solutionshw8-555-2011

solutionshw8-555-2011 - 3 1-c d Define a 2 n-1 = 1 n 3 and...

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Solutions homework 8. Page 188 Problem 6. R n 1 1 x dx = 2 n - 2 → ∞ as n → ∞ . The divergence of the series follows now from n +1 X k =1 1 n Z n 1 1 x dx. A simpler proof follows from 1 n 1 n and the fact that 1 n diverges. Page 188 Problem 7. From ln x x - 1 it follows that ln(1+ n ) (1+ n ) - 1 = n . Hence 1 ln(1+ n ) 1 n , which shows that the series 1 ln(1+ n ) diverges. Page 188 Problem 10. Multiply the first n quotients to get a 2 a 1 · a 3 a 2 ··· a n +1 a n b 2 b 1 · b 3 b 2 ··· b n +1 b n , which, by canceling common terms, is the same as a n +1 a 1 b n +1 b 1 . The convergence of a n follows now from the comparison test. Page 188 Problem 11. a. If a n +1 a n 1 ultimately, then take now b n = 1. Then b n diverges and a n +1 a n b n +1 b n ultimately implies, via the contrapositive of Problem 10, that a n diverges. b. If lim sup a n +1 a n < 1, then there exists 0 < r < 1 such that a n +1 a n < r ultimately. Take now b n = r n and apply Problem 10 to get convergence. c. Let 0 < c < 1 and define a 2 n - 1 = c n and a 2 n = 2 c n . Then a 2 n a 2 n - 1 = 2 and a 2 n +1 a 2 n = c 2 . It follows that lim sup a n +1 a n = 2 > 1, but the series converges (its sum is equal to
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Unformatted text preview: 3 1-c ). d. Define a 2 n-1 = 1 n 3 and a 2 n = 1 n 2 . Then a 2 n a 2 n-1 = n , but the series converges (it is the sum of two convergent series) Page 188 Problem 13. a n +1 a n = c n +1 , so lim sup a n +1 a n = 0, so the series converges by the Ratio test. Page 193 Problem 7. Assume | b n | ≤ M . Then | a n b n | ≤ M | a n | . Now ∑ M | a n | converges, so by the comparison test ∑ | a n b n | converges. Page 193 Problem 8. a. From Problem 7 above we know that ∑ 1 ln(1+ n ) diverges. On the other hand, by the alternating series test, ∑ (-1) n ln(1+ n ) converges, so it is conditionally convergent. b. Since lim sup n q 1 n n = 0 it follows that ∑ 1 n n converges. Now | (sin n ) n | ≤ 1, so that the given series converges absolutely by Problem 7. 1...
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This note was uploaded on 02/05/2012 for the course MATH 555 taught by Professor Staff during the Spring '07 term at South Carolina.

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