MATH 101 Quiz #6 (v.T1)
Last Name:
Thursday, March 31
First Name:
Grade:
Student-No:
Section:
Very short answer question
∞
n
X
n−1 1 + 4
1. 1 mark The series
(−1)
either: [CA] converges absolutely; [CC] converges con3 + 22n
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 + 4n
= 1, the Alternating Series Test cannot be used. Indeed,
n→∞ 3 + 22n
1 + 4n
does not exist, so the Test for Divergence shows that the
we see that lim (−1)n−1
n→∞
3 + 22n
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
n3
are needed in order to evaluate its value
Solution: Using the alternating series test:
|S − SN | ≤ bN +1 =
1
(N + 1)3
Therefore we need (N + 1)3 ≥ 106 , which is equivalent to N + 1 ≥ 100 or N ≥ 99.
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
15n2 + 5
converges or diverges.
series
n2
n=1
Solution: We note that
for all n ≥ 1. Therefore,
√
√
√
15n2 + 5
15n2
15
>
=
2
2
n
n
n
∞
X
n=1
√
∞
15n2 + 5 √ X 1
,
>
15
n2
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
2n
n=1
n3
(x − 1)n .
Solution: We compute
n+1
2 (x − 1)n+1 /(n + 1)3 an+1 = lim lim
n→∞
n→∞ an 2n (x − 1)n /n3
n3
= lim 2|x − 1|
n→∞
(n + 1)3
n3
= 2|x − 1| lim
= 2|x − 1| · 1.
n→∞ (n + 1)3
By the Ratio Test, the series converges if 2|x − 1| < 1 and diverges if 2|x − 1| > 1; in
other words, the series definitely converges when |x − 1| < 21 , or equivalently 12 < x < 32 .
Checking the right endpoint x = 32 , we see that
∞
X
2n 3
n=1
n X
∞
1
−1 =
2
n3
n=1
n3
is a convergent p-series. At the left endpoint x = 12 ,
∞
X
2n 1
n=1
n3
n X
∞
(−1)n
−1 =
2
n3
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 21 ≤ x ≤ 32 , or [ 12 , 23 ].
Marking scheme:
• 2 marks for calculating that the ratio limit is 2|x − 1|, or for calculating that the radius
of convergence is 21 . (1 mark for partial progress)
• 1 mark for deducing that the interval of convergence includes ( 12 , 23 ).
• 1 mark for showing that the left endpoint is included and 1 mark for showing that the
right endpoint is included.