Assignment 10 Solutions

# Assignment 10 Solutions - MATH 138 Calculus 2 Solutions to...

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Unformatted text preview: MATH 138 Calculus 2, Solutions to Assignment 10 1: Determine which of the following series converge. (a) ∞ X n =0 √ n 2 n 2 + 1 Solution: For n ≥ 1 we have 0 ≤ √ n 2 n 2 + 1 ≤ √ n 2 n 2 = 1 2 n 3 / 2 , and we know that X 1 2 n 3 / 2 converges ( since it is a constant multiple of the p-series with p = 3 2 ) , and so X √ n 2 n 2 + 1 converges by the C.T. (b) ∞ X n =2 1 n √ ln n Solution: Let f ( x ) = 1 x √ ln x so that a n = f ( n ). For x > 1, f ( x ) is positive, continuous and decreasing. Making the substitution u = ln x so that du = 1 x dx , we have Z ∞ 2 f ( x ) dx = Z ∞ 2 dx x √ ln x = Z ∞ ln 2 du √ u = h 2 √ u i ∞ ln 2 = ∞ . Since Z ∞ 2 f ( x ) dx diverges, the series ∞ X n =2 1 n √ ln n diverges too, by the I.T. (c) ∞ X n =1 n n 3 n n ! Solution: Let a n = n n 3 n n ! . Then a n +1 a n = ( n + 1) n +1 3 n +1 ( n + 1)! 3 n n ! n n = 1 3 n + 1 n n-→ e 3 < 1, so X a n converges by the R.T. ( If you don’t remember that lim n →∞ n + 1 n n = e , then use l’Hˆ opital’s Rule ) . 2: For each of the following series, determine whether it converges, and if so then determine whether it converges absolutely. (a) ∞ X n =1 (- 1) n √ n n + 2 Solution: Let a n = (- 1) n √ n n + 2 . Then | a n | = √ n n + 2 . Since the sequence | a n | decreases to 0, X a n converges by tha A.S.T. On the other hand, if we let b n = √ n n = 1 √ n , then we have | a n | b n-→ 1 and X b n diverges ( it is a p-series with p =...
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Assignment 10 Solutions - MATH 138 Calculus 2 Solutions to...

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