Calculus II · Unit 4A · lesson

The Squeeze Theorem for Sequences

Concept

Learning objectives

use upper and lower bounds to prove convergence when direct algebra does not reveal the limit.

The Squeeze Theorem for Sequences

Explanation

Control can matter more than an exact formula

Some sequences contain oscillatory factors that never converge by themselves. If that oscillation is multiplied by a magnitude shrinking to zero, exact values become irrelevant. Bounding the sequence between two simpler sequences can force a limit even when the internal behavior remains complicated.

The sequence version of the Squeeze Theorem works exactly like its function counterpart. The important work is choosing bounds that hold for every sufficiently large integer. Absolute-value inequalities are especially efficient: proving $|a_n-L|\le b_n$ with $b_n\to0$ proves $a_n\to L$ . This form also previews the error estimates used later for alternating and Taylor series.

Bridge

Control can replace direct computation

An oscillating sequence may be difficult to simplify directly. The Squeeze Theorem ignores the exact internal motion and traps the sequence between two simpler sequences. If both walls close onto the same limit, the middle sequence has nowhere else to go.

The method is especially useful when a bounded factor is multiplied by something that shrinks to zero. The proof concerns the shrinking envelope, not the oscillation inside it.

Proof idea

Both boundaries eventually enter the same band

Once the lower and upper sequences lie inside an $\varepsilon$ -band around $L$ , every trapped middle term lies inside as well. The formal proof simply chooses a cutoff large enough for both boundaries.

Concept

Sequence squeeze theorem

If $b_n\le a_n\le c_n$ for all sufficiently large $n$ , and $b_n\to L$ and $c_n\to L$ , then $a_n\to L$ .

Guided walkthrough

An oscillating numerator

Since $-1\le\cos n\le1$ ,

-\frac1n\le\frac{\cos n}{n}\le\frac1n.

Both bounds approach zero, so

\frac{\cos n}{n}\to0.

Worked example

Bound the absolute value first

For $a_n=\cos(n^2)/\sqrt n$ , the numerator need not converge, but

|a_n|\le\frac1{\sqrt n}\to0.

Equivalently,

-\frac1{\sqrt n}\le a_n\le\frac1{\sqrt n}.

Both bounds approach zero, so $a_n\to0$ .

Common mistake

The trap must actually close

The bounds $-1\le a_n\le1$ do not imply convergence. The sequence $(-1)^n$ satisfies them and still oscillates.

Interactive checku4a-squeeze_theorem_for_sequences-01

Evaluate $\lim_{n\to\infty}\frac{\sin n}{n}$ .

Your work stays on this device. No account or AI grader is used.

Show hint

Use $|\sin n|\le1$ .

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Exercise

Prove $(-1)^n/n^2\to0$ .

Exercise

Show $\sqrt[n]{n}\to1$ using logarithms or known bounds.

Exercise

Give an example where two bounds approach different limits and therefore do not determine the middle limit.

Exercise

Explain the phrase "for all sufficiently large $n$ .

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.

BetterGrades-original; no direct adaptation declared in the verified handoff.

Reference textbooks remain rights-separated and are not published as application assets. Any direct adaptation requires separate identification and attribution.