Calculus II · Unit 4A · lesson

Monotone and Bounded Sequences

Concept

Learning objectives

identify monotonicity and bounds and use the Monotone Convergence Theorem.

Monotone and Bounded Sequences

Explanation

A sequence can be trapped by its own direction of motion

A monotone sequence moves in only one direction: never down, or never up. Direction alone is not enough, because $a_n=n$ increases forever. A bound alone is not enough, because $(-1)^n$ stays between $-1$ and $1$ while oscillating. Together, monotonicity and boundedness prevent both escape and endless backtracking.

The Monotone Convergence Theorem is an existence theorem. It can prove that a limit exists before we know its exact value. This is particularly useful for recursively defined sequences, where algebraic formulas may be unavailable. Once existence is established, a recurrence can sometimes be passed to the limit to identify possible values, after which bounds select the valid one.

Bridge

Direction and confinement solve different problems

A sequence can increase forever and escape to infinity, or remain bounded while oscillating. Monotonicity prevents reversal; boundedness prevents escape. Together they force convergence.

For recursive sequences, prove these properties before solving a fixed-point equation. The usual pattern is induction for the bound, an inequality for monotonicity, the Monotone Convergence Theorem, and only then the equation for the limiting value.

Proof idea

The supremum supplies the target

For an increasing bounded sequence, let $L$ be its least upper bound. Since $L-\varepsilon$ is not an upper bound, some term exceeds it; every later term remains between $L-\varepsilon$ and $L$ . Hence the sequence converges to $L$ .

**Monotone plus bounded forces convergence**
The increasing sequence $a_n=1-1/n$ can never cross its upper bound $1$ , so its terms are squeezed into a narrowing vertical range.
Monotone plus bounded forces convergence. Increasing staircase beneath a least upper bound.

How to read the visual

Each term is at least the preceding term, while the dashed line blocks upward escape. The theorem guarantees a finite limit even before its exact value is calculated.

Concept

Monotone Convergence Theorem

Every increasing sequence that is bounded above converges. Every decreasing sequence that is bounded below converges.

Decision

Recursive-sequence proof order

For a recursive sequence, establish a bound, establish monotonicity, invoke the theorem to prove a limit exists, and only then pass to the limit in the recurrence. Solving the fixed-point equation first finds candidates; it does not prove that the sequence reaches any of them.

Guided walkthrough

A recursive sequence

Let $a_1=1$ and $a_{n+1}=\sqrt{2+a_n}$ . One can show inductively that $1\le a_n<2$ and that $a_{n+1}\ge a_n$ . Therefore the sequence converges. If $a_n\to L$ , then

L=\sqrt{2+L},

so $L^2-L-2=0$ . The bounds force $L=2$ , not $-1$ .

Worked example

Prove convergence before solving for the limit

Let $a_1=1$ and $a_{n+1}=\sqrt{2+a_n}$ . If $a_n<2$ , then $a_{n+1}<2$ , so induction gives an upper bound. For $0<a_n<2$ ,

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

Thus the sequence is increasing and bounded, so it converges. Its limit satisfies $L=\sqrt{2+L}$ , and the admissible solution is $L=2$ .

Optional advanced note

Suprema hiding behind the theorem

A rigorous proof uses the least-upper-bound property of the real numbers. An increasing bounded sequence has a supremum $L$ ; if the terms failed to approach $L$ , then some number below $L$ would still be an upper bound, contradicting leastness. This is one of the first places where completeness of $\mathbb R$ does real work.

Common mistake

A fixed-point equation does not prove convergence

Solving $L=g(L)$ identifies possible limits only. Establish monotonicity and boundedness first.

Interactive checku4a-monotone_and_bounded_sequences-01

Find $\lim_{n\to\infty}(1-1/n)$ .

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Show hint

The sequence is increasing and bounded above by $1$ ; direct algebra also reveals the value.

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Exercise

Show that $n/(n+1)$ is increasing and bounded above.

Exercise

Determine whether $a_n=2+(-1)^n/n$ is monotone. Does it still converge?

Exercise

For $a_{n+1}=(a_n+3/a_n)/2$ , explain why a positive limit must equal $\sqrt3$ .

Exercise

State exactly what monotonicity contributes and what boundedness contributes.

After the explanation

Use the section idea

Reading lens

Track the integer domain, late-term behavior, monotonicity, bounds, and any recurrence before asserting a limit.

Mental model

A sequence converges when every sufficiently late term remains arbitrarily close to one finite target.

Decision

Use algebraic limits when a formula is explicit; use bounds and monotonicity when a recurrence hides the formula.

Common trap

Reading a finite plot as proof or solving a recurrence's fixed-point equation before proving convergence.

Check yourself

Can you justify both the candidate limit and why the terms must approach it?

Source & rights

Original instruction with traceable references.

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