A sequence is defined by a₁ = 2, aₙ₊₁ = √(aₙ + 6). A student wants to prove it converges but cannot find a closed form. How does the Monotone Convergence Theorem help?
AMCT directly computes the limit by taking both sides to the limit and solving algebraically
BMCT requires the limit to be known in advance, so the student must first solve L = √(L + 6)
CBy verifying that the sequence is monotone and bounded, MCT guarantees convergence to some limit L; the student can then solve for L using L = √(L + 6)
DMCT applies only to decreasing sequences, so the student must first show the sequence eventually decreases
This is the key application pattern of MCT: verify monotonicity and boundedness first, conclude convergence exists (without knowing the limit), then find the limit by taking limits of both sides of the recursion. Option A is wrong — MCT guarantees existence, not a formula. Option B reverses the logical order: MCT lets you avoid computing L first. Option D is wrong — MCT applies to both increasing (bounded above) and decreasing (bounded below) sequences. The power is that existence precedes identification.
Question 2 Multiple Choice
Which pair of conditions is sufficient for the Monotone Convergence Theorem to guarantee a sequence converges?
ABounded above and eventually positive
BMonotone (increasing or decreasing) and bounded in the appropriate direction
CMonotone or bounded — either condition alone is sufficient
DMonotone, bounded, and with a known supremum that can be computed explicitly
MCT requires BOTH conditions together. A bounded sequence that is not monotone can oscillate and fail to converge (e.g., aₙ = (−1)ⁿ is bounded but diverges). A monotone sequence that is not bounded diverges to ±∞ (e.g., aₙ = n is increasing but unbounded and diverges). Option C is the most common misconception. Option D is wrong — the whole point of MCT is that you do not need to know the supremum explicitly in advance.
Question 3 True / False
The Monotone Convergence Theorem can be used to prove that the sequence aₙ = sin(nπ/4) converges, because it is bounded between −1 and 1.
TTrue
FFalse
Answer: False
Boundedness alone is not sufficient for MCT. The sequence sin(nπ/4) oscillates through values 0, √2/2, 1, √2/2, 0, −√2/2, −1, ... and is definitely not monotone. MCT requires both monotonicity and boundedness (in the appropriate direction). A bounded non-monotone sequence may or may not converge, but MCT cannot be invoked to decide. For this particular sequence, it fails to converge because it oscillates indefinitely.
Question 4 True / False
The Monotone Convergence Theorem and the Least Upper Bound property (completeness of ℝ) are logically equivalent: each can be proved from the other.
TTrue
FFalse
Answer: True
MCT is essentially the completeness axiom applied to sequences. The proof of MCT uses LUB directly: the range of a bounded increasing sequence is non-empty and bounded above, so by LUB it has a supremum, which the sequence converges to. Conversely, given MCT, you can recover LUB: for any non-empty set S bounded above, pick an increasing sequence of elements approaching sup S (possible by definition of supremum), apply MCT to get convergence, and the limit is sup S. The two are equivalent characterizations of ℝ's completeness. Neither holds in ℚ, where both break down.
Question 5 Short Answer
Explain why the Monotone Convergence Theorem allows you to prove a sequence converges without first knowing what the limit is, and why this argument would fail if you were working in ℚ instead of ℝ.
Think about your answer, then reveal below.
Model answer: MCT guarantees convergence using only two properties — monotonicity and boundedness — that can be verified without knowing the limit. The argument: the range of a bounded increasing sequence is a non-empty set bounded above, so by the Least Upper Bound property of ℝ it has a supremum L in ℝ. An epsilon argument then shows the sequence converges to L. The limit is not assumed; it is produced by completeness. In ℚ, the LUB property fails: bounded sets can have no rational supremum. For example, the sequence of rational approximations to √2 (1, 1.4, 1.41, 1.414, ...) is increasing and bounded above by 2 in ℚ, but its supremum √2 is irrational — not in ℚ. So the sequence has no limit in ℚ, and MCT fails to apply there.