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Unformatted text preview: MAT 127: Calculus C, Fall 2009 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 − ( − 1) n YES NO The sequence keeps on jumping between 0 and 2. (b) the sequence a n = 1 + cos( n ) n 2 YES NO Since cos( n ) /n 2 −→ 0 (because  cos( n )  ≤ 1), a n −→ 1 (c) the series ∞ summationdisplay n =1 cos(1 /n ) YES NO Since the sequence cos(1 /n ) −→ cos(0) = 1, not 0, the series diverges. (d) the series ∞ summationdisplay n =1 ( − 1) n √ n YES NO This series is alternating, 1 / √ n −→ 0, and 1 / √ n> 1 / √ n +1; so the Alternating Series Test applies. (e) the series ∞ summationdisplay n =1 n + sin n n 2 YES NO ( n + sin( n )) /n 2 looks like n/n 2 = 1 /n : ( n + sin( n )) /n 2 1 /n = n 2 + n sin( n ) n 2 = 1 + sin( n ) n −→ 1. Since ∞ summationdisplay n =1 1 n diverges by the pseries test ( p =1 ≤ 1), so does ∞ summationdisplay n =1 n + sin n n 2 by the Limit Comparison Test. The Comparison Test can also be used:  sin( n )  ≤ 1 = ⇒ n + sin( n ) n 2 ≥ n − 1 n 2 ≥ n/ 2 n 2 = 1 2 · 1 n if n ≥ 2 . Since 1 2 ∞ summationdisplay n =1 1 n diverges by the pseries test, so does ∞ summationdisplay n =1 n + sin n n 2 by the Comparison Test. Alter natively, ∞ summationdisplay n =1 n + sin n n 2 = ∞ summationdisplay n =1 1 n + ∞ summationdisplay n =1 sin n n 2 The first series on the righthand side diverges by the pseries test, while the last series on the right hand side converges by the pseries test, by the Comparison Test, and the Absolute Convergence Test: ∞ summationdisplay n =1 vextendsingle vextendsingle vextendsingle vextendsingle sin n n 2 vextendsingle vextendsingle vextendsingle vextendsingle = ∞ summationdisplay n =1  sin n  n 2 converges because 0 ≤  sin( n )  /n 2 ≤ 1 /n 2 and ∞ summationdisplay n =1 1 n 2 converges. Since the sum of a divergent series and a convergent series is divergent, our series diverges. Problem 2 (20pts) Find Taylor series expansions of the following functions around the given point. In each case, determine the radius of convergence of the resulting power series and its interval of convergence. (a; 10pts) f ( x ) = x 3 around x = 2 In this case, all derivatives can be computed: f ( ) ( x ) = x 3 = ⇒ f ( ) (2) = 8 , f ( 1 ) ( x ) = 3 x 2 = ⇒ f ( 1 ) (2) = 12 , f ( 2 ) ( x ) = 6 x = ⇒ f ( 2 ) (2) = 12 , f ( 3 ) ( x ) = 6 = ⇒ f ( 3 ) (2) = 6 , and f ( n ) ( x )=0 if n ≥ 4. So by the Main Taylor Formula: f ( x ) = ∞ summationdisplay n =0 f ( n ) ( a ) n !...
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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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