A sequence can be hard to understand because it gives infinitely many pieces of information one term at a time. Even if the first thousand terms look stable, the next term could jump away unless some extra structure prevents it. Monotonicity is the simplest such structure: it says the sequence is allowed to move only in one direction. Once a sequence has chosen its direction, the only remaining question is whether something stops it. The guiding picture is an approximation process. Suppose a decimal expansion is revealed by truncating more and more digits. The approximations from below never decrease, and they are trapped above by the number being approximated. The convergence does not come from a closed formula for the terms; it comes from order together with a bound. This is the prototype behind many existence arguments in analysis. [example: Decimal Truncations from Below] Let $x = 1.4142135\ldots$, and define \begin{align*} a_n = \frac{\lfloor 10^n x \rfloor}{10^n}. \end{align*} Write $m_n=\lfloor 10^n x\rfloor$. By the defining property of the floor function, \begin{align*} m_n \le 10^n x < m_n+1. \end{align*} Multiplying by $10$ gives \begin{align*} 10m_n \le 10^{n+1}x < 10m_n+10. \end{align*} Since $\lfloor 10^{n+1}x\rfloor$ is the greatest integer less than or equal to $10^{n+1}x$, the inequality $10m_n \le 10^{n+1}x$ implies \begin{align*} 10m_n \le \lfloor 10^{n+1}x\rfloor. \end{align*} Dividing by $10^{n+1}$, we get \begin{align*} a_n = \frac{m_n}{10^n} = \frac{10m_n}{10^{n+1}} \le \frac{\lfloor 10^{n+1}x\rfloor}{10^{n+1}} = a_{n+1}. \end{align*} Thus $(a_n)$ is increasing. The same floor inequality also gives the upper bound and the error estimate. Dividing \begin{align*} m_n \le 10^n x < m_n+1 \end{align*} by $10^n$ gives \begin{align*} a_n \le x < a_n+10^{-n}. \end{align*} Subtracting $a_n$ throughout yields \begin{align*} 0 \le x-a_n < 10^{-n}. \end{align*} Given $\varepsilon>0$, choose $N\in\mathbb N$ such that $10^{-N}<\varepsilon$. If $n\ge N$, then $10^{-n}\le 10^{-N}$, so \begin{align*} 0 \le x-a_n < \varepsilon. \end{align*} Hence $a_n\to x$. The truncations approach $x$ from below: monotonicity supplies the one-way motion, and the estimate $x-a_n<10^{-n}$ shows that the remaining gap vanishes. [/example] The lesson is that monotone sequences turn convergence into an order question. Instead of estimating all pairs $|a_m-a_n|$, we ask whether the sequence has a ceiling or a floor. This makes monotone sequences a bridge between the definition of a [sequence](/page/Sequence), the order structure of $\mathbb R$, and the completeness properties developed in [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). ## Definition ### Monotone Directions A general sequence may move up, down, and back again. The page topic is the special case where this never happens: after any term, the next term is constrained to lie on the same side in the order on $\mathbb R$. That single restriction is strong enough to turn many convergence questions into questions about upper and lower bounds. [definition: Monotone Sequence] Let $(a_n)_{n=1}^{\infty}$ be a sequence of [real numbers](/page/Real%20Numbers). The sequence is monotone if either \begin{align*} a_n \le a_{n+1} \end{align*} for every $n \in \mathbb N$, or \begin{align*} a_n \ge a_{n+1} \end{align*} for every $n \in \mathbb N$. [/definition] This definition is a child of the general notion of a sequence: monotonicity adds an order condition to the existing data of indexed real terms. It does not require a formula for $a_n$, and it does not assert convergence by itself. To talk efficiently about the two possible directions, we now name them separately. The upward direction is the order-theoretic version of approaching a target from below. It is the form most directly connected to suprema, because the terms climb toward the least ceiling that contains all of them. [definition: Increasing Sequence] Let $(a_n)_{n=1}^{\infty}$ be a sequence of real numbers. The sequence is increasing if \begin{align*} a_n \le a_{n+1} \end{align*} for every $n \in \mathbb N$. [/definition] Some approximation schemes begin above the desired quantity and improve by moving downward. To treat upper barriers, overestimates, and decreasing interval endpoints with the same precision, we need the reverse order condition as its own named notion. [definition: Decreasing Sequence] Let $(a_n)_{n=1}^{\infty}$ be a sequence of real numbers. The sequence is decreasing if \begin{align*} a_n \ge a_{n+1} \end{align*} for every $n \in \mathbb N$. [/definition] With this terminology, a monotone sequence is exactly a sequence that is increasing or decreasing. Many results have one proof for increasing sequences and a second proof obtained by changing signs for decreasing sequences. ### Strict Monotonicity Sometimes a sequence moves in one direction and never repeats a value. We need a stricter word for situations where the terms themselves must be distinct, such as constructing infinite ordered subsets or proving that a sequence never stabilizes. [definition: Strictly Increasing Sequence] Let $(a_n)_{n=1}^{\infty}$ be a sequence of real numbers. The sequence is strictly increasing if \begin{align*} a_n < a_{n+1} \end{align*} for every $n \in \mathbb N$. [/definition] Strict increase is not the only strict order pattern that occurs in approximation. An algorithm may produce genuinely smaller overestimates at every stage, and in such a case plateaus are ruled out from above rather than below. Naming the downward strict version keeps later examples from blurring strict decrease with ordinary decrease. [definition: Strictly Decreasing Sequence] Let $(a_n)_{n=1}^{\infty}$ be a sequence of real numbers. The sequence is strictly decreasing if \begin{align*} a_n > a_{n+1} \end{align*} for every $n \in \mathbb N$. [/definition] A strictly increasing sequence is increasing, and a strictly decreasing sequence is decreasing. The converse fails because constant stretches are allowed in monotone sequences. [example: Plateaus and Strictness] Define $(a_n)_{n=1}^{\infty}$ by \begin{align*} a_n = \lfloor \sqrt{n} \rfloor. \end{align*} We show that $(a_n)$ is increasing but not strictly increasing. For each $n\in\mathbb N$, we have \begin{align*} n \le n+1. \end{align*} Since the square root function is order-preserving on $[0,\infty)$, this gives \begin{align*} \sqrt n \le \sqrt{n+1}. \end{align*} If $x\le y$, then every integer less than or equal to $x$ is also less than or equal to $y$, so $\lfloor x\rfloor \le \lfloor y\rfloor$. Applying this with $x=\sqrt n$ and $y=\sqrt{n+1}$ gives \begin{align*} a_n=\lfloor \sqrt n\rfloor \le \lfloor \sqrt{n+1}\rfloor=a_{n+1}. \end{align*} Thus $(a_n)$ is increasing. It is not strictly increasing. Indeed, \begin{align*} 1 \le \sqrt 1 < 2, \end{align*} so $a_1=\lfloor \sqrt 1\rfloor=1$. Also \begin{align*} 1 \le \sqrt 2 < 2 \end{align*} because $1^2\le 2<2^2$, so $a_2=1$. Similarly, \begin{align*} 1 \le \sqrt 3 < 2 \end{align*} because $1^2\le 3<2^2$, so $a_3=1$. Hence \begin{align*} a_1=a_2=a_3=1. \end{align*} The sequence preserves the nondecreasing order condition, but these repeated values show that monotonicity does not imply strict monotonicity. [/example] ## Bounds and Extremal Control ### One-Sided Bounds Monotonicity alone does not stop a sequence from running away. The sequence $a_n=n$ never turns downward, but it has no finite limit. For an increasing sequence, the relevant obstruction to escape is an upper wall. The definition records a property of the whole set of terms, not merely of the early part of the sequence. [definition: Sequence Bounded Above] Let $(a_n)_{n=1}^{\infty}$ be a sequence of real numbers. The sequence is bounded above if there exists $M \in \mathbb R$ such that \begin{align*} a_n \le M \end{align*} for every $n \in \mathbb N$. [/definition] An upper bound need not be one of the terms. For decreasing sequences, the comparable obstruction is downward escape. A floor gives the sequence room to move while preventing it from passing below every real number. [definition: Sequence Bounded Below] Let $(a_n)_{n=1}^{\infty}$ be a sequence of real numbers. The sequence is bounded below if there exists $m \in \mathbb R$ such that \begin{align*} m \le a_n \end{align*} for every $n \in \mathbb N$. [/definition] A one-sided bound is tailored to a known monotone direction. Before the direction is known, or when comparing with general metric-space arguments, we often need a symmetric condition saying that the sequence cannot escape upward or downward. This is the boundedness notion used in compactness results such as Bolzano-Weierstrass. [definition: Bounded Sequence] Let $(a_n)_{n=1}^{\infty}$ be a sequence of real numbers. The sequence is bounded if there exists $C > 0$ such that \begin{align*} |a_n| \le C \end{align*} for every $n \in \mathbb N$. [/definition] For real sequences, being bounded is equivalent to being bounded above and bounded below. The two-sided formulation is convenient for comparison with metric-space boundedness, while the one-sided formulations are the ones that interact most directly with monotonicity. ### Examples and Obstructions [example: An Increasing Sequence with a Ceiling] Let \begin{align*} a_n = 1-\frac{1}{n} \end{align*} for $n\in\mathbb N$. For every $n\in\mathbb N$, \begin{align*} a_{n+1}-a_n = \left(1-\frac{1}{n+1}\right)-\left(1-\frac{1}{n}\right). \end{align*} Canceling the two copies of $1$ gives \begin{align*} a_{n+1}-a_n = -\frac{1}{n+1}+\frac{1}{n}. \end{align*} Putting the two fractions over the common denominator $n(n+1)$ gives \begin{align*} -\frac{1}{n+1}+\frac{1}{n} = \frac{-n}{n(n+1)}+\frac{n+1}{n(n+1)}. \end{align*} Therefore \begin{align*} a_{n+1}-a_n = \frac{1}{n(n+1)}. \end{align*} Since $n>0$ and $n+1>0$, we have $n(n+1)>0$, so \begin{align*} \frac{1}{n(n+1)}>0. \end{align*} Hence $a_{n+1}>a_n$ for every $n\in\mathbb N$, so $(a_n)$ is strictly increasing. The sequence is bounded above by $1$, because \begin{align*} 1-a_n = 1-\left(1-\frac{1}{n}\right). \end{align*} Canceling gives \begin{align*} 1-a_n = \frac{1}{n}. \end{align*} Since $1/n>0$, we get $a_n<1$ for every $n\in\mathbb N$. Finally, let $\varepsilon>0$. Choose $N\in\mathbb N$ such that $N>1/\varepsilon$. If $n\ge N$, then $1/n\le 1/N<\varepsilon$, and hence \begin{align*} |a_n-1| = 1-a_n = \frac{1}{n}<\varepsilon. \end{align*} Thus $a_n\to 1$. The ceiling $1$ is not reached by any term, but the gap from $a_n$ to $1$ is exactly $1/n$, so the increasing sequence climbs up to that ceiling. [/example] This example is the friendly case: the ceiling is visible, and the error can be computed. The main theorem is stronger because it gives convergence even when the exact limit is not known in advance. The failure case is equally instructive. If the wall is missing, monotonicity can organize divergence instead of preventing it. [example: Monotone but Unbounded] Let $a_n=n$ for every $n \in \mathbb N$. For each $n\in\mathbb N$, \begin{align*} a_{n+1}=n+1 \end{align*} and \begin{align*} a_{n+1}-a_n=(n+1)-n=1. \end{align*} Since $1>0$, we have $a_{n+1}>a_n$, and therefore $a_n\le a_{n+1}$ for every $n\in\mathbb N$. Thus $(a_n)$ is increasing. The sequence is not bounded above. Let $M\in\mathbb R$. By the [Archimedean property](/theorems/737), there is some $n\in\mathbb N$ such that $n>M$. For this $n$, \begin{align*} a_n=n>M. \end{align*} So no real number $M$ is an upper bound for all the terms of the sequence. Finally, $(a_n)$ cannot converge to a real number. If $a_n\to L\in\mathbb R$, then taking $\varepsilon=1$ would give some $N\in\mathbb N$ such that \begin{align*} |a_n-L|<1 \end{align*} for every $n\ge N$. This implies \begin{align*} a_n<L+1 \end{align*} for every $n\ge N$. But by the Archimedean property, we can choose $n\in\mathbb N$ with $n\ge N$ and $n>L+1$, and then \begin{align*} a_n=n>L+1, \end{align*} contradicting $a_n<L+1$. Hence $(a_n)$ is monotone but divergent, showing that monotonicity alone does not replace boundedness. [/example] ## Convergence of Monotone Sequences ### Supremum and Infimum as Limits The central theorem says that a monotone sequence converges exactly when it is bounded in the direction in which it moves. This deserves theorem status because it converts a qualitative order condition into an existence statement for a limit, and it is where the completeness of $\mathbb R$ enters the theory. [quotetheorem:743] This theorem is often the first place where order completeness proves a limit exists without naming the limit by a formula. It is a compactness-flavoured result in elementary analysis: an infinite process trapped in one direction must settle at the boundary of its range. The theorem also explains why rational numbers are not complete. A bounded increasing sequence of rational numbers may converge to an irrational real number, so the same statement fails if the ambient space is $\mathbb Q$ and limits are required to remain rational. [example: Rational Approximations to $\sqrt{2}$] Let $m_n=\lfloor 10^n\sqrt{2}\rfloor$, so \begin{align*} a_n=\frac{m_n}{10^n}. \end{align*} Since $m_n\in\mathbb Z$ and $10^n\in\mathbb N$, each $a_n$ is rational. The defining property of the floor function gives \begin{align*} m_n\le 10^n\sqrt{2}<m_n+1. \end{align*} Multiplying by $10$ gives \begin{align*} 10m_n\le 10^{n+1}\sqrt{2}<10m_n+10. \end{align*} Because $10m_n$ is an integer and $10m_n\le 10^{n+1}\sqrt{2}$, the greatest integer less than or equal to $10^{n+1}\sqrt{2}$ satisfies \begin{align*} 10m_n\le \lfloor 10^{n+1}\sqrt{2}\rfloor. \end{align*} Dividing by $10^{n+1}$ gives \begin{align*} a_n=\frac{m_n}{10^n}=\frac{10m_n}{10^{n+1}}\le \frac{\lfloor 10^{n+1}\sqrt{2}\rfloor}{10^{n+1}}=a_{n+1}. \end{align*} Thus $(a_n)$ is increasing. The same floor inequality gives the upper bound and the error estimate. Dividing \begin{align*} m_n\le 10^n\sqrt{2}<m_n+1 \end{align*} by $10^n$ gives \begin{align*} a_n\le \sqrt{2}<a_n+10^{-n}. \end{align*} Subtracting $a_n$ gives \begin{align*} 0\le \sqrt{2}-a_n<10^{-n}. \end{align*} Given $\varepsilon>0$, choose $N\in\mathbb N$ such that $10^{-N}<\varepsilon$. If $n\ge N$, then $10^{-n}\le 10^{-N}$, and hence \begin{align*} 0\le \sqrt{2}-a_n<\varepsilon. \end{align*} Therefore $a_n\to\sqrt{2}$ as a sequence of real numbers. There is no rational limit for this sequence inside $\mathbb Q$. First, $\sqrt{2}$ is irrational: if $\sqrt{2}=p/q$ with $p,q\in\mathbb N$ and $\gcd(p,q)=1$, then \begin{align*} p^2=2q^2. \end{align*} So $p^2$ is even, which implies $p$ is even. Write $p=2r$. Substituting gives \begin{align*} 4r^2=2q^2. \end{align*} Dividing by $2$ gives \begin{align*} 2r^2=q^2. \end{align*} Thus $q^2$ is even, so $q$ is even, contradicting $\gcd(p,q)=1$. Hence $\sqrt{2}\notin\mathbb Q$. If the rational sequence $(a_n)$ had a limit $q\in\mathbb Q$, then it would also have the real limit $q$. Since we already proved that its real limit is $\sqrt{2}$, [uniqueness of limits](/theorems/625) would force $q=\sqrt{2}$. Equivalently, if $q\ne\sqrt{2}$, taking $\varepsilon=|q-\sqrt{2}|/2$ would eventually give both $|a_n-q|<\varepsilon$ and $|a_n-\sqrt{2}|<\varepsilon$, so \begin{align*} |q-\sqrt{2}|\le |q-a_n|+|a_n-\sqrt{2}|<2\varepsilon=|q-\sqrt{2}|, \end{align*} a contradiction. Thus the sequence converges in $\mathbb R$ but has no limit in $\mathbb Q$, showing that the completeness of $\mathbb R$ is essential. [/example] ### Nested Intervals A single monotone sequence gives a moving lower or upper approximation. Many existence arguments use two moving barriers at once: a left endpoint that moves right and a right endpoint that moves left. To discuss that construction, we first name the interval pattern where each stage refines the previous one. [definition: Nested Closed Intervals] A sequence of closed intervals $([a_n,b_n])_{n=1}^{\infty}$ in $\mathbb R$ is nested if \begin{align*} [a_{n+1},b_{n+1}] \subset [a_n,b_n] \end{align*} for every $n \in \mathbb N$. [/definition] Nestedness turns two endpoint sequences into a trap. The obstruction in such arguments is that no single interval endpoint need equal the desired point: one only knows that every stage still contains it. A nested-interval principle supplies the missing existence statement by converting two compatible monotone endpoint sequences into a nonempty limiting intersection. [quotetheorem:624] Nested intervals are one of the main ways monotone sequences appear without being named. Bisection methods, decimal expansions, compactness arguments, and existence proofs all use moving lower and upper bounds to trap an unknown object. ## Recursive Sources of Monotone Real Sequences The main theorem is most useful when monotonicity is discovered rather than built into a formula. A recursive rule may produce an increasing or decreasing real sequence after an invariant interval has been found. We use fixed point iteration here only as a source of monotone sequences, not as a full development of fixed point theory. [definition: Fixed Point Iteration] Let $I \subset \mathbb R$ be an interval, let $f: I \to I$ be a function, and let $a_1 \in I$. The fixed point iteration generated by $f$ from $a_1$ is the sequence $(a_n)_{n=1}^{\infty}$ defined by \begin{align*} a_{n+1} &= f(a_n) \end{align*} for every $n \in \mathbb N$. [/definition] A recursive rule can oscillate or diverge, so monotone convergence enters only after an order condition has been identified. The useful condition is not just that $f$ maps the interval into itself, but that it respects the interval order: once two possible inputs satisfy an inequality, applying $f$ should not reverse that inequality. Without such a condition, the comparison between consecutive iterates can be destroyed at the next step, and the recursion gives no stable monotone sequence to which the convergence theorem can be applied. This obstruction belongs to the function itself, not merely to one chosen starting value. We therefore separate out the order condition that makes comparisons survive repeated application of the same rule. [definition: Order-Preserving Function] Let $I \subset \mathbb R$ be an interval. A function $f: I \to I$ is order-preserving if for all $x,y \in I$, \begin{align*} x \le y \implies f(x) \le f(y). \end{align*} [/definition] Order preservation propagates a [first inequality](/theorems/2897) through the whole recursive sequence. The theorem below records a standard monotone-sequence template: prove the iterates move monotonically, use the closed interval as a bound, and use continuity only at the final step to identify the limit. [quotetheorem:8264] This theorem is included because it shows how the elementary [monotone convergence theorem](/theorems/509) is actually used. The fixed point language is secondary: the decisive real-sequence facts are monotonicity, one-sided trapping, and passage to the limit after convergence has already been obtained. [example: Square Root by a Decreasing Iteration] Let $c>0$, choose $a_1>\sqrt{c}$, and define \begin{align*} a_{n+1} = \frac{1}{2}\left(a_n+\frac{c}{a_n}\right). \end{align*} We first show by induction that every term stays above $\sqrt{c}$. The initial inequality $a_1>\sqrt{c}$ is part of the choice of $a_1$. Suppose $a_n>\sqrt{c}$. Then $a_n>0$, so \begin{align*} a_{n+1}-\sqrt{c}=\frac{1}{2}\left(a_n+\frac{c}{a_n}\right)-\sqrt{c}. \end{align*} Putting the terms over the denominator $2a_n$ gives \begin{align*} a_{n+1}-\sqrt{c}=\frac{a_n^2+c-2a_n\sqrt{c}}{2a_n}. \end{align*} Since $c=(\sqrt{c})^2$, the numerator factors as \begin{align*} a_n^2+c-2a_n\sqrt{c}=(a_n-\sqrt{c})^2. \end{align*} Therefore \begin{align*} a_{n+1}-\sqrt{c}=\frac{(a_n-\sqrt{c})^2}{2a_n}. \end{align*} Here $(a_n-\sqrt{c})^2>0$ and $2a_n>0$, so $a_{n+1}-\sqrt{c}>0$. Hence $a_{n+1}>\sqrt{c}$, and induction gives \begin{align*} a_n>\sqrt{c} \end{align*} for every $n\in\mathbb N$. Now compute the change from one term to the next: \begin{align*} a_{n+1}-a_n=\frac{1}{2}\left(a_n+\frac{c}{a_n}\right)-a_n. \end{align*} Combining the two $a_n$ terms gives \begin{align*} a_{n+1}-a_n=\frac{c-a_n^2}{2a_n}. \end{align*} Using $c=(\sqrt{c})^2$, the numerator factors as \begin{align*} c-a_n^2=(\sqrt{c}-a_n)(\sqrt{c}+a_n). \end{align*} Since $a_n>\sqrt{c}>0$, we have $\sqrt{c}-a_n<0$, $\sqrt{c}+a_n>0$, and $2a_n>0$. Thus \begin{align*} a_{n+1}-a_n<0. \end{align*} So $(a_n)$ is decreasing, and the already proved inequality $a_n>\sqrt{c}$ shows that it is bounded below by $\sqrt{c}$. By the *[Monotone Convergence Theorem for Sequences](/theorems/626)*, there is some real number $L$ such that $a_n\to L$. Since $a_n\ge \sqrt{c}$ for every $n$, the limit satisfies $L\ge \sqrt{c}$, so $L>0$. The shifted sequence has the same limit, so $a_{n+1}\to L$. Also $a_n\to L$ and $L>0$ imply $c/a_n\to c/L$ by the quotient law for limits. Taking limits in \begin{align*} a_{n+1}=\frac{1}{2}\left(a_n+\frac{c}{a_n}\right) \end{align*} therefore gives \begin{align*} L=\frac{1}{2}\left(L+\frac{c}{L}\right). \end{align*} Multiplying by $2L$ gives \begin{align*} 2L^2=L^2+c. \end{align*} Subtracting $L^2$ gives \begin{align*} L^2=c. \end{align*} Since $L>0$ and $\sqrt{c}$ is the positive square root of $c$, we get \begin{align*} L=\sqrt{c}. \end{align*} Thus the iteration remains above $\sqrt{c}$, decreases at every step, and converges down to $\sqrt{c}$. [/example] The example shows a typical workflow: prove an invariant bound, prove monotonicity, invoke the monotone convergence theorem, and then identify the limit from the defining relation. ## Subsequences and Order Patterns Not every sequence is monotone, but every real sequence contains some monotone order pattern if we pass to a subsequence. To make that statement precise, we first need the formal language of extraction: keeping infinitely many terms while preserving their original order. [definition: Subsequence] Let $(a_n)_{n=1}^{\infty}$ be a sequence of real numbers. A subsequence of $(a_n)$ is a sequence $(a_{n_k})_{k=1}^{\infty}$, where $(n_k)_{k=1}^{\infty}$ is a strictly increasing sequence of natural numbers. [/definition] A subsequence can discard oscillations, but it cannot reorder the original data. The remaining question is whether the retained terms themselves move in one direction when read in their inherited order. This distinction matters because extraction alone only selects terms, while monotone extraction selects terms whose values satisfy the same increasing or decreasing inequalities as an ordinary monotone sequence. [definition: Monotone Subsequence] Let $(a_n)_{n=1}^{\infty}$ be a sequence of real numbers. A monotone subsequence of $(a_n)$ is a subsequence $(a_{n_k})_{k=1}^{\infty}$ that is monotone as a sequence in $k$. [/definition] The existence theorem below is a key ingredient in the [Bolzano-Weierstrass theorem](/theorems/628). It matters because it says monotone structure is unavoidable in real sequences: even a wildly oscillating sequence contains an infinite ordered trace. [quotetheorem:8266] The extracted monotone subsequence need not converge if it is unbounded, so compactness requires one more ingredient. Boundedness supplies that ingredient by giving every subsequence the same two-sided control as the original sequence. Combining boundedness with the [monotone subsequence theorem](/theorems/8266) produces the central [sequential compactness](/page/Sequential%20Compactness) result for real sequences. [quotetheorem:171] The monotone subsequence theorem is useful precisely because the original sequence may have no visible order. Oscillation does not prevent ordered structure from appearing after extraction. [example: Oscillation with Monotone Subsequences] Let \begin{align*} a_n &= (-1)^n+\frac{1}{n}. \end{align*} The full sequence is neither increasing nor decreasing. Its first three terms are \begin{align*} a_1 &= (-1)^1+\frac{1}{1}=-1+1=0. \end{align*} Also \begin{align*} a_2 &= (-1)^2+\frac{1}{2}=1+\frac{1}{2}=\frac{3}{2}. \end{align*} And \begin{align*} a_3 &= (-1)^3+\frac{1}{3}=-1+\frac{1}{3}=-\frac{2}{3}. \end{align*} Since $a_1<a_2$, the sequence is not decreasing. Since $a_2>a_3$, the sequence is not increasing. Thus the full sequence is not monotone. Now restrict to even indices. For $k\in\mathbb N$, \begin{align*} a_{2k} &= (-1)^{2k}+\frac{1}{2k}=1+\frac{1}{2k}. \end{align*} For each $k\in\mathbb N$, \begin{align*} a_{2(k+1)}-a_{2k} &= \left(1+\frac{1}{2k+2}\right)-\left(1+\frac{1}{2k}\right). \end{align*} Canceling the two copies of $1$ gives \begin{align*} a_{2(k+1)}-a_{2k} &= \frac{1}{2k+2}-\frac{1}{2k}. \end{align*} Putting the fractions over the common denominator $(2k+2)(2k)$ gives \begin{align*} a_{2(k+1)}-a_{2k} &= \frac{2k-(2k+2)}{(2k+2)(2k)}. \end{align*} Thus \begin{align*} a_{2(k+1)}-a_{2k} &= \frac{-2}{(2k+2)(2k)}<0. \end{align*} So the even subsequence $(a_{2k})$ is strictly decreasing. Moreover, \begin{align*} |a_{2k}-1| &= \left|1+\frac{1}{2k}-1\right|=\frac{1}{2k}. \end{align*} Given $\varepsilon>0$, choose $K\in\mathbb N$ such that $K>1/(2\varepsilon)$. If $k\ge K$, then \begin{align*} |a_{2k}-1|=\frac{1}{2k}\le \frac{1}{2K}<\varepsilon. \end{align*} Hence $a_{2k}\to 1$. Now restrict to odd indices. For $k\in\mathbb N$, \begin{align*} a_{2k-1} &= (-1)^{2k-1}+\frac{1}{2k-1}=-1+\frac{1}{2k-1}. \end{align*} For each $k\in\mathbb N$, \begin{align*} a_{2(k+1)-1}-a_{2k-1} &= \left(-1+\frac{1}{2k+1}\right)-\left(-1+\frac{1}{2k-1}\right). \end{align*} Canceling the two copies of $-1$ gives \begin{align*} a_{2(k+1)-1}-a_{2k-1} &= \frac{1}{2k+1}-\frac{1}{2k-1}. \end{align*} Putting the fractions over the common denominator $(2k+1)(2k-1)$ gives \begin{align*} a_{2(k+1)-1}-a_{2k-1} &= \frac{(2k-1)-(2k+1)}{(2k+1)(2k-1)}. \end{align*} Therefore \begin{align*} a_{2(k+1)-1}-a_{2k-1} &= \frac{-2}{(2k+1)(2k-1)}<0. \end{align*} So the odd subsequence $(a_{2k-1})$ is strictly decreasing. Also, \begin{align*} |a_{2k-1}-(-1)| &= \left|-1+\frac{1}{2k-1}+1\right|=\frac{1}{2k-1}. \end{align*} Given $\varepsilon>0$, choose $K\in\mathbb N$ such that $2K-1>1/\varepsilon$. If $k\ge K$, then $2k-1\ge 2K-1$, so \begin{align*} |a_{2k-1}-(-1)|=\frac{1}{2k-1}\le \frac{1}{2K-1}<\varepsilon. \end{align*} Hence $a_{2k-1}\to -1$. The original sequence oscillates between two regions and has no monotone direction, but its even and odd subsequences each recover a clear decreasing order pattern. [/example] ## Eventual Monotonicity Many natural sequences have a few irregular early terms before settling into a consistent direction. Since convergence ignores finitely many initial terms, it would be too rigid to demand monotonicity from $n=1$ in every application. The increasing tail condition records the moment after which the upward order has stabilized. [definition: Eventually Increasing Sequence] Let $(a_n)_{n=1}^{\infty}$ be a sequence of real numbers. The sequence is eventually increasing if there exists $N \in \mathbb N$ such that \begin{align*} a_n \le a_{n+1} \end{align*} for every $n \ge N$. [/definition] The threshold $N$ is part of the existence statement, not fixed in advance. Downward tail behaviour is equally common, especially for error terms, upper approximations, and sequences produced by decreasing barriers. We therefore need a tail version of decreasing that ignores the same kind of finite initial disturbance. [definition: Eventually Decreasing Sequence] Let $(a_n)_{n=1}^{\infty}$ be a sequence of real numbers. The sequence is eventually decreasing if there exists $N \in \mathbb N$ such that \begin{align*} a_n \ge a_{n+1} \end{align*} for every $n \ge N$. [/definition] In many arguments, the direction of the tail is less important than the fact that a direction exists. The combined notion below lets convergence arguments discard the finite initial disturbance and apply the ordinary monotone theorem to the remaining tail. [definition: Eventually Monotone Sequence] Let $(a_n)_{n=1}^{\infty}$ be a sequence of real numbers. The sequence is eventually monotone if it is eventually increasing or eventually decreasing. [/definition] Eventual monotonicity should have the same convergence power as monotonicity, because limits are unaffected by changing or removing finitely many terms. The only possible obstruction is escape on the tail: after the finite initial segment is ignored, the remaining monotone sequence still needs the appropriate bound. With boundedness in place, the ordinary monotone convergence argument applies to the tail and then transfers the same limit back to the original sequence. [quotetheorem:8268] The boundedness hypothesis is stated two-sided because the eventual direction may not be known from the theorem statement. In applications, a one-sided bound on the monotone tail is enough once the direction has been identified. [example: Separate Monotone Subsequences Do Not Force a Monotone Tail] Let \begin{align*} a_n = \frac{n+(-1)^n}{n+1}. \end{align*} For even indices, $n=2k$, and $(-1)^{2k}=1$, so \begin{align*} a_{2k}=\frac{2k+1}{2k+1}=1. \end{align*} Thus the even subsequence is constant, hence both increasing and decreasing. For odd indices, $n=2k-1$, and $(-1)^{2k-1}=-1$, so \begin{align*} a_{2k-1}=\frac{(2k-1)-1}{2k}=\frac{2k-2}{2k}. \end{align*} Dividing numerator and denominator by $2$ gives \begin{align*} a_{2k-1}=\frac{k-1}{k}=1-\frac{1}{k}. \end{align*} For each $k\in\mathbb N$, \begin{align*} a_{2(k+1)-1}-a_{2k-1}=\left(1-\frac{1}{k+1}\right)-\left(1-\frac{1}{k}\right). \end{align*} Canceling the two copies of $1$ gives \begin{align*} a_{2(k+1)-1}-a_{2k-1}=-\frac{1}{k+1}+\frac{1}{k}. \end{align*} Putting the fractions over the common denominator $k(k+1)$ gives \begin{align*} a_{2(k+1)-1}-a_{2k-1}=\frac{-k}{k(k+1)}+\frac{k+1}{k(k+1)}=\frac{1}{k(k+1)}. \end{align*} Since $k(k+1)>0$, this difference is positive, so the odd subsequence is strictly increasing. The full sequence is not eventually increasing. Let $N\in\mathbb N$ be arbitrary, and choose $k\in\mathbb N$ with $2k\ge N$. Then \begin{align*} a_{2k}=1. \end{align*} Also \begin{align*} a_{2k+1}=a_{2(k+1)-1}=1-\frac{1}{k+1}. \end{align*} Since $1/(k+1)>0$, we have \begin{align*} a_{2k+1}<1=a_{2k}. \end{align*} Thus the increasing inequality fails at the index $2k\ge N$. The full sequence is not eventually decreasing either. For the same arbitrary $N$, choose $k\in\mathbb N$ with $2k-1\ge N$. Then \begin{align*} a_{2k-1}=1-\frac{1}{k}<1=a_{2k}. \end{align*} Thus the decreasing inequality fails at the index $2k-1\ge N$. Hence the sequence has monotone even and odd subsequences, but no monotone tail; monotonicity along separate subsequences does not imply eventual monotonicity of the full sequence. [/example] ## Pointwise Monotone Function Sequences as a Boundary The page has so far concerned real-valued terms. There is one nearby extension worth isolating because it uses the same index-order idea: a sequence whose terms are functions may be monotone point by point. This section is a boundary marker, meant to prevent confusion with monotone functions rather than to turn the page into a measure theory chapter. [definition: Pointwise Increasing Sequence of Functions] Let $E$ be a set, and let $(f_n)_{n=1}^{\infty}$ be a sequence of functions $f_n: E \to \mathbb R$. The sequence is pointwise increasing if \begin{align*} f_n(x) \le f_{n+1}(x) \end{align*} for every $x \in E$ and every $n \in \mathbb N$. [/definition] This definition says nothing about whether each function $f_n$ is increasing as a function of $x$. Approximation from above requires the opposite pointwise comparison, for instance when closed sets shrink or upper barriers decrease to a target. [definition: Pointwise Decreasing Sequence of Functions] Let $E$ be a set, and let $(f_n)_{n=1}^{\infty}$ be a sequence of functions $f_n: E \to \mathbb R$. The sequence is pointwise decreasing if \begin{align*} f_n(x) \ge f_{n+1}(x) \end{align*} for every $x \in E$ and every $n \in \mathbb N$. [/definition] Pointwise monotonicity lets us apply the real monotone convergence theorem separately at each point. Measure theory asks a stronger downstream question: if functions increase point by point, does integration respect the limiting process? The following theorem is included as a signpost to that later subject, with the real sequence idea still visible at its core. [quotetheorem:509] This theorem is deeper than the real sequence version because it exchanges a limit with an integral. For the present page, its role is mainly diagnostic: it shows how the same monotone-index structure survives when the terms are no longer numbers. [example: Increasing Indicators] Let $E=[0,1]$ with [Lebesgue measure](/page/Lebesgue%20Measure) $\mathcal L^1$, and define \begin{align*} f_n(x)=\mathbb{1}_{[0,1-1/n]}(x). \end{align*} For every $n\in\mathbb N$, \begin{align*} \left(1-\frac{1}{n+1}\right)-\left(1-\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}>0. \end{align*} Hence \begin{align*} 1-\frac{1}{n}\le 1-\frac{1}{n+1}. \end{align*} Therefore \begin{align*} [0,1-1/n]\subset [0,1-1/(n+1)]. \end{align*} If $x\in [0,1-1/n]$, then $f_n(x)=1$ and the inclusion gives $f_{n+1}(x)=1$. If $x\notin [0,1-1/n]$, then $f_n(x)=0\le f_{n+1}(x)$. Thus \begin{align*} f_n(x)\le f_{n+1}(x) \end{align*} for every $x\in E$ and every $n\in\mathbb N$, so $(f_n)$ is pointwise increasing. We now identify the pointwise limit. If $0\le x<1$, then $1-x>0$, so by the Archimedean property we can choose $N\in\mathbb N$ with \begin{align*} N>\frac{1}{1-x}. \end{align*} Then \begin{align*} \frac{1}{N}<1-x. \end{align*} So \begin{align*} x<1-\frac{1}{N}. \end{align*} For every $n\ge N$, \begin{align*} \frac{1}{n}\le \frac{1}{N}, \end{align*} and hence \begin{align*} x<1-\frac{1}{N}\le 1-\frac{1}{n}. \end{align*} Thus $f_n(x)=1$ for all $n\ge N$, so $f_n(x)\to 1$. If $x=1$, then $1\notin [0,1-1/n]$ for every $n$, so $f_n(1)=0$ for every $n$ and $f_n(1)\to 0$. Therefore \begin{align*} f_n(x)\to \mathbb{1}_{[0,1)}(x) \end{align*} for every $x\in [0,1]$. Since the integral of an indicator is the measure of its set, \begin{align*} \int_E f_n\,d\mathcal L^1=\mathcal L^1([0,1-1/n]). \end{align*} Lebesgue measure gives the length of an interval, so \begin{align*} \mathcal L^1([0,1-1/n])=\left(1-\frac{1}{n}\right)-0=1-\frac{1}{n}. \end{align*} Similarly, \begin{align*} \int_E \mathbb{1}_{[0,1)}\,d\mathcal L^1=\mathcal L^1([0,1))=1. \end{align*} Finally, \begin{align*} 1-\left(1-\frac{1}{n}\right)=\frac{1}{n}. \end{align*} Since $1/n\to 0$, the numbers $\int_E f_n\,d\mathcal L^1=1-1/n$ increase to $1$, which is exactly the integral of the pointwise limit. [/example] The function version also clarifies a common ambiguity. A sequence of functions can be pointwise increasing even when no individual function is increasing in its input variable. [example: Pointwise Monotone but Not Monotone in Space] Let $E=\mathbb R$ and define \begin{align*} f_n(x)=\left(1-\frac{1}{n}\right)\sin^2 x. \end{align*} We show that the sequence $(f_n)$ is increasing in the index $n$ pointwise. For every $n\in\mathbb N$, \begin{align*} \left(1-\frac{1}{n+1}\right)-\left(1-\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{n+1}=\frac{(n+1)-n}{n(n+1)}=\frac{1}{n(n+1)}. \end{align*} Since $n(n+1)>0$, this difference is positive, so \begin{align*} 1-\frac{1}{n}\le 1-\frac{1}{n+1}. \end{align*} Also $\sin^2 x\ge 0$ for every $x\in\mathbb R$. Multiplying the last inequality by $\sin^2 x$ gives \begin{align*} \left(1-\frac{1}{n}\right)\sin^2 x\le \left(1-\frac{1}{n+1}\right)\sin^2 x. \end{align*} Thus \begin{align*} f_n(x)\le f_{n+1}(x) \end{align*} for every $x\in\mathbb R$ and every $n\in\mathbb N$, so $(f_n)$ is pointwise increasing. This does not mean that the individual functions are increasing as functions of $x$. For every $n\ge 2$, we have \begin{align*} 0<1-\frac{1}{n}. \end{align*} Now $\pi/2<\pi$, and \begin{align*} f_n(\pi/2)=\left(1-\frac{1}{n}\right)\sin^2(\pi/2)=\left(1-\frac{1}{n}\right)\cdot 1=1-\frac{1}{n}. \end{align*} Also \begin{align*} f_n(\pi)=\left(1-\frac{1}{n}\right)\sin^2(\pi)=\left(1-\frac{1}{n}\right)\cdot 0=0. \end{align*} Hence \begin{align*} f_n(\pi/2)>f_n(\pi) \end{align*} even though $\pi/2<\pi$. Therefore $f_n$ is not increasing on $\mathbb R$ for every $n\ge 2$. The example separates monotonicity in the sequence index from monotonicity in the spatial variable. [/example] ## Beyond and Connected Topics Monotone sequences are an entry point into completeness. In standard treatments of the real numbers, the monotone convergence theorem, the least upper bound property, the nested interval theorem, and Cauchy completeness are closely linked formulations of the same completeness phenomenon, once the surrounding order and field structure have been fixed. These connections are developed naturally alongside real numbers, suprema, and limits in [Cambridge IA Numbers and Sets](/page/Cambridge%20IA%20Numbers%20and%20Sets) and [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). They also lead directly to compactness. The monotone subsequence theorem helps prove Bolzano-Weierstrass for bounded real sequences, which is one of the first sequential compactness results. From there the same ideas expand into metric spaces, complete spaces, compact sets, and the topology of function spaces. In measure theory and integration, monotone approximation becomes a construction principle. Nonnegative [measurable functions](/page/Measurable%20Functions) are built from increasing sequences of simple functions, and the monotone convergence theorem for functions is one of the core tools behind Fatou's lemma, dominated convergence, and $L^p$ spaces. These themes connect elementary sequence arguments to the analytic machinery used in [Cambridge II Analysis of Functions](/page/Cambridge%20II%20Analysis%20of%20Functions). In functional analysis, monotone sequences reappear through ordered Banach spaces, weak compactness arguments, approximation schemes, and fixed point methods. The order idea is elementary, but the habit of constructing an object as the limit of controlled approximations persists throughout [Cambridge III Functional Analysis](/page/Cambridge%20III%20Functional%20Analysis). ## References Androma, [Cambridge IA Numbers and Sets](/page/Cambridge%20IA%20Numbers%20and%20Sets). Androma, [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). Androma, [Cambridge II Analysis of Functions](/page/Cambridge%20II%20Analysis%20of%20Functions). Androma, [Cambridge III Functional Analysis](/page/Cambridge%20III%20Functional%20Analysis). Tom M. Apostol, *Mathematical Analysis* (1974). Walter Rudin, *Principles of Mathematical Analysis* (1976). Terence Tao, *Analysis I* (2016).