Exam 3, Fall '04, Solutions

# Exam 3, Fall '04, Solutions - Tufts University Math 12...

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

Unformatted text preview: Tufts University Math 12 Department of Mathematics Exam 3 Solutions 1. (a) a n = 8 n n x 3 n . Therefore, lim n →∞ ¯ ¯ ¯ ¯ a n +1 a n ¯ ¯ ¯ ¯ = lim n →∞ ¯ ¯ ¯ ¯ 8 n +1 x 3( n +1) n + 1 · n 8 n x 3 n ¯ ¯ ¯ ¯ = 8 | x | 3 lim n →∞ n n + 1 = 8 | x | 3 . By the ratio test, ∞ X n =1 a n converges if 8 | x | 3 < 1 ⇐⇒ | x | < 1 2 . Therefore, the radius of convergence is 1 2 . If x = 1 2 , ∞ X n =1 a n = ∞ X n =1 1 n , the divergent harmonic series. If x =- 1 2 , ∞ X n =1 a n = ∞ X n =1 (- 1) n n , the convergent alternating harmonic series. Therefore, the interval of convergence is [- 1 2 , 1 2 ) . (b) a n = (2 n +1)! 3 n ( x + 2) n . Therefore, lim n →∞ ¯ ¯ ¯ ¯ a n +1 a n ¯ ¯ ¯ ¯ = lim n →∞ ¯ ¯ ¯ ¯ (2( n + 1) + 1)!( x + 2) n +1 3 n +1 · 3 n (2 n + 1)!( x + 2) n ¯ ¯ ¯ ¯ = | x + 2 | lim n →∞ (2 n + 3)(2 n + 2) 3 = ∞ if x 6 =- 2 ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

Exam 3, Fall '04, Solutions - Tufts University Math 12...

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

View Full Document
Ask a homework question - tutors are online