# EXAM3 - Version 013 Exam 3 cheng(58520 This print-out...

This preview shows pages 1–3. Sign up to view the full content.

Version 013 – Exam 3 – cheng – (58520) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Determine iF the sequence { a n } converges when a n = (2 n 1)! (2 n + 1)! , and iF it converges, fnd the limit. 1. converges with limit = 4 2. converges with limit = 1 4 3. converges with limit = 0 correct 4. converges with limit = 1 5. does not converge Explanation: By defnition, m ! is the product m ! = 1 . 2 . 3 . . . . . m oF the frst m positive integers. When m = 2 n 1, thereFore, (2 n 1)! = 1 . 2 . 3 . . . . (2 n 1) , while (2 n + 1)! = 1 . 2 . 3 . . . . . (2 n 1)2 n (2 n + 1) . when m = 2 n + 1. But then, (2 n 1)! (2 n + 1)! = 1 2 n (2 n + 1) . Consequently, the given sequence converges with limit = 0 . 002 10.0 points Determine iF the sequence { a n } converges, and iF it does, fnd its limit when a n = 4 n + ( 1) n 2 n + 4 . 1. converges with limit = 5 2 2. converges with limit = 2 3 3. converges with limit = 2 correct 4. converges with limit = 3 2 5. sequence does not converge Explanation: AFter division by n we see that a n = 4 + ( 1) n n 2 + 4 n . But ( 1) n n , 4 n −→ 0 as n → ∞ , so a n 2 as n → ∞ . Conse- quently, the sequence converges and has limit = 2 . 003 10.0 points ±ind all values oF x For which the series s n = 0 (3 x 5) n 4 n converges. 1. converges only on p 1 3 , 1 3 P 2. converges only on p 3 , 3 P 3. converges only on p 3 , 1 3 P

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

View Full Document
Version 013 – Exam 3 – cheng – (58520) 2 4. converges only on p 1 3 , 3 P correct 5. converges only on p 5 3 , 5 3 P Explanation: A geometric series n = 0 ar n (i) converges when | r | < 1, and (ii) diverges when | r | ≥ 1, Now for the given series, a = 1 , r = 3 x 5 4 , so it converges exactly when 1 < 3 x 5 4 < 1 . By solving these inequalities we thus see that the given series converges only on the interval p 1 3 , 3 P . 004 10.0 points First ±nd a n so that s n =1 a n = 6 + 3 2 + 2 3 + 3 4 + 6 5 5 + . . . and then determine whether the series con- verges or diverges. 1. a n = 3 2 n 3 / 2 , series converges 2. a n = 6 n 3 / 2 , series converges correct 3. a n = 3 2 n 3 / 2 , series diverges 4. a n = 6 n 1 / 2 , series diverges 5. a n = 6 n 3 / 2 , series diverges 6. a n = 6 n 1 / 2 , series converges Explanation: By inspection a n = 6 n 3 / 2 . To test for convergence we use the Integral test with f ( x ) = 6 x 3 / 2 . This is a positive, continuous, decreasing function on [1 , ). Furthermore, i 1 f ( x ) dx = lim n →∞ i n 1 6 x 3 / 2 dx = lim n →∞ b 12 x 1 / 2 B n 1 = 12 , so the improper integral i 1 f ( x ) dx is convergent, which by the Integral test means that the series converges .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 11

EXAM3 - Version 013 Exam 3 cheng(58520 This print-out...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online