Spring 2006 - Terras' Class - Exam 1 (Version B)

Spring 2006 - Terras' Class - Exam 1 (Version B) -...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
1 Solutions to Exam #1 - Version α 1) a) n=1 1 n(n + 5) converges by the comparison test since 2 11 n(n 5) n + and = 1 n 2 n 1 converges by the integral test. b) 2 n 2 n=1 n+ 1 (-1) 2n +1 diverges since the terms don’t approach 0 as n goes to infinity. The absolute value of a term is 2 2 1 21 n + + which approaches 1/2 by l’Hôpital’s rule and thus the actual term has no limit. Then even terms go to 1/2 and the odd ones to -1/2. c) 1 (2 )! = converges by comparison with e = = 1 n n! 1 , since ! 1 ! 1 ) 1 ( ) 1 2 )( 2 ( n n n n n n n n + = " " (2n)! n n . e = = 1 n n! 1 converges by the ratio test as 1 !0 (1 ) ! 1 a an + == ++ i , as n goes to infinity. You could also use the ratio test on the original series, but that is a little harder. 2) f(x)= n n n n=1 x (-1) 5 is a geometric series. If we substitute u=-x/5, then f(x)= n n=1 1 u=u 1-u , converging for |u|<1. a) The radius of convergence of the series for f(x) is R=5, since we need |u|=|x/5|<1 or |x|<5. Or we can use the ratio test to find the radius of convergence.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 3

Spring 2006 - Terras' Class - Exam 1 (Version B) -...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online