227_sample_exam_3

227_sample_exam_3 - : 1 : (c) (5 pts) Find the Taylor...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 227 Sections 4 and 5 Sample Exam on Series 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 n 2 n 2 1 (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 ! 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 : (b) (5 pts) Find a bound on the error of approximation in (a) on the interval & 0 : 1 ± x ±
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: : 1 : (c) (5 pts) Find the Taylor series for f: 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. 1...
View Full Document

Ask a homework question - tutors are online