242hw8solns - 1 2 not 0 6 ∞ X n =1-1 n(1 1 n n 4 DIVERGES...

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Math 242 Homework 8 Solutions (may be somewhat terse) In Questions 1 to 4, determine whether the series converges or diverges. Give reasons. 1. X n =1 n - 1 n · 4 n CONVERGES. Try basic comparison with b n = n n · 1 4 n = 1 4 n , or limit comparison with the same b n . Or try the ratio test. 2. X n =1 ( - 1) n sin ± π n ² CONVERGES. This is an alternating series, and when n → ∞ , we have π n 0, so sin π n 0. 3. X n =1 1 + sin n 10 n CONVERGES. Basic comparison using 0 10 n 1 + sin n 10 n 2 10 n . 4. X n =1 n n 3 + 2 DIVERGES. Limit comparison is probably easiest. Can compare with b n = n n 3 / 2 = 1 n 1 / 2 . In Questions 5 to 8, determine whether the series converges absolutely, con- verges conditionally, or diverges. Give reasons. 5. X n =1 ( - 1) n n 1 + 2 n 1
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DIVERGES. Using the “divergence test”, note that n 1 + 2 n approaches
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Unformatted text preview: 1 2 , not 0. 6. ∞ X n =1 (-1) n (1 . 1) n n 4 DIVERGES. In fact, (1 . 1) n n 4 approaches ∞ . Easiest for this problem is probably to use the ratio test. 7. ∞ X n =1 (-1) n n √ n 4 + 2 CONDITIONALLY CONVERGES. The “positive part” is less than b n = n √ n 4 = n n 2 = 1 n . We can use the limit comparison test to show that the sum of the absolute values is divergent. 8. ∞ X n =1 (-1) n n 2 2 n n ! ABSOLUTELY CONVERGES. Use the ratio test. Note that ( n + 1) 2 2 n +1 ( n + 1)! · n ! n 2 2 n = ( n + 1) 2 n 2 · 1 n + 1 · 2 1 , a product of three “chunks” that approach 1, 0, and 2 respectively. 2...
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This note was uploaded on 12/07/2011 for the course MATH 242 taught by Professor Wang during the Spring '08 term at University of Delaware.

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242hw8solns - 1 2 not 0 6 ∞ X n =1-1 n(1 1 n n 4 DIVERGES...

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