# calPostClass 11.6-solutions - harris(tlh2479 PostClass 11.6...

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This preview shows page 1 out of 6 pages. Unformatted text preview: harris (tlh2479) – PostClass 11.6 – morales – (56815) This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – ﬁnd all choices before answering. 001 10.0 points Determine whether the following series ∞ n=1 5n (n + 2) 42n+1 is absolutely convergent, conditionally convergent, or divergent. 1. conditionally convergent correct 2. divergent 3. absolutely convergent Explanation: Since ∞ (−1) n=1 n−1 1 e1/n =− 4n 4 ∞ (−1)n n=1 2. conditionally convergent 3. absolutely convergent correct Explanation: With an = ∞ (−1)n n=1 e1/n , n we have to decide if the series 1. divergent e1/n n is absolutely convergent, conditionally convergent, or divergent. First we check for absolute convergence. Now, since e1/n ≥ 1 for all n ≥ 1, 1 e1/n ≥ > 0. 4n 4n 5n (n + 2) 42n+1 But by the p-series test with p = 1, the series we see that an+1 an 1 = 5n+1 (n + 2) 42n+1 ((n + 1) + 2) 42(n+1)+15n = n+2 n+3 5 16 lim an+1 an = Thus n→∞ 5 < 1, 16 so by the Ratio Test the given series is ∞ n=1 1 4n diverges, and so by the Comparison Test, the series ∞ e1/n 4n n=1 too diverges; in other words, the given series is not absolutely convergent. To check for conditional convergence, consider the series absolutely convergent . ∞ (−1)n f (n) 002 Determine whether the series ∞ (−1)n−1 n=1 n=1 10.0 points e1/n 4n is absolutely convergent, conditionally convergent or divergent. where e1/x . 4x Then f (x) > 0 on (0, ∞). On the other hand, f ( x) = 1+x 1 1/x e1/x . f (x) = − 3 e − 2 = −e1/x 4x 4x 4 x3 ′ harris (tlh2479) – PostClass 11.6 – morales – (56815) Thus f ′ (x) < 0 on (0, ∞), so f (n) > f (n + 1) for all n. Finally, since 2 Consequently, by the Ratio test, the given series is lim e1/x = 1 , absolutely convergent . x→∞ we see that f (n) → 0 as n → ∞. By the Alternating Series Test, therefore, the series 004 10.0 points ∞ To apply the ratio test to the inﬁnite series an , the value (−1)n f (n) n=1 n is convergent. Consequently, the given series is λ = lim n→∞ conditionally convergent . has to be determined. Compute λ for the series ∞ keywords: 003 an+1 an n=1 10.0 points 6n . 4n6 + 8 1. λ = 0 Determine whether the following series ∞ n=1 2. λ = (−7)n n! is absolutely convergent, conditionally convergent, or divergent. 3 2 3. λ = 3 4 4. λ = 6 correct 1. conditionally convergent 5. λ = 2. divergent 3. absolutely convergent correct Explanation: By algebra, Explanation: The given series has the form ∞ an = an , n=1 (−7)n . n! But then an+1 an = 7(n!) 7 = , (n + 1)! n+1 lim an+1 6n+1 = an 6n 4n6 + 8 4(n + 1)6 + 8 4n6 + 8 = 4(n + 1)6 + 8 4+ an+1 an = 0 < 1. . But 4 n+1 n 8 n6 6 8 +6 n Since in which case n→∞ 1 2 n+1 n 6 = 1 1+ n 6 →1 . harris (tlh2479) – PostClass 11.6 – morales – (56815) as n → ∞, we see that lim n→∞ converges if the series an+1 an ∞ n→∞ k=1 8 n6 6 4+ = lim 3 6 n+1 4 n 8 +6 n = 6. Consequently, 1 4k converges. But this last series is a geometric series with 1 < 1, r= 4 hence convergent. Consequently, the given series is λ=6. absolutely convergent . 005 10.0 points Determine whether the series ∞ k=1 006 sin2 (k ) (−1)k−1 4k 10.0 points Determine whether the following series is absolutely convergent, conditionally convergent or divergent ∞ n=1 1. divergent is absolutely convergent, conditionally convergent, or divergent. 2. conditionally convergent 3. absolutely convergent correct Explanation: To check for absolute convergence we have to decide if the series ∞ k=1 sin2 (k ) 4k is convergent. For this we can use the Comparison Test with ak sin2 (k ) = , 4k bk = 1 . 4k 1. absolutely convergent 2. divergent correct 3. conditionally convergent Explanation: Since the nth term of the series has the form (−1)n n n , an = 6 36 we use the Root Test, for then |an |1/n = For then 0 ≤ a k ≤ bk , k=1 sin2 (k ) 4k 1 61/n n →∞ 36 as n → ∞. The Root Test thus ensures that the series is since 0 ≤ sin2 (k ) ≤ 1. Thus the series ∞ (−n)n 62 n+1 divergent . 007 10.0 points harris (tlh2479) – PostClass 11.6 – morales – (56815) Consequently, for the given series, To apply the ratio test to the inﬁnite series an , the value λ=1. n λ = lim n→∞ an+1 an 008 has to be determined, 1 sin 5n + 7 n=1 10.0 points Decide whether the series Compute λ for the series ∞ 4 ∞ 1 n n=1 . 1 n+1 4n n n2 converges or diverges. 1. λ = 1 7 1. diverges 1 12 2. converges correct 2. λ = Explanation: The given series has the form 3. λ = 1 correct ∞ an , 1 4. λ = 5 |an |1/n = Explanation: By algebra, sin 1 n+1 1 n . 1 n+1 1 n+1 · 1 n sin 1 n · n . n+1 so sin lim n→∞ 1 n+1 1 n+1 while lim n→∞ 1 n sin 1 n lim |an |1/n = n→∞ n→∞ , n n+1 n+1 n n = lim n→∞ 1+ 1 n n = e < 3. = 1, converges . =1 009 5n + 7 5n + 5 + 7 10.0 points To apply the root test to an inﬁnite series an the value of 5n + 7 5(n + 1) + 7 n = lim n→∞ n + 1 e < 1, 4 Consequently, the Root Test ensures that the given series also. Thus n→∞ n since lim sin(x) lim = 1, x→0 x λ = lim 1 n+1 4 n in which case sin = But n2 But then 5. λ = 0 sin an n=1 1 n+1 =n 4 n n ρ = lim |an |1/n n→∞ . has to be determined. harris (tlh2479) – PostClass 11.6 – morales – (56815) Compute the value of ρ for the series ∞ n=1 6n + 5 n n 7 2 Explanation: The given series has the form . ∞ n=1 |an |1/n = 4 3. ρ = 21 lim |an |1/n = n→1 2 7 5. ρ = . n−1 n n , in which case 10 7 4. ρ = n2 But then 7 correct 2 2. ρ = n−2 n a n = 4n an , 35 2 1. ρ = 5 4 < 1, e2 since lim Explanation: After division, n→∞ 6n + 5 5 = 6 1+ n 6n n−2 n = lim n→∞ , n 1− 2 n n = 1 1 < 2. 2 e 2 Consequently, the Root Test ensures that the given series so 1/n (an ) = 1/n 5 6 1+ 6n 7 . 2 converges . But lim 6 5 1+ 6n 1/n n→∞ 1/n =1 011 Determine whether the series as n → ∞. Consequently, ρ= 010 ∞ 7 . 2 n=0 10.0 points 4 n n=1 n−2 n converges or diverges. is absolutely convergent, conditionally convergent, or divergent. n2 2. conditionally convergent 3. divergent Explanation: We use the Ratio Test with 1. diverges 2. converges correct (−2)n (2n)! 1. absolutely convergent correct Decide whether the series ∞ 10.0 points an = (−2)n . (2n)! harris (tlh2479) – PostClass 11.6 – morales – (56815) For then an+1 an for all x > 1. Thus f (x) is positive and decreasing for all x > 1. But the improper integral (−2)n+1 (2n)! (2n + 2)! (−2)n = ∞ (−2)n+1 −2 (2n)! . = = n (2n + 2)! (−2) (2n + 1)(2n + 2) Thus lim n→∞ an+1 an 2 = 0 < 1. (2n + 1)(2n + 2) = lim n→∞ 6 ∞ f (x) dx = 2 2 1 dx x ln(x) does not converge. By the Integral Test, therefore, the given series is not absolutely convergent. It could still converge conditionally, however. To use the Alternating Series Test to show that the series ∞ (−1)n f (n) n=2 Consequently, the series is absolutely convergent . is convergent, we have to check that (i) f (n) > f (n + 1) for n ≥ 2, 012 10.0 points Determine whether the series ∞ (−1)n n=2 6 n ln(n) converges conditionally, converges absolutely, or diverges. 3. series converges absolutely Explanation: The given series can be rewritten as ∞ n n=2 (−1)n f (n) , n=2 where f ( x) = 1 = (x ln(x))−1 > 0 x ln(x) for all x > 1. On the other hand, by the Chain and Product Rules, f ′ ( x) = − Since f is decreasing for all x ≥ 2, however, we see that n≥2 =⇒ 1 (ln(x) + 1) < 0 (x ln(x))2 f (n) > f (n + 1) . On the other hand, x→∞ 2. series converges conditionally correct 6 (−1) =6 n ln(n) x→∞ lim 1. series diverges ∞ (ii) lim f (x) = 0 . 1 = 0. x ln(x) Consequently, by the Alternating series Test, the given series is conditionally convergent . ...
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