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Unformatted text preview: Problems graded: 14.6 (10 points), 15.6 (15 points), 17.5 (10 points), 17.13 (10 points); the other problems were graded on completion to be worth five points each 14.2. (a) For n ≥ 2, ( n − 1) ≥ . 5 n so n 1 n 2 ≥ 1 2 n . However, Σ 1 n diverges to ∞ by Formula 2 of Example 1 of this section ( p = 1); therefore, so does Σ 1 2 n . By the Comparison Test (and the fact that the convergence of a series is independent of its first term), Σ( n − 1) /n 2 diverges to ∞ . (e) We note that the ratio of the ( n + 1)th term to the n th term is ( n +1) 2 / ( n +1)! n 2 /n ! = (( n +1) /n ) 2 n +1 which approaches 0 as n approaches ∞ (by the limit rules of Section 9; n +1 n = 1 + n 1 which approaches 1 as n approaches ∞ ) so the Ratio Test tells us Σ n 2 /n ! converges. (g) We note that the ratio of the ( n + 1)th term to the n th term is ( n +1) / 2 n +1 n/ 2 n = 1+ n 1 2 which approaches 1 2 as n approaches ∞ (by the limit rules of Section 9) so the Ratio Test tells us Σ n/ 2 n converges. 14.6. (a) Suppose Σ  a n  converges; by Theorem 14.4, it satisfies the Cauchy criterion. To show Σ a n b n converges we let ǫ > 0 and B > 0 be an upper bound for  b n  ; we seek to show Σ a n b n satisfies the Cauchy criterion. As Σ  a n  satisfies this criterion, there exists M such that for n > m > M , Σ n k = m  a k  < ǫ/B . Therefore, for such m and n ,  Σ n k = m a k b k  ≤ Σ n k = m  a k  b k  (by the triangle inequality) ≤ B Σ n k = m  a k  < Bǫ/B = ǫ so we note that Σ a n b n satisfies the Cauchy criterion and therefore converges by Theorem 14.4. (b) Corollary 14.7 states that absolutely convergent series are convergent; letting Σ a n be an absolutely convergent series, this is a special case of part (a) where b n = 1 for each n . 14.7. If Σ a n is a convergent series of nonnegative numbers and p > 1 the Cauchy criterion tells us that for each ǫ > 0 there exists M > 0 such that for n ≥ m > N ,  Σ n j = m a j  < ǫ . Supposing ǫ < 1, the a j are nonnegative so the absolute value sign is redundant; this tells us a j ∈ [0 , 1) for j > N and therefore a p j ≤ a j so  Σ n j = m a p j  ≤ Σ n j = m a j < ǫ giving the desired convergence. 14.10. We let a n = 2 n ( 1) n ; we note that the n th root of a n is equal to 2 for even n and . 5 for odd n so the limit superior of  a n  1 /n is 2; therefore, the series converges by the Root Test. However, a n +1 a n = 2 2 n +1 for n odd and 2 (2 n +1) for n even; as 2 2 n +1 diverges to infinity and 2 (2 n +1) converges to zero, we have that lim sup  a n +1 /a n  = ∞ and lim inf  a n +1 /a n  = 0 so the Ratio Test gives no information. (NOTE: Answers may vary.) 1 14.12. We choose the sequence of positive integers { n k } as follows....
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This note was uploaded on 10/21/2010 for the course MATH 131A 262447201 taught by Professor Neiman during the Spring '10 term at UCLA.
 Spring '10
 NEIMAN

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