Questions: Sequences and Convergence

Question 1

The sequence {(-1)ⁿ} satisfies |(-1)ⁿ| = 1 for all n, so it is bounded. What can we conclude?

• A. It converges, since all bounded sequences converge.
• B. It converges to 0, since terms alternate around 0.
• C. It diverges, since being bounded does not guarantee convergence.
• D. The monotone convergence theorem guarantees convergence.

Answer: (C)

{(-1)ⁿ} oscillates between -1 and 1 indefinitely and never settles near any single value. Boundedness is a necessary but not sufficient condition for convergence — the monotone convergence theorem requires both boundedness AND monotonicity. Since {(-1)ⁿ} is not monotone, the theorem does not apply.

Question 2

If the sequence {aₙ} converges to 0, then the series Σaₙ converges.

• T. True
• F. False

Answer: False

The harmonic series Σ(1/n) is the classic counterexample: the sequence {1/n} converges to 0, yet the series diverges (grows without bound). Sequence convergence and series convergence are different questions — sequence convergence asks whether individual terms approach a limit, while series convergence asks whether the running total of those terms approaches a finite sum.

Question 3

Give an example of a sequence that is bounded but does not converge, and explain why the monotone convergence theorem does not apply to it.

Model answer: The sequence {(-1)ⁿ} is bounded (|(-1)ⁿ| = 1 for all n) but diverges by oscillation. The monotone convergence theorem requires the sequence to be both bounded and monotone (non-decreasing or non-increasing). Since {(-1)ⁿ} alternates, it is not monotone, so the theorem does not apply.

This question targets a very common misconception. Boundedness alone only guarantees that subsequences converge (Bolzano-Weierstrass), not the sequence itself. Monotonicity is the second ingredient that rules out oscillation and forces convergence.