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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.
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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.