quiz06_s103_solns

quiz06_s103_solns - lim n ( | (-1) n (8 1 /n 2-1) n | ) 1...

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Name: Math 1B Quiz 6 Solutions Section 103 October 13, 2009 1. (4 points) Test the series for convergence or divergence: X n =1 tan(1 /n ) n Use the Limit Comparison Test with the convergent p-series 1 n 3 / 2 . Calculate: lim n →∞ tan(1 /n ) n 1 n n = lim x →∞ tan(1 /x ) 1 x = lim x →∞ ( - 1 /x 2 ) sec 2 (1 /x ) ( - 1 /x 2 ) = sec 2 (0) = 1 . Since the limit is strictly between 0 and (0 < 1 < ), the LCT applies. Therefore, since 1 n 3 / 2 converges, tan(1 /n ) n converges. 2. (4 points) Test the series to determine whether it diverges, converges conditionally, or converges absolutely. X n =1 ( - 1) n (8 1 /n 2 - 1) n Use the root test. Calculate:
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Unformatted text preview: lim n ( | (-1) n (8 1 /n 2-1) n | ) 1 /n = lim n 8 1 /n 2-1 = 0 . Since 0 &lt; 1, the root test applies, and we can conclude that the series converges absolutely. 3. (2 points) (a) Find an x such that the power series n =1 n ! x n converges. Solution: x = 0. (b) Find an x such that the power series n =1 ( 1 2 ) n x n diverges. For any x with | x | &gt; 2, the terms ( x/ 2) n diverge, so the power series diverges. 1...
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This note was uploaded on 03/25/2011 for the course MATH 1B taught by Professor Reshetiken during the Fall '08 term at University of California, Berkeley.

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