solution_pdf2 - suleimenov(bs26835 HW 02 rusin(55565 This...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
suleimenov (bs26835) – HW 02 – rusin – (55565) 1 This print-out should have 20 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 = 1 n ln p 5 2 n + 4 P , and iF it does, fnd its limit. 1. limit = ln 5 6 2. limit = ln 5 2 3. limit = ln 2 4. limit = 0 correct 5. the sequence diverges Explanation: AFter division by n we see that 5 2 n + 4 = 5 n 2 + 4 n , so by properties oF logs, a n = 1 n ln 5 n 1 n ln p 2 + 4 n P . But by known limits (or use L’Hospital), 1 n ln 5 n , 1 n ln p 2 + 4 n P −→ 0 as n → ∞ . Consequently, the sequence { a n } converges and has limit = 0 . 002 10.0 points Determine iF { a n } converges when a n = 5 n 3 2 n 2 , and iF it does, fnd its limit. 1. limit = + 2 3 2. limit = 2 3. limit = 5 4. sequence diverges correct 5. limit = 3 Explanation: AFter simplifcation, a n = 5 n n 2 n = n ( 5 n 2 ) , so a n → ∞ as n → ∞ . But then the sequence is unbounded, and hence diverges . 003 10.0 points Determine iF the sequence { a n } converges, and iF it does, fnd its limit when a n = ± n + 1 n 1 ² 3 n . 1. limit = 1 2. does not converge 3. limit = e 3 4. limit = e 6 5. limit = e 6 correct 6. limit = e 3 Explanation: By the Laws oF Exponents, a n = ± n 1 n + 1 ² 3 n = ± 1 1 n ² 3 n ± 1 + 1 n ² 3 n .
Background image of page 1

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

View Full Document Right Arrow Icon
suleimenov (bs26835) – HW 02 – rusin – (55565) 2 But lim n →∞ p 1 + x n P n = e x . Consequently, by Properties of Limits the given limit exists and limit = e 6 . 004 10.0 points Determine whether the sequence { a n } con- verges or diverges when a n = ( 1) n 1 n n 2 + 5 , and if it converges, Fnd the limit. 1. converges with limit = 1 5 2. sequence diverges 3. converges with limit = 5 4. converges with limit = 0 correct 5. converges with limit = 1 5 6. converges with limit = 5 Explanation: After division, a n = ( 1) n 1 n n 2 + 5 = ( 1) n 1 n + 5 n . Consequently, 0 ≤ | a n | = 1 n + 5 n 1 n . But 1 /n 0 as n → ∞ , so by the Squeeze theorem, lim n | a n | = 0 . But −| a n | ≤ a n ≤ | a n | , so by the Squeeze theorem again the given sequence { a n } converges and has limit = 0 . keywords: 005 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 = 0 correct 2. converges with limit = 1 3. converges with limit = 1 4 4. converges with limit = 4 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 .
Background image of page 2
suleimenov (bs26835) – HW 02 – rusin – (55565) 3 006 10.0 points Determine whether the sequence { a n } con- verges or diverges when a n = 2 + cos 2 n 5 + 2 n . 1. diverges 2. converges with limit = 0 correct 3. converges with limit = 1 3 4. converges with limit = 2 5 5. converges with limit = 2 Explanation: Since 0 cos 2 n 1 , we see that 2 5 + 2 n a n 3 5 + 2 n .
Background image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 10

solution_pdf2 - suleimenov(bs26835 HW 02 rusin(55565 This...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online