Common Errors In Monotone Sequence Proofs

Why Monotone-Sequence Proofs Need Care

A sequence may look simple when its first few terms are written out, but a proof must establish what happens for every index. In arguments about monotone sequences, several tempting shortcuts can leave a decisive gap: boundedness does not by itself imply convergence, a recurrence does not automatically produce a monotone sequence, and a fixed-point equation does not by itself identify the limit.

This tutorial focuses on recognizing those gaps and repairing them. The central habit is to keep separate the claims that must be proved: monotonicity, boundedness, convergence, and identification of a limit.

Definition: A sequence \((a_n)\) is nondecreasing if \(a_{n+1}\ge a_n\) for every \(n\in\mathbb N\), and nonincreasing if \(a_{n+1}\le a_n\) for every \(n\in\mathbb N\). It is monotone if it is either nondecreasing or nonincreasing. It is bounded above if there is a real number \(M\) such that \(a_n\le M\) for every \(n\), and bounded below if there is a real number \(m\) such that \(a_n\ge m\) for every \(n\).

These definitions use inequalities that must hold for all indices. Checking several terms can suggest which inequality might be true, but it does not establish the universal claim.

Error 1: Concluding Convergence from Boundedness Alone

The monotone convergence theorem requires both monotonicity and the appropriate bound. A bounded sequence can fail to converge when it keeps oscillating.

Theorem: Every nondecreasing sequence of real numbers that is bounded above converges. Every nonincreasing sequence of real numbers that is bounded below converges. The monotonicity and boundedness hypotheses are both required for these conclusions.

Worked Example: Bounded Does Not Mean Convergent

Consider \(a_n=(-1)^n\), for \(n\in\mathbb N\). Its first terms are

a_1=-1,\qquad a_2=1,\qquad a_3=-1,\qquad a_4=1.

For every \(n\), \(a_n\) is either \(-1\) or \(1\), so \(-1\le a_n\le1\). Thus the sequence is bounded. But its odd-indexed terms are all \(-1\), while its even-indexed terms are all \(1\). If the whole sequence converged to \(L\), each of these subsequences would also have to converge to \(L\). The odd-indexed subsequence has limit \(-1\), and the even-indexed subsequence has limit \(1\), which is impossible because \(-1\ne1\). Therefore \((a_n)\) does not converge.

Check: The bounds hold for both possible term values, and the two constant subsequences have different limits. This verifies boundedness and proves nonconvergence.

A common incorrect argument would be: “The terms stay between \(-1\) and \(1\), so the sequence converges.” The displayed sequence disproves that inference. To use the monotone convergence theorem, first establish the relevant monotonicity inequality; boundedness alone is not enough.

Error 2: Assuming a Recurrence Makes Its Sequence Monotone

A recurrence specifies how to calculate the next term from the current term. It does not, by itself, specify whether the next term is larger or smaller. For a recurrence \(x_{n+1}=f(x_n)\), monotonicity must be checked along the actual sequence, for example by determining the sign of \(x_{n+1}-x_n=f(x_n)-x_n\) for every \(n\).

Worked Example: A Recurrence That Oscillates

Let \(x_1=0\) and define \(x_{n+1}=2-x_n\). Calculate successive terms:

x_1=0,\qquad x_2=2-x_1=2,\qquad x_3=2-x_2=0,\qquad x_4=2-x_3=2,\qquad x_5=2-x_4=0.

In fact, \(x_{2k-1}=0\) and \(x_{2k}=2\) for every \(k\in\mathbb N\): applying the recurrence to \(0\) gives \(2\), and applying it to \(2\) gives \(0\). Hence the sequence alternates and is not monotone. It is bounded between \(0\) and \(2\), but it does not converge, since its odd-indexed terms are \(0\) and its even-indexed terms are \(2\).

Check: Substitution confirms both recurring values: \(2-0=2\) and \(2-2=0\). The recurrence therefore produces the alternating sequence claimed, not a monotone one.

Even if a recurrence map preserves order between inputs, that property alone does not show that the sequence moves in one direction. The initial comparison between \(x_2\) and \(x_1\) must be established, and the recurrence must preserve the comparison needed at every subsequent step. A proof may use induction for this purpose, but it must state the inductive claim and verify both its initial case and its preservation.

Error 3: Treating Computed Terms as a Proof

Numerical calculations are useful for forming a conjecture, locating a possible error, or suggesting which inequality to prove. But a finite list of terms cannot establish a claim about every term of an infinite sequence. The next example makes that limitation explicit.

Worked Example: Five Increasing Terms Do Not Prove Monotonicity

Define a sequence by

a_n= \begin{cases} n, & 1\le n\le5,\\ 0, & n\ge6. \end{cases}

The first terms are \(a_1=1\), \(a_2=2\), \(a_3=3\), \(a_4=4\), and \(a_5=5\). They increase at each of the first four comparisons. But \(a_6=0\), so

$$ a_6-a_5=0-5=-5<0. $$

Thus the sequence is not nondecreasing. It is not nonincreasing either, since \(a_2-a_1=2-1=1>0\). Consequently it is not monotone, despite the increasing pattern in its first five terms.

Check: The definition gives \(a_5=5\) and \(a_6=0\), so the claimed decrease occurs exactly at the sixth term. The earlier computed values cannot rule it out.

The correction is not to avoid numerical work. It is to use it appropriately: calculations may reveal a plausible statement, while a proof must verify the defining inequality for arbitrary indices. For a recursive sequence, this often means proving an invariant bound and then using it to establish the sign of \(x_{n+1}-x_n\). An observed pattern is a guide to the proof, not a substitute for it.

Why the Monotone Convergence Theorem Works

The theorem is not an additional pattern to memorize; its conclusion follows from the completeness of the real numbers. Here is the proof for a nondecreasing sequence bounded above. The proof also shows exactly where each hypothesis is used.

Form the set of terms.
Let \(S=\{a_n:n\in\mathbb N\}\). The sequence is bounded above, so \(S\) has a real supremum \(L=\sup S\).

Use the supremum property.
For any \(\varepsilon>0\), \(L-\varepsilon\) cannot be an upper bound for \(S\), because it is smaller than the least upper bound \(L\). Therefore there is an index \(N\) such that \(a_N>L-\varepsilon\).

Use monotonicity and the upper bound.
For every \(n\ge N\), nondecreasingness gives \(a_n\ge a_N>L-\varepsilon\). Since \(L\) is an upper bound, \(a_n\le L\). Thus \(L-\varepsilon<a_n\le L\), so \(|a_n-L|<\varepsilon\).

Conclude convergence.
For every \(\varepsilon>0\), all \(n\ge N\) satisfy \(|a_n-L|<\varepsilon\). This is the definition of \(a_n\to L\), so the sequence converges to \(L\).

The argument depends on boundedness to obtain a finite supremum and on monotonicity to ensure every later term stays above the selected \(a_N\). If either step is missing, this proof does not apply. The nonincreasing, bounded-below case follows by applying the same argument to \((-a_n)\).

Error 4: Choosing a Fixed Point Without Justification

Suppose a sequence is defined recursively by \(x_{n+1}=f(x_n)\), and suppose it has already been proved that \(x_n\to L\). If \(f\) is continuous at \(L\), then taking limits in the recurrence gives \(L=f(L)\). This identifies a necessary equation for the limit. It does not prove convergence, and if the equation has several solutions, it does not by itself select which one is the limit.

Worked Example: Use the Sequence's Bounds to Select a Fixed Point

Let \(x_1=\frac12\) and \(x_{n+1}=x_n^2\). The fixed-point equation is \(L=L^2\), which has two solutions. We must prove the sequence converges and justify which solution applies.

Calculate the first terms without skipping the substitutions:

x_1=\frac12,\qquad x_2=\left(\frac12\right)^2=\frac14,\qquad x_3=\left(\frac14\right)^2=\frac1{16},\qquad x_4=\left(\frac1{16}\right)^2=\frac1{256},\qquad x_5=\left(\frac1{256}\right)^2=\frac1{65536}.

To establish the pattern for all \(n\), suppose \(0<x_n\le\frac12\). Then \(x_{n+1}=x_n^2>0\), and because \(x_n\le\frac12<1\), multiplication by the positive \(x_n\) gives \(x_n^2<x_n\). Thus \(0<x_{n+1}<x_n\le\frac12\). Since \(0<x_1\le\frac12\), induction proves these bounds and strict decrease for every term.

The sequence is decreasing and bounded below by \(0\), so the monotone convergence theorem gives a limit \(L\). The function \(f(x)=x^2\) is continuous, so the recurrence implies \(L=L^2\). Rearranging gives \(L(L-1)=0\), hence \(L=0\) or \(L=1\). But every term satisfies \(x_n\le\frac12\), so the limit also satisfies \(L\le\frac12\). This excludes \(L=1\), leaving \(L=0\).

Check: The selected limit satisfies the fixed-point equation, since \(0^2=0\). The decreasing positive terms and their upper bound also rule out the other fixed point, \(1\). Both convergence and fixed-point selection have therefore been justified.

A common faulty argument stops after solving \(L=f(L)\) and announces one of the roots without referring to the sequence. The correction is to establish convergence first, pass to the limit under a valid continuity condition, and then use information about the sequence—such as an invariant interval or a bound—to exclude roots the limit cannot equal.

A Practical Proof Check

Before accepting a monotone-sequence argument, check that each claim has its own justification. These questions help expose gaps without replacing the proof itself:

Is the sequence actually shown to be nondecreasing or nonincreasing for every index?
Is the required upper or lower bound stated and verified for every term?
Does the cited convergence theorem apply with both hypotheses established?
If a recurrence is used, is the sign of \(x_{n+1}-x_n\) justified on the values the sequence actually takes?
If a fixed-point equation has multiple solutions, what fact about the sequence rules out the unwanted ones?

The common thread is that a plausible pattern, a bound, and a fixed-point equation answer different questions. A sound proof states which question each fact addresses and supplies the missing reasoning between them.

Check Your Understanding

For each question, identify the hypothesis or justification that a complete proof needs.

Why does the sequence \(a_n=(-1)^n\) show that boundedness alone does not imply convergence?
For \(x_{n+1}=f(x_n)\), what comparison can be checked to test whether the sequence is increasing or decreasing?
What does a finite list of increasing terms establish, and what does it fail to establish?
In the proof using \(L=\sup\{a_n:n\in\mathbb N\}\), where are boundedness and monotonicity each used?
If a convergent recurrence has two fixed points, what additional information can justify choosing one of them as the limit?