B If an infinite series n a n converges then lim n a n 0 But for the given

B if an infinite series n a n converges then lim n a

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( B ) If an infinite series n a n converges, then lim n → ∞ a n = 0 . But for the given series in ( B ), lim n → ∞ a n = lim n → ∞ n - 8 n + 4 = 1 . Consequently, the series in ( B ) does not con- verge. ( C ) After division, 8 n + 4 7 n - 6 = 8 + 4 n 7 - 6 n , so the inequality 8 n + 4 7 n - 6 n 8 7 n holds for all n . But the series n = 1 8 7 n is a geometric series whose common ratio r = 8 7 . Now r > 1, so this geometric series does not converge. Hence by the comparison test the series in ( C ) does not converge. Consequently, of the given infinite series, only A converges. 015 10.0 points Which of the following series ( A ) n = 1 (5 sin n ) 2 6 n 2 + 2 ( B ) n = 2 5 n 8 n 2 ln n + 2 ( C ) n = 1 5 sin n π 6 n + 8 diverge(s)? 1. B only correct 2. A only 3. B and C 4. A, B, and C 5. A and C Explanation: ( A ) Since (sin θ ) 2 1, the inequality 0 (5 sin n ) 2 6 n 2 + 2 < 25 1 n 2
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cadena (jc59484) – HW11 – lawn – (55930) 11 holds for all n . But by the Integral test, the infinite series n = 1 1 n 2 converges. Consequently, the comparison test ensures that the series n = 1 (5 sin n ) 2 6 n 2 + 2 converges. ( B ) After division, 5 n 8 n 2 ln n + 2 = 5 8 n ln n + 2 n . But n ln n 5 8 n ln n + 2 n -→ 5 8 > 0 , so n ln n 5 n 8 n 2 ln n + 2 -→ 5 8 > 0 . On the other hand, by the Integral test the infinite series n = 2 1 n ln n does not converge, hence by the limit compari- son test, the given series in ( B ) does converge. ( C ) As the Integral test shows, the series n = 1 5 6 n + 8 diverges. However, sin n π = 0 for all n . Hence every term in the given series in ( C ) reduces to 0, i.e. , the series becomes 0 + 0 + 0 + . . . + 0 + . . . , which obviously converges (and has sum 0). Hence the given series in ( C ) actually con- verges. Consequently, of the given infinite series, only B diverges. 016 10.0 points Find all values of p for which the infinite series n = 1 7 n 5 8 n 8 + 1 p converges. 1. p < - 3 4 2. p > 3 3. p > 3 4 4. p > 1 3 correct 5. p < - 1 3 Explanation: After division, 7 n 5 8 n 8 + 1 = 7 8 n 3 + 1 n 5 , so n 3 7 n 5 8 n 8 + 1 -→ 7 8 > 0 as n → ∞ . Thus by the Limit Comparison test, the infinite series n = 1 7 n 5 8 n 8 + 1 p will converge if and only if the infinite series n = 1 1 n 3 p converges. But by the Integral test we know that the series n = 1 1 n p
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cadena (jc59484) – HW11 – lawn – (55930) 12 converges if and only if p > 1. Consequently, the given series will converge if and only if 3 p > 1, i.e. , when p > 1 3 . 017 10.0 points Determine all values of p for which the series n = 4 3 n p ln n converges. 1. p > 2 2. p < 0 3. p > 0 4. p 1 5. p > 1 correct 6. p < - 2 Explanation: By the Divergence Test, the series n = 4 3 n p ln n diverges when p < 0 because then lim n → ∞ 1 n p ln n = . Now suppose p 0. If 0 p 1, then 3 n ln n 3 n p ln n , so by the Comparison Test the given series diverges because, by the Integral Test with f ( x ) = 3 x ln x , 4 x < , the series n = 4 3 n ln n diverges.
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