This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Mathematics 1B. Fall Semester 2006 Professor: Daniel Tataru Midterm 2 Solutions 1 (20) . Determine the interval of convergence of the following series. Do they converge at endpoints ? a ) X n =1 ( x 1) 2 n n 4 n Solution: Using the ratio test we compute lim n ( x 1) 2( n +1) n + 1 4 n +1 ( x 1) 2 n n 4 n = lim n ( x 1) 2 4 n n + 1 = ( x 1) 2 4 The limit is less than 1 if  x 1  < 2. Hence the radius of convergence is R = 2. At the endpoints we have x 1 = 2 and the series becomes X n =1 1 n which is a divergent pseries. Thus the interval of convergence is ( 1 , 3). b ) X n =2 ln n + 1 n 1 x n Solution: Using the Taylor expansion for ln(1 + x ) we write ln n + 1 n 1 = ln 1 + 2 n 1 = 2 n 1 2 ( n 1) 2 + Then for the ratio test we compute lim n ln n + 2 n x n +1 ln n + 1 n 1 x n = x lim n 2 n 2 n 2 + 2 n 1 2 ( n 1) 2 + = x The limit is less than 1 if  x  < 1. Hence the radius of convergence is R = 1. At the endpoint x = 1 we obtain the alternating series X n =2 ( 1) n ln 1 + 2 n 1 Due to the expansion above we have ln n + 1 n 1 & 0 as n therefore the series converges by the alternating test. At the endpoint x = 1 we obtain the series X n =2 ln 1 + 2 n 1 Due to the expansion above this is comparable to the harmonic series X n =2...
View
Full
Document
This note was uploaded on 02/19/2012 for the course MATH 1 taught by Professor Wilkening during the Spring '08 term at University of California, Berkeley.
 Spring '08
 WILKENING

Click to edit the document details