MATH 101 Quiz #6 (v.M1)
Last Name:
Friday, April 1
First Name:
Grade:
Student-No:
Section:
Very short answer question
∞
X
(−1)n
1. 1 mark Determine whether the series
is absolutely convergent, conditionally conn + n2
n=1
vergent, or divergent.
Answer: absolutely convergent
∞ ∞
X
X
(−1)n 1
1
1
Solution: Since
converges by the
≤ 2 , the series
=
2
2
2
n+n
n
n+n
n +n
n=1
n=1
∞
X
1
Comparison Test with the convergent p-series
.
2
n
n=1
Marking scheme: 1 for a correct answer in the box.
Short answer questions—you must show your work
2. 2 marks Determine whether the series
∞
X
n2 + 3
is convergent or divergent. You must
3 + 4n
n
n=1
justify your answer.
Solution: For large n, the numerator n2 + 3 ≈ n2 and the denominator n3 + 4n ≈ n3 ,
2
so the nth term is approximately nn3 = n1 . So we apply the Limit Comparison Test with
2 +3
an = nn3 +4n
and bn = n1 . Since
1 + n32
an
(n2 + 3)/(n3 + 4n)
= 1,
lim
= lim
= lim
n→∞ bn
n→∞
n→∞ 1 + 4n
1/n
n3
P
P
the given series ∞
an converges if and only if the series ∞
n=1
n=1 bn converges. Since the
P∞
P∞ 1
series n=1 bn = n=1 n is a divergent p-series, both series diverge .
Marking scheme: 2 marks for a correct argument showing that the series diverges. 1 mark
for giving the answer “divergent” or “diverges” without a correct justification; alternatively 1
mark for mentioning the Comparison or Limiting Comparison Test or computing a limit as in
the solution above, but not giving a full argument.
N
X
(−1)n
3. 2 marks What is the smallest value of N such that the partial sum
approximates
n
n
·
10
n=1
∞
X
(−1)n
within an accuracy of 10−6 ?
n
n
·
10
n=1
Solution: By the alternating series test, the error introduced when we approximate the
∞
N
X
X
(−1)n
(−1)n
1
series
by
is at most the first omitted term
. By trial
n
n
(N +1)
n
·
10
n
·
10
(N
+
1)10
n=1
n=1
and error, we find that this expression becomes smaller than 10−6 when N + 1 ≥ 6. So the
smallest allowable value is N = 5 .
Marking scheme:
• 1 mark for the alternating series test truncation error bound
• 1 mark for N = 5
Long answer question—you must show your work
4. 5 marks Find all values x for which the series
∞
X
(2x − 3)n
n=1
Solution: We apply the Ratio Test with an =
n2
(2x−3)n
.
n2
converges.
Since
(2x−3)n+1 a n2
1
n+1 (n+1)2 =
lim
lim =
lim
|2x − 3| = lim
|2x − 3| = |2x − 3|
(2x−3)n 2
n→∞
n→∞ n→∞ (1 + 1/n)2
n→∞ (n + 1)
an
2
n
we have convergence for
|2x − 3| < 1 ⇐⇒ −1 < 2x − 3 < 1 ⇐⇒ 2 < 2x < 4 ⇐⇒ 1 < x < 2
and divergence for |2x − 3| > 1. For |2x − 3| = 1, i.e. for 2x − 3 = ±1, i.e. for x = 1, 2,
P∞ 1
P
(±1)n
the series reduces to ∞
n=1 n2 , which converges absolutely, because
n=1 np converges
for p = 2 > 1. So the given series converges if and only if 1 ≤ x ≤ 2 . Equivalently, the
series has interval of convergence [1, 2] .
Marking scheme:
• 1 mark for applying the Ratio Test
• 2 marks for getting the correct limit |2x − 3|
• 1 mark for showing absolute convergence when |2x − 3| = 1
• 1 mark for the answer is any of the forms |2x − 3| ≤ 1 or 1 ≤ x ≤ 2 or [1, 2]