MATH 101 Quiz #6 (v.T3)
Last Name:
Thursday, March 31
First Name:
Grade:
Student-No:
Section:
Very short answer question
∞
X
+ 4n3
1. 1 mark The series
(−1)
either: [CA] converges absolutely; [CC] converges con2 + 3n3
n=1
ditionally; [D] diverges; or [N] none of the above. Determine which answer is correct, and put
its abbreviation in the box.
n1
Answer: D
1 + 4n3
4
= , the Alternating Series Test cannot be used. Indeed,
3
n→∞ 2 + 3n
3
3
1
+
4n
does not exist, so the Test for Divergence shows that the
we see that lim (−1)n
n→∞
2 + 3n3
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 ≤ 10−6 ?
∞
X
(−1)n−1
n=1
n2
are needed in order to evaluate its value
Solution: Using the alternating series test:
|S − SN | ≤ bN +1 =
1
(N + 1)2
Therefore we need (N + 1)2 ≥ 106 , which is equivalent to N + 1 ≥ 1000 or N ≥ 999.
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
7n4 + 2
converges or diverges.
series
n3
n=1
Solution: We note that
for all n ≥ 1. Therefore,
√
7n4 + 2
>
n3
∞
X
n=1
√
√
7n4
7
=
3
n
n
√
∞
7n4 + 2 √ X 1
,
>
7
n3
n
n=1
which is a divergent harmonic series. By the Comparison Test, the original series also
diverges.
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
3n
n=1
n2
(x − 2)n .
Solution: We compute
n+1
3 (x − 2)n+1 /(n + 1)2 an+1 = lim lim
n→∞
n→∞ an 3n (x − 2)n /n2
n2
= lim 3|x − 2|
n→∞
(n + 1)2
n2
= 3|x − 2| lim
= 3|x − 2| · 1.
n→∞ (n + 1)2
By the Ratio Test, the series converges if 3|x − 2| < 1 and diverges if 3|x − 2| > 1; in
other words, the series definitely converges when |x − 2| < 31 , or equivalently 53 < x < 73 .
Checking the right endpoint x = 73 , we see that
∞
X
3n 7
n=1
n X
∞
1
−2 =
3
n2
n=1
n2
is a convergent p-series. At the left endpoint x = 53 ,
∞
X
3n 5
n=1
n2
n X
∞
(−1)n
−2 =
3
n2
n=1
converges as well (either by the Alternating Series Test, or because the right-endpoint
calculation shows that it converges absolutely). Therefore the interval of convergence of
the original series is 35 ≤ x ≤ 73 , or [ 53 , 37 ].
Marking scheme:
• 2 marks for calculating that the ratio limit is 3|x − 2|, or for calculating that the radius
of convergence is 31 . (1 mark for partial progress)
• 1 mark for deducing that the interval of convergence includes ( 53 , 37 ).
• 1 mark for showing that the left endpoint is included and 1 mark for showing that the
right endpoint is included.