solution_pd3f - suleimenov(bs26835 – HW03 – rusin...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: suleimenov (bs26835) – HW03 – rusin – (55565) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Sorry to be late getting these questions posted – I took my own advice to go play in the snow, and I hope you did too! Due date is still Wednesday evening. 001 10.0 points Determine whether the following series ( A ) ∞ summationdisplay k =1 4 ln(2 k ) k 2 , ( B ) ∞ summationdisplay k =1 1 + sin(2 k ) k 2 + 1 converge or diverge. 1. A diverges, B converges 2. both series converge correct 3. both series diverge 4. A converges, B diverges Explanation: ( A ) The function f ( x ) = 4 ln 2 x x 2 is continous and positive on [1 , ∞ ); in addi- tion, since f ′ ( x ) = 4 parenleftbigg 1 − 2 ln2 x x 3 parenrightbigg < on [1 , ∞ ), f is also decreasing on this inter- val. This suggests applying the Integral Test. Now, after Integration by Parts, we see that integraldisplay t 1 f ( x ) dx = 4 bracketleftBig − ln(2 x ) x − 1 x bracketrightBig t 1 , and so integraldisplay ∞ 1 f ( x ) dx = 4(1 + ln 2) . The Integral Test thus ensures that series ( A ) converges . ( B ) Note first that the inequalities ≤ 1 + sin(2 k ) k 2 + 1 ≤ 2 k 2 + 1 ≤ 2 k 2 hold for all n ≥ 1. On the other hand, by the p-series test the series ∞ summationdisplay k = 1 1 k 2 is convergent since p = 2 > 1. Thus, by the comparison test, series ( B ) converges . keywords: 002 10.0 points Which of the following series are convergent: A . ∞ summationdisplay n = 1 2 n 2 / 3 B . 1 + 1 √ 2 + 1 √ 3 + 1 2 + 1 √ 5 + . . . C . ∞ summationdisplay n =1 3 n 2 + 1 1. A and C only 2. A only 3. A and B only 4. B and C only suleimenov (bs26835) – HW03 – rusin – (55565) 2 5. none of them 6. all of them 7. C only correct 8. B only Explanation: By the Integral test, if f ( x ) is a positive, decreasing function, then the infinite series ∞ summationdisplay n =1 f ( n ) converges if and only if the improper integral integraldisplay ∞ 1 f ( x ) dx converges. Thus for the three given series we have to use an appropriate choice of f . A. Use f ( x ) = 2 x 2 / 3 . Then integraldisplay ∞ 1 f ( x ) dx is divergent. B. Use f ( x ) = 1 x 1 / 2 . Then integraldisplay ∞ 1 f ( x ) dx is divergent. C. Use f ( x ) = 3 x 2 + 1 . Then integraldisplay ∞ 1 f ( x ) dx is convergent (tan − 1 integral). keywords: convergent, Integral test, 003 10.0 points Determine whether the series ∞ summationdisplay n = 2 n 5(ln n ) 2 is convergent or divergent. 1. divergent correct 2. convergent Explanation: By the Divergence Test, a series ∞ summationdisplay n = N a n will be divergent for each fixed choice of N if lim n →∞ a n negationslash = 0 since it is only the behaviour of a n as n → ∞ that’s important. Now, for the given series, N = 2 and a n = n 5(ln n ) 2 ....
View Full Document

This note was uploaded on 03/20/2011 for the course M 408 C taught by Professor Treisman during the Spring '07 term at University of Texas.

Page1 / 11

solution_pd3f - suleimenov(bs26835 – HW03 – rusin...

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

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