MATH 101 Quiz #6 (v.M3)
Last Name:
Friday, April 1
First Name:
Grade:
Student-No:
Section:
Very short answer question
1. 1 mark Determine whether the series
∞
X
(−1)2n+1
n=1
1+n
is absolutely convergent, conditionally
convergent, or divergent.
Answer: divergent
Solution: Note that (−1)2n+1 = (−1) · (−1)2n = −1, we can simplify
∞
X
(−1)2n+1
n=1
1+n
=−
∞
X
n=1
1
1+n
∞
1 X 1
1
≥
,
diverges by the Comparison Test with the divergent harmonic
Since
1+n
2n n=1 1 + n
series. Also, the extra factor of −1 in the original series does not change the conclusion of
divergence.
Marking scheme: 1 for a correct answer in the box.
Short answer questions—you must show your work
∞
X
n3 − 4
is convergent or divergent. You must
2. 2 marks Determine whether the series
5 − 6n
2n
n=1
justify your answer.
Solution: For large n, the numerator n3 − 4 ≈ n3 and the denominator 2n5 − 6n ≈ 2n5 ,
1
n3
so the nth term is approximately 2n
5 = 2n2 . So we apply the Limit Comparison Test with
3 −4
an = 2nn5 −6n
and bn = n12 . Since
1 − n43
an
(n3 − 4)/(2n5 − 6n)
1
lim
= lim
= lim
6 = ,
2
n→∞ bn
n→∞
n→∞
1/n
2
2 − n4
P∞
P∞
the given
series
a
converges
if
and
only
if
the
series
n
n=1
n=1 bn converges. Since the
P∞
P∞ 1
series n=1 bn = n=1 n2 is a convergent p-series, both series converge .
Marking scheme: 2 marks for a correct argument showing that the series converges. 1 mark
for giving the answer “convergent” or “converges” 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.
3. 2 marks What is the smallest value of N such that the partial sum
∞
X
(−1)n−1
n=1
n · 10n
N
X
(−1)n−1
n=1
n · 10n
approximates
within an accuracy of 10−7 ?
Solution: By the alternating series test, the error introduced when we approximate the
∞
N
X
X
(−1)n−1
(−1)n−1
1
series
by
is at most the first omitted term
. By
n
n
(N +1)
n
·
10
n
·
10
(N
+
1)10
n=1
n=1
trial and error, we find that this expression becomes smaller than 10−7 when N + 1 ≥ 7.
So the smallest allowable value is N = 6 .
Marking scheme:
• 1 mark for the alternating series test truncation error bound
• 1 mark for N = 6
Long answer question—you must show your work
4. 5 marks Find all values x for which the series
∞
X
(2x + 5)n
n=1
Solution: We apply the Ratio Test with an =
n3
(2x+5)n
.
n3
converges.
Since
(2x+5)n+1 a n3
1
n+1 (n+1)3 =
lim
lim =
lim
|2x + 5| = lim
|2x + 5| = |2x + 5|
(2x+5)n 3
n→∞
n→∞ n→∞ (1 + 1/n)3
n→∞ (n + 1)
an
3
n
we have convergence for
|2x + 5| < 1 ⇐⇒ −1 < 2x + 5 < 1 ⇐⇒ −6 < 2x < −4 ⇐⇒ −3 < x < −2
and divergence for |2x + 5| > 1. For |2x + 5| = 1, i.e. for 2x + 5 = ±1, i.e. for x = −2, −3,
P∞ 1
P
(±1)n
the series reduces to ∞
n=1 n3 , which converges absolutely, because
n=1 np converges
for p = 3 > 1. So the given series converges if and only if −3 ≤ x ≤ −2 . Equivalently,
the series has interval of convergence [−3, −2] .
Marking scheme:
• 1 mark for applying the Ratio Test
• 2 marks for getting the correct limit |2x + 5|
• 1 mark for showing absolute convergence when |2x + 5| = 1
• 1 mark for the answer is any of the forms |2x + 5| ≤ 1 or −3 ≤ x ≤ −2 or [−3, −2]