227_sample_exam_3_spring_2009[1]

# 227_sample_exam_3_spring_2009[1] - ± x ± 1(c(10 pts Find...

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Math 227.04 Sample Exam on Series Due as hw on Monday, May 4 Name Instructions. You must show your work in order to receive full credit. Please display your answers and erase or cross out any work you wish to be ignored. P 1 n =1 2( 1 3 ) n . 2. (10 pts each) Test the following series for convergence/divergence. In each case, state clearly which test you are using. If you are applying a comparison test, declare your benchmark series. (a) 1 X n =1 ( 1) n n 2 n 2 1 1

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(b) 1 X n =1 ( 2) n ( n + 1) n (c) 1 X n =2 1 n ln n (d) 1 X n =1 3 n n ! 2
3. (10 pts) Find the basic interval of convergence for the following power series. Do not bother to test the endpoints. 1 X n =0 n ( x 2) n ( n 2 + 1)2 n : 4. Let f ( x ) = ln(1 x ) : (a) (10 pts) Find the third order Taylor polynomial for f centered at a = 0 : 3

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(b) (5 pts) Find a bound on the error of approximation in (a) on the interval & 0 : 1
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Unformatted text preview: ± x ± : 1 : (c) (10 pts) Find the Taylor series for f: 4 5. (10 pts) True or False. Each correct answer is worth 2 points, each incorrect answer is worth negative 2 points. Each unanswered question is worth 0 points. Your total for this problem will be rounded up to zero if it is negative. Assume below that a n and b n are all positive. (a) If 1 P n =1 a n and 1 P n =1 b n converge, then 1 P n =1 a n b n must also converge. (b) If lim n !1 a n +1 =a n ! L < 1 , then lim n !1 a n = 0 : (c) If 1 P n =1 a n and 1 P n =1 b n both diverge, then so must 1 P n =1 a n b n : (d) If 1 P n =1 ( & 1) n a n is conditionally covergent, then a n +1 =a n ! 1 as n ! 1 : (e) If lim n !1 a n b n = 0 and 1 P n =1 b n diverges, then 1 P n =1 a n must also diverge. 5...
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227_sample_exam_3_spring_2009[1] - ± x ± 1(c(10 pts Find...

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