# A carefully determine the convergence of the series n

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Chapter 11 / Exercise 12
Calculus
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9.(1 point) (a)Carefullydeterminetheconvergenceoftheseriesn. The series isSolution:SOLUTIONNote that for both of these series the terms alternate in sign,n=1(-1)n5A. absolutely convergentB. conditionally convergentC. divergent(b)Carefullydeterminetheconvergenceoftheseriesn=1(-1)n5n. The series isA. absolutely convergentB. conditionally convergentC. divergentand the magnitudes of successive terms decrease, so that we areable to apply the alternating series test.(a)Forn=1(-1)n5n, we have limn15n=0. Thus, by the alternat-ing series test, we know that this series converges (it is at leastconditionally convergent). Becausen=115ndiverges (because itis a constant times ap-series withp=1), is only conditionally,not absolutely, convergent.(b)Forn=1(-1)n5n, we have limn15n=0. Thus, by the alter-nating series test, we know that this series converges (it is atleast conditionally convergent). Becausen=115nalso converges(by comparison to ap-series or by the integral test), our originalseries must be not only conditionally convergent but absolutelyconvergent.Correct Answers:BA 10.(1 point) For each of the series below select the letterfrom a to c that best applies and the letter from d to k that bestapplies. A possible answer is af, for example. 1.n=16+sin(n)n2.n=1(2n+2)!(n!)23.n=1cos(nπ)nπ4.n=11nn5.n=1cos2(nπ)nπ6.n=21nlog(4+n)Correct Answers:cgcjbdae 3
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Chapter 11 / Exercise 12
Calculus
Stewart Expert Verified
ceck 11.(1 point) Match each of the following with the correctstatement. 1.n=1(n+1)(62-1)n62n2.n=1(-1)nnn+23.n=1(-1)n4n+64.n=1(-6)nn65.n=1sin(6n)n2Correct Answers:ACCDA
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