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Unformatted text preview: MAT 127: Calculus C, Fall 2010 Solutions to Final Exam Problem 1 (10pts) Determine whether each of the following sequences or series converges or not. In each case, clearly circle either YES or NO , but not both. Each correct answer is worth 2 points. (a) the sequence a n = 1 + cos 3 n n YES NO Since cos 3 ( n ) /n 0 (because  cos 3 ( n )  1), a n 1 (b) the sequence a n = n 2 (1 e 1 /n ) YES NO lim n a n = lim n 1 e 1 /n (1 /n ) 2 = lim x + 1 e x x 2 = lim x + e x 2 x = 1 2 lim x + 1 x = . The third equality uses lHospital, which is applicable here because (1 e x ) , x 2 0 as x 0. Alternatively, using e x = summationdisplay n =0 x n n ! , we find that a n = n 2 ( 1 (1 + 1 1! n + 1 2! n 2 + 1 3! n 3 + . . . ) ) = n 1 2 1 6 n . . . where . . . involve 1 /n 2 , 1 /n 3 , and so on. As n , a n thus approaches and so diverges. (c) the series summationdisplay n =1 n + ( 1) n n 2 + 1 YES NO n + ( 1) n n 2 + 1 looks like n/n 2 = 1 /n : ( n + ( 1) n ) / ( n 2 + 1) 1 /n = n 2 + ( 1) n n n 2 + 1 = 1 + ( 1) n /n 1 + 1 /n 2 1 Since summationdisplay n =1 1 n diverges by the pseries test ( p =1 1) and both series are nonnegative, by the Limit Comparison Test our series also diverges. Alternatively, summationdisplay n =1 n + ( 1) n n n 2 + 1 = summationdisplay n =1 n n 2 + 1 + summationdisplay n =1 ( 1) n n 2 + 1 The last series on RHS converges by the Alternating Series Test (it is alternating and approaching 0, and the absolute values of its terms are decreasing). The first series on RHS diverges by Limit Comparison to the pseries summationdisplay n =1 1 n , Comparison to summationdisplay n =1 n n 2 + n 2 = 1 2 summationdisplay n =1 1 n , or by the Integral Test with f ( x ) = x/ ( x 2 + 1). Since the sum of a divergent series and a convergent series is divergent, our series diverges. (d) the series summationdisplay n =1 ( 1) n n 2 n + 1 YES NO Since the sequence ( 1) n n 2 n + 1 = ( 1) n 1 2 + 1 /n 1 2 does not approach 0, our series diverges by the Test for Divergence. (e) the series summationdisplay n =1 2 n 3 n + 5 n YES NO 2 n 3 n +5 n looks like 2 n 5 n = parenleftbigg 2 5 parenrightbigg n : 2 n / 3 n +5 n (2 / 5) n = 1 3 n +5 n / 5 n = 1 radicalbig (3 / 5) n +(5 / 5) n 1 Since 2 &lt; 5, the geometric series summationdisplay n =1 parenleftbigg 2 5 parenrightbigg n converges. Since both series are nonnegative, by the Limit Comparison Test our series also converges. Problem 2 (10pts) Answer Only. Put your answer to each question in the corresponding box in the simplest possible form. No credit will be awarded if the answer in the box is wrong; partial credit may be awarded if the answer in the box is correct, but not in the simplest possible form....
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This note was uploaded on 01/25/2012 for the course MAT 127 taught by Professor Guanyushi during the Fall '07 term at SUNY Stony Brook.
 Fall '07
 GuanYuShi
 Calculus

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