MATH 101 Quiz #6 (v.M2)
Last Name:
Friday, April 1
First Name:
Grade:
Student-No:
Section:
Very short answer question
∞
X
(−1)n
√ is absolutely convergent, conditionally con1. 1 mark Determine whether the series
1+ n
n=1
vergent, or divergent.
Answer: conditionally convergent
1
√ is clearly nonnegative, decreasing and
1+ n
approaches 0 as n → ∞, hence it converges by the Alternating Series Test.
∞
∞ X
(−1)n X
1
1
1
√ ≥ √ , the series with absolute values
√ =
√
However, since
1+ n
2 n
1+ n
1+ n
Solution: The series is alternating and
n=1
n=1
∞
X
1
√ .
diverges by the Comparison Test with the divergent p-series (p = 1/2 ≤ 1)
n
n=1
Therefore, the given series converges conditionally.
Marking scheme: 1 for a correct answer in the box.
Short answer questions—you must show your work
∞
X
n+3
is convergent or divergent. You must
2. 2 marks Determine whether the series
n2 + 2n
n=1
justify your answer.
Solution: For large n, the numerator n + 3 ≈ n and the denominator n2 + 2n ≈ n2 ,
so the nth term is approximately nn2 = n1 . So we apply the Limit Comparison Test with
1
an = nn+3
2 +2n and bn = n . Since
1 + n3
an
(n + 3)/(n2 + 2n)
lim
= lim
= lim
= 1,
n→∞ bn
n→∞
n→∞ 1 + 2
1/n
n
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 10−4 ?
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 10−4 when N + 1 ≥ 4. So the
smallest allowable value is N = 3 .
Marking scheme:
• 1 mark for the alternating series test truncation error bound
• 1 mark for N = 3
Long answer question—you must show your work
4. 5 marks Find all values x for which the series
∞
X
(x + 2)n
n=1
Solution: We apply the Ratio Test with an =
n2
(x+2)n
.
n2
converges.
Since
(x+2)n+1 a n2
1
n+1 (n+1)2 =
lim
lim =
lim
|x + 2| = lim
|x + 2| = |x + 2|
(x+2)n 2
n→∞
n→∞ n→∞ (1 + 1/n)2
n→∞ (n + 1)
an
2
n
we have convergence for
|x + 2| < 1 ⇐⇒ −1 < x + 2 < 1 ⇐⇒ −3 < x < −1
and divergence for |x + 2| > 1. For |x + 2| = 1, i.e. for x + 2 = ±1, i.e. for x = −3, −1,
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 −3 ≤ x ≤ −1 . Equivalently,
the series has interval of convergence [−3, −1] .
Marking scheme:
• 1 mark for applying the Ratio Test
• 2 marks for getting the correct limit |x + 2|
• 1 mark for showing absolute convergence when |x + 2| = 1
• 1 mark for the answer is any of the forms |x + 2| ≤ 1 or −3 ≤ x ≤ −1 or [−3, −1]