# 116 - MATH 153 Selected Solutions for 11.6 Exercise 10...

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

Selected Solutions for § 11.6 Exercise 10 : Determine whether the series is absolutely convergent, condi- tionally convergent or divergent. X n =1 ( - 1) n n n 3 + 2 Solution : First, we can say that the series does not converge absolutely by comparison with n n 3 / 2 = 1 n 1 / 2 : lim n →∞ n n 3 +2 1 n 1 / 2 = lim n →∞ n · n 1 / 2 n 3 + 2 · 1 /n 3 / 2 1 /n 3 / 2 = lim n →∞ 1 p 1 + 2 /n 3 = 1 1 + 0 = 1 . Thus, since 1 /n 1 / 2 diverges, so does n n 3 +2 . The series does pass the Alternating Series Test, though; let b n = n n 3 +2 . Then b n 0 as n → ∞ . It remains to show that b n decreases. We know that d dx x x 3 + 2 = x 3 + 2 - x ( 1 2 3 x 2 x 3 +2 ) x 3 + 2 . As the denominator is positive for all x > 0, if we can show the numerator is negative we’ll be done. But x 3 + 2 - x ± 1 2 3 x 2 x 3 + 2 ² < 0 x 3 + 2 - 3 2 x 3 < 0 2 - 1 2 x 3 < 0 3 4 < x. So the

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 ]}

### Page1 / 2

116 - MATH 153 Selected Solutions for 11.6 Exercise 10...

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

View Full Document
Ask a homework question - tutors are online