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Unformatted text preview: Calculus Math 21C, Fall 2010 Midterm I: Solutions 1. [20%] Do the following sequences { a n } converge or diverge as n ? If a sequence converges, find its limit. Justify your answers. (a) a n = 2 + ( 1) n ; (b) a n = n e n ; (c) a n = 1 + 2 n n . Solution. (a) The sequence diverges since it oscillates between 1 and 3. For example, if = 1, there is no number L such that  a n L  < for all sufficiently large n , since then we would have both  1 L  < 1 (or < L < 2) and  3 L  < 1 or (2 < L < 4), which is impossible. So there is no L that satisfies the definition of the limit. (b) Both the numerator n and the denominator e n are differentiable functions that diverge to as n . Differentiating the numerator and denominator with respect to n , we get by lH ospitals rule that lim n n e n = lim n 1 e n = 0 . Thus, the sequence converges to 0. (c) The sequence converges to e 2 . To show this, let b n = ln a n = n ln 1 + 2 n . Then, writing n = 1 /x , we have lim n b n = lim x ln(1 + 2 x ) x . The fraction is an indeterminate limit (0 / 0) of differentiable functions, so by lH ospitals rule lim x ln(1 + 2 x ) x = lim x 2 / (1 + 2 x ) 1 = lim x 2 1 + 2 x = 2 . 1 Thus, b n 2 as n . Since a n = e b n and e x is a continuous function, it follows that lim n a n = e lim n b n = e 2 . More generally, for any < x < , we have the limit e x = lim n 1 + x n n 2 2. [20%] Determine whether the following series converge or diverge. You can use any appropriate test provided that you explain your answer....
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