This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Rehman (aar638) – HW13 – sachse – (56620) 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine whether the following series ( A ) ∞ summationdisplay n =1 2 ln(4 n ) n 2 , ( B ) ∞ summationdisplay n = 1 1 + cos(4 n ) n 2 + 1 converge or diverge. 1. A diverges, B converges 2. both series diverge 3. A converges, B diverges 4. both series converge correct Explanation: ( A ) The function f ( x ) = 2 ln 4 x x 2 is continous and positive on [ 1 2 , ∞ ); in addi tion, since f ′ ( x ) = 2 parenleftbigg 1 − 2 ln4 x x 3 parenrightbigg < on [ 1 2 , ∞ ), 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 = 2 bracketleftBig − ln(4 x ) x − 1 x bracketrightBig t 1 , and so integraldisplay ∞ 1 f ( x ) dx = 2(1 + ln 4) . The Integral Test thus ensures that series ( A ) converges . ( B ) Note first that the inequalities ≤ 1 + cos(4 n ) n 2 + 1 ≤ 2 n 2 + 1 ≤ 2 n 2 hold for all n ≥ 1. On the other hand, by the pseries test the series ∞ summationdisplay n = 1 1 n 2 is convergent since p = 2 > 1. Thus, by the comparison test, series ( B ) converges . keywords: 002 10.0 points Determine the convergence or divergence of the series ( A ) ∞ summationdisplay k =2 k 2(ln k ) 2 , and ( B ) ∞ summationdisplay k = 1 tan − 1 k 1 + k 4 . 1. A diverges, B converges correct 2. both series converge 3. both series diverge 4. A converges, B diverges Explanation: ( A ) By the Divergence Test, a series ∞ summationdisplay k = N a k Rehman (aar638) – HW13 – sachse – (56620) 2 will be divergent for each fixed choice of N if lim k →∞ a k negationslash = 0 since it is only the behaviour of a k as k → ∞ that’s important. Now, for the given series, N = 2 and a k = k 2(ln k ) 2 . But by L’Hospital’s Rule applied twice, lim x →∞ x (ln x ) 2 = lim x →∞ 1 (2 ln x ) /x = lim x →∞ x 2 ln x = lim x →∞ 1 2 /x = ∞ . By the Divergence Test, therefore, series ( A ) diverges . ( B ) We apply the Limit Comparison Test with a k = tan − 1 k 1 + k 4 , b k = 1 k 4 . For lim k →∞ k 4 parenleftBig tan − 1 k 1 + k 4 parenrightBig = lim k →∞ tan − 1 k = π 2 . Thus the given series ∞ summationdisplay k = 1 tan − 1 k 1 + k 4 is convergent if and only if the series ∞ summationdisplay k = 1 1 k 4 is convergent. But by the pseries test, this last series converges because p = 4 > 1. Con sequently, series ( B ) converges . 003 10.0 points If a m , b m , and c m satisfy the inequalities < a m ≤ c m ≤ b m , for all m , what can we say about the series ( A ) : ∞ summationdisplay m =1 a m , ( B ) : ∞ summationdisplay m = 1 b m if we know that the series ( C ) : ∞ summationdisplay m =1 c m is divergent but know nothing else about a m and b m ?...
View
Full
Document
This note was uploaded on 02/01/2011 for the course MATH 408L taught by Professor Gogolev during the Fall '09 term at University of Texas.
 Fall '09
 GOGOLEV
 Calculus, Inequalities

Click to edit the document details