Unformatted text preview: 3 1c ). d. Deﬁne a 2 n1 = 1 n 3 and a 2 n = 1 n 2 . Then a 2 n a 2 n1 = 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.
 Spring '07
 Staff

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