CST 8244 Final Exam Questions and
Answers | 2026 Update | 100% Correct -
Algonquin College.
: Sequences and Series
1. Give an example of two convergent series ∑aₙ and ∑bₙ such that ∑aₙbₙ diverges.
A) aₙ = 1/n², bₙ = n²
B) aₙ = 1/n, bₙ = 1/n
C) aₙ = (-1)ⁿ/n, bₙ = (-1)ⁿ/n
D) aₙ = 1/n², bₙ = 1/n
Correct Answer: A
Rationale: If aₙ = 1/n² and bₙ = n², then ∑aₙ converges by the p-series test (p=2>1), but
∑bₙ diverges, and aₙbₙ = 1, whose series diverges. Note that for ∑aₙbₙ to diverge while
both original series converge, one series must have terms that "cancel" the convergence
of the other.
2. Find a divergent sequence {cₙ} for which the sequence of averages {aₙ}
converges, where aₙ = (c₁ + c₂ + ... + cₙ)/n.
A) cₙ = n
B) cₙ = (-1)ⁿ
C) cₙ = (-1)ⁿn
D) cₙ = n²
Correct Answer: C
Rationale: This is the Cesàro mean. For cₙ = (-1)ⁿn, the sequence diverges, but the
average of the first n terms converges to 0. The partial sums grow like (-1)ⁿn/2, which
divided by n gives a convergent sequence .
, 3. Give an example of a sequence that has a strictly increasing subsequence, a
strictly decreasing subsequence, and a constant subsequence.
A) aₙ = (-1)ⁿ
B) aₙ = sin(n)
C) aₙ = (-1)ⁿ + 1/n
D) aₙ = cos(πn)
Correct Answer: B
Rationale: The sequence sin(n) oscillates and is dense in [-1,1], allowing selection of
increasing, decreasing, and constant subsequences. Option A only has two values (-1
and 1) .
4. Find a sequence {bₙ} such that for all p ∈ ℕ, there exists a subsequence of {bₙ}
that converges to p.
A) bₙ = 1/n
B) bₙ = n
C) An enumeration of the rationals
D) bₙ = sin(n)
Correct Answer: C
Rationale: An enumeration of the rationals has every rational number as a subsequential
limit point, including all natural numbers .
5. Give an example of two divergent sequences {aₙ} and {bₙ} such that {aₙbₙ} is a
Cauchy sequence.
A) aₙ = n, bₙ = 1/n
B) aₙ = (-1)ⁿ, bₙ = 1/n
C) aₙ = n, bₙ = 1/n²
D) aₙ = (-1)ⁿ, bₙ = n
Correct Answer: A
Rationale: aₙ = n diverges and bₙ = 1/n diverges (since it converges to 0, it diverges from
any non-zero limit). Their product aₙbₙ = 1, which is a constant sequence and hence
Cauchy .
6. Let {pₙ} be a Cauchy sequence that has a convergent subsequence. Prove that
{pₙ} converges.
A) True, by definition of Cauchy sequences
Answers | 2026 Update | 100% Correct -
Algonquin College.
: Sequences and Series
1. Give an example of two convergent series ∑aₙ and ∑bₙ such that ∑aₙbₙ diverges.
A) aₙ = 1/n², bₙ = n²
B) aₙ = 1/n, bₙ = 1/n
C) aₙ = (-1)ⁿ/n, bₙ = (-1)ⁿ/n
D) aₙ = 1/n², bₙ = 1/n
Correct Answer: A
Rationale: If aₙ = 1/n² and bₙ = n², then ∑aₙ converges by the p-series test (p=2>1), but
∑bₙ diverges, and aₙbₙ = 1, whose series diverges. Note that for ∑aₙbₙ to diverge while
both original series converge, one series must have terms that "cancel" the convergence
of the other.
2. Find a divergent sequence {cₙ} for which the sequence of averages {aₙ}
converges, where aₙ = (c₁ + c₂ + ... + cₙ)/n.
A) cₙ = n
B) cₙ = (-1)ⁿ
C) cₙ = (-1)ⁿn
D) cₙ = n²
Correct Answer: C
Rationale: This is the Cesàro mean. For cₙ = (-1)ⁿn, the sequence diverges, but the
average of the first n terms converges to 0. The partial sums grow like (-1)ⁿn/2, which
divided by n gives a convergent sequence .
, 3. Give an example of a sequence that has a strictly increasing subsequence, a
strictly decreasing subsequence, and a constant subsequence.
A) aₙ = (-1)ⁿ
B) aₙ = sin(n)
C) aₙ = (-1)ⁿ + 1/n
D) aₙ = cos(πn)
Correct Answer: B
Rationale: The sequence sin(n) oscillates and is dense in [-1,1], allowing selection of
increasing, decreasing, and constant subsequences. Option A only has two values (-1
and 1) .
4. Find a sequence {bₙ} such that for all p ∈ ℕ, there exists a subsequence of {bₙ}
that converges to p.
A) bₙ = 1/n
B) bₙ = n
C) An enumeration of the rationals
D) bₙ = sin(n)
Correct Answer: C
Rationale: An enumeration of the rationals has every rational number as a subsequential
limit point, including all natural numbers .
5. Give an example of two divergent sequences {aₙ} and {bₙ} such that {aₙbₙ} is a
Cauchy sequence.
A) aₙ = n, bₙ = 1/n
B) aₙ = (-1)ⁿ, bₙ = 1/n
C) aₙ = n, bₙ = 1/n²
D) aₙ = (-1)ⁿ, bₙ = n
Correct Answer: A
Rationale: aₙ = n diverges and bₙ = 1/n diverges (since it converges to 0, it diverges from
any non-zero limit). Their product aₙbₙ = 1, which is a constant sequence and hence
Cauchy .
6. Let {pₙ} be a Cauchy sequence that has a convergent subsequence. Prove that
{pₙ} converges.
A) True, by definition of Cauchy sequences
