \( a_1 = 2 \ge \sqrt{2} \). For \( n \ge 1 \), by AM–GM on the positive numbers \( a_n \) and \( \frac{2}{a_n} \): \[ a_{n+1} = \frac{1}{2}\left( a_n + \frac{2}{a_n} \right) \ge \sqrt{ a_n \cdot \frac{2}{a_n} } = \sqrt{2}. \] (AM–GM for two terms follows from \( \left( \sqrt{x} - \sqrt{y} \right)^2 \ge 0 \).)

   Decreasing from the first term. For every \( n \), \[ a_n - a_{n+1} = a_n - \frac{a_n}{2} - \frac{1}{a_n} = \frac{a_n^2 - 2}{2 a_n} \ge 0, \] since \( a_n \ge \sqrt{2} \) gives \( a_n^2 \ge 2 \), and \( a_n \gt 0 \). So \( (a_n) \) is decreasing.

   Convergence. \( (a_n) \) is decreasing and bounded below by \( \sqrt{2} \), so by the monotone convergence theorem \( L := \lim a_n \) exists and \( L \ge \sqrt{2} \gt 0 \).

   Value. Since \( L \neq 0 \), the map \( x \mapsto \frac{1}{2}\left(x + \frac{2}{x}\right) \) is continuous at \( L \), so letting \( n \to \infty \) in the recursion gives \( L = \frac{1}{2}\left( L + \frac{2}{L} \right) \), i.e. \( 2L^2 = L^2 + 2 \), i.e. \( L^2 = 2 \). With \( L \gt 0 \) this forces \( L = \sqrt{2} \).

3. Let \( e_n = \left( 1 + \dfrac{1}{n} \right)^n \). Prove that \( (e_n) \) is increasing and bounded above by \( 3 \), and conclude that \( \lim_{n\to\infty} e_n \) exists (this limit is the definition of \( e \)).
   Solution

   Increasing. By the binomial theorem, \[ e_n = \sum_{k=0}^{n} \binom{n}{k} \frac{1}{n^k} = \sum_{k=0}^{n} \frac{1}{k!} \prod_{j=1}^{k-1} \left( 1 - \frac{j}{n} \right), \] using \( \binom{n}{k} \frac{1}{n^k} = \frac{1}{k!} \cdot \frac{n(n-1)\cdots(n-k+1)}{n^k} = \frac{1}{k!} \prod_{j=1}^{k-1} \left(1 - \frac{j}{n}\right) \) (empty product \( = 1 \) for \( k = 0, 1 \)). Replacing \( n \) by \( n+1 \) increases every factor \( 1 - \frac{j}{n} \) to \( 1 - \frac{j}{n+1} \) (each factor lies in \( (0,1] \) and strictly increases for \( j \ge 1 \)), and also appends the extra non-negative term \( k = n+1 \). Hence every summand weakly increases and a new one is added, so \( e_{n+1} \ge e_n \); in fact \( e_{n+1} \gt e_n \) for \( n \ge 1 \).

   Bounded above. Each product factor is at most \( 1 \), so \[ e_n \le \sum_{k=0}^{n} \frac{1}{k!} \le 1 + 1 + \sum_{k=2}^{n} \frac{1}{2^{k-1}} \lt 1 + 1 + 1 = 3, \] where \( k! \ge 2^{k-1} \) for \( k \ge 1 \) (induction: true for \( k=1 \); \( (k+1)! = (k+1)k! \ge 2 \cdot 2^{k-1} \) since \( k + 1 \ge 2 \)) and the geometric tail sums to \( \sum_{k=2}^{n} 2^{-(k-1)} = 1 - 2^{-(n-1)} \lt 1 \).

   Conclusion. \( (e_n) \) is increasing and bounded above by \( 3 \); by the monotone convergence theorem it converges. Its limit, which lies in \( (2, 3] \) since \( e_1 = 2 \) and the sequence strictly increases, is by definition \( e \).

4. (Nested interval theorem.) Let \( I_n = [x_n, y_n] \) be closed intervals with \( I_{n+1} \subseteq I_n \) for all \( n \). Using the monotone convergence theorem, prove \( \bigcap_{n=1}^{\infty} I_n \neq \emptyset \). Show additionally that if \( y_n - x_n \to 0 \) the intersection is a single point, and give an example showing the conclusion fails for nested open intervals.
   Solution

   Setup. \( I_{n+1} \subseteq I_n \) means \( x_n \le x_{n+1} \le y_{n+1} \le y_n \). Hence \( (x_n) \) is increasing and \( (y_n) \) is decreasing. Moreover every \( x_m \le y_n \) for all \( m, n \): if \( m \le n \) then \( x_m \le x_n \le y_n \); if \( m \gt n \) then \( x_m \le y_m \le y_n \). In particular \( (x_n) \) is bounded above by \( y_1 \) and \( (y_n) \) is bounded below by \( x_1 \).

   Convergence. By the monotone convergence theorem, \( x := \lim x_n = \sup_n x_n \) and \( y := \lim y_n = \inf_n y_n \) exist. Since \( x_m \le y_n \) for all \( m,n \), taking \( m \to \infty \) gives \( x \le y_n \) for each \( n \) (limits preserve non-strict inequalities), then taking \( n \to \infty \) gives \( x \le y \).

   Non-empty intersection. For each \( n \): \( x_n \le \sup_m x_m = x \le y = \inf_m y_m \le y_n \), so \( [x, y] \subseteq I_n \) for every \( n \); hence \( \emptyset \neq [x,y] \subseteq \bigcap_n I_n \). (In fact \( \bigcap_n I_n = [x,y] \): any \( t \) in the intersection satisfies \( x_n \le t \le y_n \) for all \( n \), so \( x \le t \le y \).)

   Shrinking case. If \( y_n - x_n \to 0 \), then \( y - x = \lim (y_n - x_n) = 0 \) by the algebra of limits, so the intersection is the single point \( \{x\} \).

   Open intervals fail. \( J_n = \left( 0, \frac{1}{n} \right) \) satisfies \( J_{n+1} \subseteq J_n \), but \( \bigcap_n J_n = \emptyset \): any \( t \gt 0 \) is excluded from \( J_n \) once \( \frac{1}{n} \le t \) (Archimedean property), and \( t \le 0 \) is in no \( J_n \). Closedness of the intervals is essential.

5. (Bolzano–Weierstrass via monotone subsequences.) Call \( m \in \mathbb{N} \) a peak index of \( (a_n) \) if \( a_m \ge a_n \) for all \( n \gt m \). (a) Prove that every real sequence has a monotone subsequence. (b) Deduce the Bolzano–Weierstrass theorem: every bounded real sequence has a convergent subsequence.
   Solution

   (a) Two cases according to the set \( P \) of peak indices.

   Case 1: \( P \) is infinite. List its elements in increasing order \( m_1 \lt m_2 \lt m_3 \lt \cdots \). Since each \( m_k \) is a peak index and \( m_{k+1} \gt m_k \), we get \( a_{m_k} \ge a_{m_{k+1}} \) for every \( k \). Thus \( (a_{m_k})_k \) is a decreasing subsequence.

   Case 2: \( P \) is finite. Choose \( n_1 \) larger than every element of \( P \) (possible: \( P \) finite has a maximum, or is empty and any \( n_1 \) works). We build \( n_1 \lt n_2 \lt n_3 \lt \cdots \) recursively with \( a_{n_1} \lt a_{n_2} \lt \cdots \) — more precisely, with \( a_{n_{k+1}} \gt a_{n_k} \): since \( n_k \gt \max P \), the index \( n_k \) is not a peak index, so by definition there exists \( n \gt n_k \) with \( a_n \gt a_{n_k} \); set \( n_{k+1} \) equal to such an \( n \). Each \( n_{k+1} \gt n_k \gt \max P \) is again not a peak index, so the recursion continues indefinitely and produces a strictly increasing subsequence. In either case a monotone subsequence exists.

   (b) Let \( (a_n) \) be bounded, say \( |a_n| \le M \) for all \( n \). By (a) it has a monotone subsequence \( (a_{n_k}) \), which inherits the bound \( |a_{n_k}| \le M \): it is monotone and bounded (above and below, hence in particular on the side opposing its direction of motion). By the monotone convergence theorem \( (a_{n_k}) \) converges. Hence every bounded real sequence has a convergent subsequence, which is the Bolzano–Weierstrass theorem.