{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

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

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

1 Solutions to Exam #1 - Version α 1) a) 2 n 2 n=1 n +1 (-1) 3n +1 diverges since the terms don’t approach 0 as n goes to infinity. The absolute value of a term is 1 3 1 2 2 + + n n which approaches 1/3 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. b) n=1 1 n(n +2) converges by the comparison test since 2 n 1 2) n(n 1 + and = 1 n 2 n 1 converges by the integral test. c) 1 (2 )! n n n n = 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 1 1 ! 0 ( 1)! 1 n n a n a n n + = = + + 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) 3 is a geometric series. If we substitute u=-x/3, 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=3, since we need |u|=|x/3|<1 or |x|<3. Or we can use the ratio test to find the radius of convergence.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern