Midterm Exam 2 Solutions for Math 231: Convergence of Series and Improper Integrals - Prof, Exams of Calculus

The solutions to problem 1-5 of the midterm exam 2 for math 231, a college-level mathematics course focusing on the convergence of series and improper integrals. The solutions include the determination of the convergence or divergence of various series and improper integrals using different tests such as the alternating series test, the integral test, the comparison test, and the root test.

Typology: Exams

2010/2011

Uploaded on 06/27/2011

koofers-user-ce2
koofers-user-ce2 🇺🇸

10 documents

1 / 4

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Math 231 Midterm Exam 2 (Solutions)
Prof. I.Kapovich October 19, 2009
Problem 1.[20 points]
For each of the following statements indicate whether it is true or false.
You DO NOT need to provide explanations for your answers in this problem.
(1) If limn→∞ an= 0 then the series P
n=1 anconverges.
(2) If a series P
n=1 andiverges then the series P
n=1 |an|also diverges.
(3) If P
n=1(an+bn) converges then the series P
n=1 anand P
n=1 bn
both converge.
(4) Whenever anis a monotone increasing sequence of real numbers then
there exists ARsuch that limn→∞ an=A.
(5) If an0 for all n1 and the series P
n=1 anconverges, then the
series P
n=1 ansin(an) also converges.
Answers:
(1) FALSE. For example, limn→∞
1
n= 0 but the series P
n=1 1
ndiverges.
(2) TRUE. If P
n=1 |an|converges, then, by Theorem 3 in Ch 11.6, the
series P
n=1 analso converges, contrary to our assumption that P
n=1 an
diverges.
(3) FALSE. For example, P
n=1(n+ (n)) = P
n=1 0 converges but the
series P
n=1 nand P
n=1(n) both diverge.
(4) FALSE. For example, the sequence an=nis positive and monotone
increasing, but limn→∞ n= 6∈ R.
(5) TRUE. Indeed, |ansin(an)| |an|=an. Therefore by the Comparison
Test the series P
n=1 |ansin(an)|converges and hence, by Theorem 3 in Ch
11.6, the series P
n=1 ansin(an) also converges.
Problem 2.[20 points]
For each of the following series determine if it converges conditionally,
converges absolutely or diverges. Show all the details of your work.
(a)
X
n=2
(1)n
nln n
(b)
X
n=1
(1)n(n+ 1)n
(2n+ 1)n
Solution.
(a) Since the sequence bn=1
nln nsatisfies bn0, limn→∞
1
nln n= 0 and
is bnmonotone decreasing for n2, the series P
n=2
(1)n
nln nconverges by
the Alternating Series Test. We have |(1)n
nln n|=1
nln n0 for n2. To
1
pf3
pf4

Partial preview of the text

Download Midterm Exam 2 Solutions for Math 231: Convergence of Series and Improper Integrals - Prof and more Exams Calculus in PDF only on Docsity!

Math 231 Midterm Exam 2 (Solutions) Prof. I.Kapovich October 19, 2009

Problem 1.[20 points] For each of the following statements indicate whether it is true or false. You DO NOT need to provide explanations for your answers in this problem.

(1) If limn→∞ an = 0 then the series

n=1 an^ converges. (2) If a series

n=1 an^ diverges then the series^

n=1 |an|^ also diverges. (3) If

n=1(an^ +^ bn) converges then the series^

n=1 an^ and^

n=1 bn both converge. (4) Whenever an is a monotone increasing sequence of real numbers then there exists A ∈ R such that limn→∞ an = A. (5) If an ≥ 0 for all n ≥ 1 and the series

n=1 an^ converges, then the series

n=1 an^ sin(an) also converges. Answers: (1) FALSE. For example, limn→∞ (^1) n = 0 but the series

n=

1 n diverges. (2) TRUE. If

n=1 |an|^ converges, then, by Theorem 3 in Ch 11.6, the series

n=1 an^ also converges, contrary to our assumption that^

n=1 an diverges.

(3) FALSE. For example,

n=1(n^ + (−n)) =^

n=1 0 converges but the series

n=1 n^ and^

n=1(−n) both diverge. (4) FALSE. For example, the sequence an = n is positive and monotone increasing, but limn→∞ n = ∞ 6 ∈ R.

(5) TRUE. Indeed, |an sin(an)| ≤ |an| = an. Therefore by the Comparison Test the series

n=1 |an^ sin(an)|^ converges and hence, by Theorem 3 in Ch 11.6, the series

n=1 an^ sin(an) also converges.

Problem 2.[20 points] For each of the following series determine if it converges conditionally, converges absolutely or diverges. Show all the details of your work. (a) ∑∞

n=

(−1)n n ln n

(b) ∑∞

n=

(−1)n^

(n + 1)n (2n + 1)n

Solution. (a) Since the sequence bn = (^) n ln^1 n satisfies bn ≥ 0, limn→∞ (^) n ln^1 n = 0 and

is bn monotone decreasing for n ≥ 2, the series

n=

(−1)n n ln n converges by the Alternating Series Test. We have | (−1)

n n ln n |^ =^

1 n ln n ≥^ 0 for^ n^ ≥^ 2.^ To 1

investigate the convergence or divergence of the series

n=

1 n ln n , we apply the Integral Test:

∫ (^) ∞

2

x ln x

dx = lim t→∞

∫ (^) t

2

x ln x

dx = lim t→∞

∫ (^) t

2

d ln x ln x

lim t→∞

∫ (^) t

2

[ln | ln x|]t 2 = lim t→∞

ln(ln t) − ln(ln 2)

so that the integral diverges.∑ Therefore, by the Integral Test, the series ∞ n=

1 n ln n also diverges. Thus the series

n=

(−1)n n ln n converges conditionally. (b) We apply the Root Test:

lim n→∞

n

∣∣(−1)n^ (n^ + 1)

n (2n + 1)n

∣∣ = lim n→∞

n + 1 2 n + 1

= lim n→∞

1 + (^1) n 2 + (^1) n

Since 12 < 1, the Root Test implies that the series

n=1(−1)

n (n+1)n (2n+1)n^ con- verges absolutely.

Problem 3.[20 points] For each of the following series determine if it converges or diverges. Show all the details of your work. (a) ∑∞

n=

2 n + 1 n + 1 (b) ∑∞

n=

n^2 + 3 n!

Solution. (a) We have

lim n→∞

2 n + 1 n + 1

= lim n→∞

2 + (^) n^1 1 + (^) n^1

Since 2 6 = 0 and the n-th term of the series does not converge to 0, Theorem 6 in Ch. 11.2 implies that the series

n=

2 n+ n+1 diverges. (b) We apply the Ratio Test:

nlim→∞

∣ an+ an

∣ (^) = lim n→∞

(n + 1)^2 + 3 (n + 1)!

n! n^2 + 3

= lim n→∞

n + 1

n^2 + 2n + 4 n^2 + 3

Since 0 < 1, the Ratio Test implies that the series

n=

n^2 + n! converges.

Problem 4.[20 points]

(b) Take, for example, an = 1 + (−1)n. Then an = 0 for odd n ≥ 1 and an = 2 for even n ≥ 1. Thus an ≥ 0 and the value of an alternates between 0 and 2. Therefore the limit limn→∞ an does not exist.