Homework 12 Solutions - First-Year Interest Group Seminar | N 1, Assignments of Health sciences

Download Homework 12 Solutions - First-Year Interest Group Seminar | N 1 and more Assignments Health sciences in PDF only on Docsity! Van Ligten (hlv63) – HW12 – Gilbert – (56650) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points If the nth partial sum of ∑ ∞ n = 1 an is Sn = 4n − 5 n + 1 , (i) what is a1? 1. a1 = 1 2 2. a1 = 9 2 3. a1 = 4 4. a1 = − 9 2 5. a1 = − 1 2 correct Explanation: Since a1 = S1, a1 = − 1 2 . 002 (part 2 of 3) 10.0 points (ii) What is an for n > 1? 1. an = 9 n2 2. an = 1 n(n + 1) 3. an = 9 n(n − 1) 4. an = 9 n(n + 1) correct 5. an = 1 n(n − 1) 6. an = 1 n2 Explanation: Since Sn = a1 + a2 + · · · + an , we see that an = Sn − Sn−1 . But Sn = 4 n − 5 n + 1 = 4(n + 1) − 9 n + 1 = 4 − 9 n + 1 . Consequently, an = 9 n − 9 n + 1 = 9 n(n + 1) for all n > 1. 003 (part 3 of 3) 10.0 points (iii) What is the sum ∑ ∞ n =1 an? 1. sum = 5 2. sum = 3 3. sum = 4 correct 4. sum = 1 5. sum = 2 Explanation: By definition sum = lim n→∞ Sn = lim n→∞ (4 n − 5 n + 1 ) . Thus sum = 4 . 004 10.0 points Determine whether the series ∞ ∑ n=1 5−n6n+1 is convergent or divergent. If it is convergent, find its sum. 1. 42 Van Ligten (hlv63) – HW12 – Gilbert – (56650) 2 2. divergent correct 3. 30 4. 38 5. 17 Explanation: ∞ ∑ n=1 5−n6n+1 = ∞ ∑ n=1 6 ( 6 5 )n . Since 6 5 > 1, the series is divergent. 005 10.0 points Find the sum of the series ∞ ∑ n =0 (3x + 5)n 4n for those values of x for which it converges. 1. sum = 9 − 3 x 4 2. sum = −1 + 3 x 4 3. sum = − 4 1 + 3x correct 4. sum = 4 9 − 3x 5. sum = − 4 1 − 3x 6. sum = 4 9 + 3x Explanation: When the geometric series ∑ ∞ n=0 ar n con- verges it has sum = a 1 − r . Now for the given series, a = 1 , r = 3x + 5 4 , Consequently, it has sum = 1 1 − 3x + 5 4 = − 4 1 + 3x . 006 10.0 points Find the sum of the infinite series tan2 θ − tan4 θ + tan6 θ + . . . + (−1)n−1 tan2n θ + . . . whenever the series converges. 1. sum = tan2 θ 2. sum = − sin2 θ 3. sum = cos2 θ 4. sum = − cos2 θ 5. sum = sin2 θ correct Explanation: For general θ the series tan2 θ − tan4 θ + tan6 θ + . . . + (−1)n−1 tan2n θ + . . . is an infinite geometric series whose common ratio is − tan2 θ. Since the initial term in this series is tan2 θ, its sum is thus given by tan2 θ 1 + tan2 θ = tan2 θ sec2 θ . Consequently sum = sin2 θ . 007 10.0 points A business executive realizes that he is out of shape so he begins an exercise program in Van Ligten (hlv63) – HW12 – Gilbert – (56650) 5 1. diverges correct 2. converges with sum = ln ( 6 5 ) 3. converges with sum = ln ( 12 5 ) 4. converges with sum = ln ( 5 6 ) 5. converges with sum = ln ( 5 12 ) Explanation: An infinite series ∑ an diverges when lim n→∞ an 6= 0 . For the given series an = ln ( 5n 6n + 6 ) . But lim n→∞ ( 5n 6n + 6 ) = lim n→∞ ( 5 6 + 6n ) = 5 6 , so lim n→∞ ln ( 5n 6n + 6 ) = ln 5 6 6= 0 . Consequently, the given infinite series diverges . 012 10.0 points Determine whether the infinite series ∞ ∑ n = 1 3(n + 1)2 n(n + 2) converges or diverges, and if converges, find its sum. 1. converges with sum = 3 4 2. converges with sum = 3 2 3. converges with sum = 3 4. converges with sum = 3 8 5. diverges correct Explanation: By the Divergence Test, an infinite series ∑ n an diverges when lim n→∞ an 6= 0 . Now, for the given series, an = 3(n + 1)2 n(n + 2) = 3n2 + 6n + 3 n2 + 2n . But then, lim n→∞ an = 3 6= 0 . Consequently, the Divergence Test says that the given series diverges . keywords: infinite series, Divergence Test, ra- tional function 013 10.0 points Determine whether the infinite series 4 − 2 + 1 − 1 2 + 1 4 · · · is convergent or divergent, and if convergent, find its sum. 1. convergent with sum 1 8 2. convergent with sum 8 3 correct 3. convergent with sum 3 8 Van Ligten (hlv63) – HW12 – Gilbert – (56650) 6 4. divergent 5. convergent with sum 8 Explanation: The infinite series 4 − 2 + 1 − 1 2 + 1 4 · · · = ∞ ∑ n = 1 a rn−1 is an infinite geometric series with a = 4 and r = −12 . But a geometric series ∞ ∑ n =1 a rn−1 (i) converges with sum a 1 − r when |r| < 1, (ii) and diverges when |r| ≥ 1 . Consequently, the given series is convergent with sum 8 3 . 014 10.0 points Determine whether the series ∞ ∑ n=0 ( 4 5 )n/2 is convergent or divergent, and if convergent, find its sum. 1. divergent 2. convergent with sum = √ 5 − 2√ 5 3. convergent with sum = √ 5√ 5 − 2 correct 4. convergent with sum = √ 5 − 2 2 5. convergent with sum = 2√ 5 − 2 Explanation: The infinite series ∞ ∑ n=0 ( 4 5 )n/2 is an infinite geometric series ∑ ∞ n=0 ar n with a = 1 and r = 2/ √ 5. But ∑ ∞ n=0 ar n is (i) convergent with sum a 1 − r when |r| < 1, and (ii) divergent when |r| ≥ 1 . So the given series is convergent with sum = √ 5√ 5 − 2 . 015 10.0 points Rewrite the series 2 ( 4 5 )2 sin 3 5 − 2 ( 4 5 )3 sin 4 6 + 2 ( 4 5 )4 sin 5 7 + . . . using summation notation. 1. sum = ∞ ∑ k =3 ( −4 5 )k−1 2 sin k k + 2 correct 2. sum = 20 ∑ k = 3 ( −4 5 )k−1 2 sin k k + 2 3. sum = 50 ∑ k = 2 ( 4 5 )k 2 sin(k + 1) k + 3 4. sum = ∞ ∑ k = 1 ( −4 5 )k 2 sin(k + 2) k + 4 5. sum = ∞ ∑ k = 3 ( −4 5 )k−1 2 sin k k + 1 Van Ligten (hlv63) – HW12 – Gilbert – (56650) 7 Explanation: The given series is an infinite series, so two of the answers must be incorrect because they are finite series written in summation notation. Starting summation at k = 3 we see that the general term of the infinite series is ak = 2 ( −4 5 )k−1 sin k k + 2 . Consequently, sum = ∞ ∑ k =3 ( −4 5 )k−1 2 sin k k + 2 . 016 10.0 points If the nth partial sum of an infinite series is Sn = 5n2 − 1 3n2 + 2 , what is the sum of the series? 1. sum = 23 12 2. sum = 11 6 3. sum = 7 4 4. sum = 5 3 correct 5. sum = 19 12 Explanation: By definition sum = lim n→∞ Sn = lim n→∞ (5n2 − 1 3n2 + 2 ) . Thus sum = 5 3 . 017 10.0 points Compare the values of the series A = ∞ ∑ n =2 6 n2.3 and the improper integral B = ∫ ∞ 1 6 x2.3 dx . 1. A > B 2. A = B 3. A < B correct Explanation: In the figure 1 2 3 4 5 . . .a2 a3 a4 a5 the bold line is the graph of the function f(x) = 6 x2.3 on [1, ∞) and the area of each of the rectan- gles is one of the values of an = 6 n2.3 . Clearly from this figure we see that a2 = f(2) < ∫ 2 1 f(x) dx, a3 = f(3) < ∫ 3 2 f(x) dx , Van Ligten (hlv63) – HW12 – Gilbert – (56650) 10 since it is only the behaviour of an as n → ∞ that’s important. Now, for the given series, N = 3 and an = n 3(lnn)2 . But by L’Hospital’s Rule applied twice, lim x→∞ x (lnx)2 = lim x→∞ 1 (2 lnx)/x = lim x→∞ x 2 lnx = lim x→∞ 1 2/x = ∞ . Consequently, the series diverges . (B) The function f(x) = 4 ln x x is continous and positive on [3, ∞); in addi- tion, since f ′(x) = 4 ( 1 − ln x x2 ) < 0 on [3, ∞), f is also decreasing on this inter- val. This suggests applying the Integral Test. Now, after substitution u = lnx, we see that ∫ t 3 f(x) dx = 2 [ (lnx)2 ]t 3 . Thus the improper integral ∫ ∞ 3 f(x) dx does not converge, and so the Integral Test ensures that the given series diverges . keywords: 022 10.0 points If the improper integral ∫ ∞ 1 1 xp dx converges, which of the following statements is (are) always true? (A) ∑ n 1 np converges; (B) ∑ n 1 np+1 diverges; (C) ∑ n 1 np−1 converges; (D) ∑ n 1 np−1 diverges; (E) ∑ n 1 np+1 converges. 1. A and E only correct 2. A, D and E 3. B and D only 4. A only 5. A, C and E only Explanation: To apply the Integral test we need to start with a function f which is positive, continuous and decreasing on [1, ∞). Then the integral test says that the improper integral ∫ ∞ 1 f(x) dx converges if and only if the infinite series ∞ ∑ n =1 f(n) converges. In the given example f(x) = 1 xp , Van Ligten (hlv63) – HW12 – Gilbert – (56650) 11 which is a function both continuous and pos- itive on [1, ∞). It will also be decreasing on [1, ∞) if f ′(x) < 0 for all x > 1. But f ′(x) = − p xp+1 , so f will be decreasing provided p > 0. On the other hand, the improper integral ∫ ∞ 1 1 xp dx converges if and only if lim n→∞ ∫ n 1 1 xp dx . exists. But ∫ n 1 1 xp dx = [ − 1 pxp−1 ]n 1 = 1 p ( 1 − 1 np−1 ) . Consequently, the improper integral ∫ ∞ 1 1 xp dx converges if and only if p > 1. Hence by the Integral test, the infinite series ∞ ∑ n = 1 1 xp converges if and only if p > 1. Now we can check which of the statements is (are) always true. (A) This is always true because of the Inte- gral test. (B), (E) Since p > 1 =⇒ p + 1 > 1, the Integral test ensures that ∑ n 1 np+1 converges. Thus (B) is false and (E) is true. (C), (D) The series ∑ n 1 np−1 converges if and only if p − 1 > 1, i.e., when p > 2. Since the convergence of the improper integral ∫ ∞ 1 1 xp dx guarantees only that p > 1, we see that state- ments (C) and (D) are true for some values of p and false for others. Consequently, of the statements, only A and E are always true.