MATH 101 Quiz #6 (v.T2)
Last Name:
Thursday, March 31
First Name:
Grade:
Student-No:
Section:
Very short answer question
∞
2
X
n+1 1 + 3n
1. 1 mark The series
(−1)
either: [CA] converges absolutely; [CC] converges con3 + 2n2
n=1
ditionally; [D] diverges; or [N] none of the above. Determine which answer is correct, and put
its abbreviation in the box.
Answer: D
1 + 3n2
3
= , the Alternating Series Test cannot be used. Indeed,
2
n→∞ 3 + 2n
2
2
1
+
3n
does not exist, so the Test for Divergence shows that the
we see that lim (−1)n+1
n→∞
3 + 2n2
series diverges. (Note that [N] can’t possibly be the correct answer, since every series either
converges absolutely, converges conditionally, or diverges.)
Solution: Since lim
Marking scheme: 1 mark for the correct answer.
Short answer questions—you must show your work
2. 2 marks How many terms of the series
within an error that is ≤ 10−4 ?
∞
X
(−1)n−1
n=1
n4
are needed in order to evaluate its value
Solution: Using the alternating series test:
|S − SN | ≤ bN +1 =
1
(N + 1)4
Therefore we need (N + 1)4 ≥ 104 , which is equivalent to N + 1 ≥ 10 or N ≥ 9.
Marking scheme:
minor errors.
2 marks for the correct answer; 1 mark for a correct procedure with
3. 2 marks Use the Comparison Test (not the Limit Comparison Test) to show whether the
∞ √
X
3n2 − 7
converges or diverges.
series
n3
n=2
Solution: We note that
for all n ≥ 2. Therefore,
√
3n2 − 7
<
n3
∞
X
n=2
√
√
3n2
3
= 2
3
n
n
√
∞
3n2 − 7 √ X 1
3
<
,
2
n3
n
n=2
which is a convergent p-series. By the Comparison Test, the original series also converges.
Marking scheme: 2 marks for the correct answer with valid justification; 1 mark for a
correct procedure with minor errors.
Long answer question—you must show your work
4. 5 marks Find the interval of convergence for
∞
X
4n
n=1
n
(x − 1)n .
Solution: We compute
n+1
4 (x − 1)n+1 /(n + 1) an+1 = lim lim
n→∞
n→∞ an 4n (x − 1)n /n
n
= lim 4|x − 1|
n→∞
n+1
n
= 4|x − 1| lim
= 4|x − 1| · 1.
n→∞ n + 1
By the Ratio Test, the series converges if 4|x − 1| < 1 and diverges if 4|x − 1| > 1; in
other words, the series definitely converges when |x − 1| < 41 , or equivalently 34 < x < 54 .
Checking the right endpoint x = 54 , we see that
∞
X
4n 5
n=1
n X
∞
1
−1 =
4
n
n=1
n
is the divergent harmonic series. At the left endpoint x = 34 ,
∞
X
4n 3
n=1
n
n X
∞
(−1)n
−1 =
4
n
n=1
converges by the Alternating Series Test. Therefore the interval of convergence of the
original series is 43 ≤ x < 54 , or [ 34 , 54 ).
Marking scheme:
• 2 marks for calculating that the ratio limit is 4|x − 1|, or for calculating that the radius
of convergence is 41 . (1 mark for partial progress)
• 1 mark for deducing that the interval of convergence includes ( 34 , 45 ).
• 1 mark for showing that the left endpoint is included and 1 mark for showing that the
right endpoint is not included.