quiz_sl1

# quiz_sl1 - M408D Quiz Solution You must show your work to...

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

M408D Quiz Solution You must show your work to get full credit. 1. Determine whether the series X n =1 ( - 1) n n n 3 + 2 is absolutely convergent, condi- tionally convergent or divergent. We always check absolutely convergence rst because absolutely convergence implies con- vergence. Let a n = ( - 1) n n n 3 + 2 , then | a n | = ( - 1) n n n 3 + 2 = n n 3 + 2 n n 3 + 3 n 3 = 1 2 n = 1 2 n 1 / 2 . Since X 1 n 1 / 2 is divergent ( p -series with p = 1 / 2 1 ), X 1 2 n 1 / 2 is also divergent. Therefore by comparison test, the series X | a n | is divergent. The series a n is an alternating series with b n = n n 3 + 2 . We can use the alternating series test to check convergence. lim n →∞ b n = lim n →∞ n n 3 + 2 = lim n →∞ 1 / n p 1 + 2 /n 3 = 0 Let f ( x ) = x 2 x 3 + 2 . Then f 0 ( x ) = 2 x 4 + 2 x - 3 x 4 ( x 3 + 2) 2 = - x 4 + 2 x ( x 3 + 2) 2 < 0 , x 2 Since square root is an increasing function and

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.

## This note was uploaded on 04/02/2010 for the course M 408d taught by Professor Sadler during the Fall '07 term at University of Texas.

### Page1 / 2

quiz_sl1 - M408D Quiz Solution You must show your work to...

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

View Full Document
Ask a homework question - tutors are online