A The Ratio Test is the better one to use because a n 1 a n 3 n n 1 n 1 n 1 n n

# A the ratio test is the better one to use because a n

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( A ) The Ratio Test is the better one to use because a n +1 a n = 3 n ! ( n + 1)! ( n + 1) n +1 n n . Now n ! ( n + 1)! = 1 n + 1 , while ( n + 1) n +1 n n = ( n + 1) n + 1 n n . Thus a n +1 a n = 3 n + 1 n n -→ 3 e > 1 as n → ∞ , so series ( A ) diverges. ( B ) The Root Test is the better one to apply because | a n | 1 /n = 6 n + 2 -→ 0 as n → ∞ , so series ( B ) converges. ( C ) Again the Root Test is the better one to apply because of the n th powers. For then | a n | 1 /n = 6 5 3 n 2 n + 4 -→ 9 5 > 1 as n → ∞ , so series (C) diverges. Consequently, of the given infinite series, only A and C diverge. 019 10.0 points Decide which of the following series diverge. ( A ) n = 1 n 7 + 8 n + 2 2 7 n ( B ) n = 1 7 n + 8 n 2 + 7 n ( C ) n = 1 n - 7 n + 8 8 7 n 1. B only 2. C only correct 3. all of them 4. A only 5. A and B 6. A and C 7. B and C 8. none of them Explanation: We compute one of lim n → ∞ a n +1 a n , lim n → ∞ ( a n ) 1 /n for each of the given series. ( A ) The ratio test is the better one to use: a n +1 a n = 2 7 ( n + 1) 7 + 8 n 7 + 8 n + 2 ( n + 1) + 2 . But ( n + 1) 7 + 8 n 7 + 8 -→ 1 as n → ∞ , while n + 2 ( n + 1) + 2 -→ 1 . Thus lim n → ∞ a n +1 a n = 2 7 , so series ( A ) converges.
rabinovich (ahr454) – HW12 – um – (55915) 11 ( B ) The root test is the better one to use: ( a n ) 1 /n = 7 n + 8 n 2 + 7 -→ 0 as n, → ∞ , so series ( B ) converges. ( C ) Again the ratio test is the better one to use: a n +1 a n = 8 7 n + 1 - 7 n - 7 n + 8 n + 1 + 8 . But n + 1 - 7 n - 7 , n + 8 n + 1 + 8 -→ 1 as n → ∞ . Thus series ( C ) diverges. Consequently, of the given infinite series, only C diverges . 020 10.0 points Which of the following infinite series con- verges conditionally? 1. n = 1 - 7 6 n 2. n = 1 7( - n ) n ( n + 3) n 3. m = 1 8 7 m - 6 4. n = 1 cos( n π ) 3 n 8 n + 6 n 5. k = 1 ( - 1) k 6 8 k ln( k ) + 7 correct Explanation: The question requires us to find which of the five series has the property that n a n converges but n | a n | does not converge. (i) In the case a k = ( - 1) k 6 8 k ln( k ) + 7 the alternating series test applies since 6 8 k ln( k ) + 7 6 8( k + 1) ln( k + 1) + 7 and lim k → ∞ 6 8 k ln( k ) + 7 = 0 . Thus k = 1 ( - 1) k 6 8 k ln( k ) + 7 converges. On the other hand, if we set b k = 1 k ln( k ) , then lim k → ∞ b k | a k | = 8 k ln( k ) + 7 6 k ln( k ) = 4 3 = 0 , while by the integral test, k = 2 b k = k = 2 1 k ln( k ) diverges. The Limit Comparison test thus says that k = 1 | a k | diverges. Consequently, the series k = 1 ( - 1) k 6 8 k ln( k ) + 7 converges conditionally but not absolutely. (ii) In the case a n = 7( - n ) n ( n + 3) n
rabinovich (ahr454) – HW12 – um – (55915) 12 we see that lim n → ∞ | a n | = 7 e - 3 = 0 ; but then a n oscillates between - 7 e - 3 and 7 e - 3 as n → ∞ , so cannot converge to 0. Consequently, neither of n = 1 a n , n = 1 | a n | converges. (iii) Since | a m | = a m in the case a m = 8 7 m - 6 , both series converge or both do not converge. But by the Comparison and p -series tests with p = 1, the series m = 1 8 7 m - 6 does not converge, so neither of m = 1 a m , m = 1 | a m | converges. (iv) In the case a n = cos( n π ) 3 n 8 n + 6 n = ( - 1) n 3 n 8 n + 6 n we apply the root test to the series n | a n | .

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• Fall '11
• Gramlich
• Accounting, Mathematical Series, lim