tapia (jat4858) – HW09 – clark – (52990) 1This print-out should have 22 questions.Multiple-choice questions may continue o nthe next column or page – find all choicesbefore answering.001 10.0 pointsDetermine whether the series∞Xn = 1(−1)n+1e1/n4nis absolutely convergent, conditionally con-vergent or divergent.1. divergent2. conditional ly convergent correct3. absolutely convergentExplanation:Since∞Xn = 1(−1)n+1e1/n4n= −14∞Xn = 1(−1)ne1/nn,we have to decide if the series∞Xn = 1(−1)ne1/nnis absolutely convergent, conditionally con-vergent, or divergent.First we check for absolute convergence.Now, since e1/n≥ 1 for all n ≥ 1,e1/n4n≥14n> 0 .But by the p-series test with p = 1, the series∞Xn = 114ndiverges, and so by the Comparison Test, theseries∞Xn = 1e1/n4ntoo diverges; in other words, the given seriesis not absolutely convergent.To check for conditional convergence, con-sider the series∞Xn = 1(−1)nf(n)wheref(x) =e1/x4x.Then f(x) > 0 on (0, ∞). On the other hand,f′(x) = −14x3e1/x−e1/x4x2= −e1/x1 + x4x3.Thus f′(x) < 0 on (0, ∞), so f(n) > f(n + 1)for all n. Finally, sincelimx → ∞e1/x= 1 ,we see that f(n) → 0 as n → ∞. By theAlternating Series Test, therefore, the series∞Xn = 1(−1)nf(n)is convergent.Consequently, the given series isconditionally convergent .keywords:002 10.0 pointsTo apply the ratio test to the infinite seriesXnan, the value ofλ = limn → ∞an+1anhas to be determined.Compute λ for the series∞Xn=19nn!nn.tapia (jat4858) – HW09 – clark – (52990) 21. λ =92 e2. λ = 03. λ = 94. λ =9ecorrect5. λ =92Explanation:Since(n + 1)!n!=1 ·2 · . . . · n · (n + 1)1 · 2 · . . . ·n= n + 1 ,while(n + 1)n+1nn=n + 1nn(n + 1)=1 +1nn(n + 1) ,we see thatan+1an= 9n + 1(n + 1)1 +1nn=91 +1nnButlimn → ∞1 +1nn= e .Consequently, for the given series,λ =9e.003 10.0 pointsDetermine whether the series∞Xn = 1(−1)n+24√nis absolutely convergent, conditionally con-vergent, or divergent.1. absolutely convergent2. conditionally convergent correct3. divergentExplanation:By the Alternating Series test, the series∞Xn = 1(−1)n+24√nconverges. On the other hand, by the p-seriestest with p =14≤ 1, the series∞Xn=114√nis divergent. Consequently, the series isconditionally convergent .004 10.0 pointsDetermine whether the series∞Xn = 3(−1)n−1ln n6n − 1converges conditionally, converges absolutel y,or diverges.1. diverges2. converges conditionally correct3. converges absolutelyExplanation:The given series can be written as∞Xn = 3(−1)n−1ln n6n − 1= −∞Xn = 3(−1)nf(n) ,wheref(x) =ln x6x − 1.tapia (jat4858) – HW09 – clark – (52990) 3Butlimx → ∞xf(x)ln x=16,so by the Limit Comparison Test and theIntegral Test applied to the seriesXn = 3ln nn,the given series is not absolutely convergent.On the other hand, by the Alternating SeriesTest, it will converge conditionally if(i) f(n) > f(n + 1) for n ≥ 3,(ii) limx→∞f(x) = 0 .Nowf′(x) =6x − 1x− 6 l n x(6x −1)2=6(1 −ln x) −1x(6x − 1)2.When x > 3, therefore,f′(x) = −1 + 6x(ln x − 1)x(6x − 1)2< 0 ,so f(x) is decreasing for all x ≥ 3. Thusn ≥ 3 =⇒ f(n) > f(n + 1) .To determine the limit of f(x) as x → ∞ weuse L’Hospital’s Rule; for thenlimx → ∞ln x6x − 1= limx → ∞16x= 0 ,solimx → ∞f(x) = 0 .Consequently, the given series isconditionally convergent .005 10.0 pointsDetermine whether the seriesI.∞Xn = 1(n!)2(2n)!3n,II.∞Xn = 112nn + 1nn2,converge or diverge.1. only series II converges2. both series converge3. only series I converges correct4. both series divergeExplanation:I. Since the series has the form∞Xn = 1an, an=(n!)2(2n)!3n,we apply the Ratio Test. For thenan+1an=3(n + 1)! (n + 1)! (2n) !n! n! (2n + 2)!=3(n + 1 )(n + 1)(2n + 1 )(2n + 2)=3(n2+ 2n + 1)4n2+ 6n + 2,in which caselimn → ∞an+1an=34< 1 .By the Ratio Test, therefore, the given seriesconverges .II. Since the series has the form∞Xn = 1an, an=12nn + 1nn2,we apply the Root Test. For then(an)1/n=12n + 1nn.tapia (jat4858) – HW09 – clark – (52990) 4Butlimn → ∞n + 1nn= limn → ∞1 +1nn= e ,in which case,limn → ∞(an)1/n=e2> 1 .By the Roo t Test, therefore, the given seriesdiverges .keywords: RatioRootTest, RatioRoo tTes-tExam,006 10.0 pointsDetermine whether the series1 −2!1 · 3+3!1 · 3 · 5−4!1 · 3 · 5 · 7+···+(−1)n−1n!1 · 3 · 5 ···(2n − 1)+ ···is absolutely convergent, conditionally con-vergent, or divergent.1. divergent2. conditionally convergent3. absolutely convergent correctExplanation:The given series is of the form∞Xn = 1anwherean= (−1)n−1n!1 · 3 · 5 ···(2n − 1).But thenan+1= (−1)n(n + 1)!1 · 3 · 5 ···(2(n + 1) − 1),and soan+1an= −(n + 1)!n!×1 · 3 · 5 ···(2n − 1)1 · 3 · 5 ···(2n −1) ·(2(n + 1) − 1)= −n + 12n + 1.In this case,limn→∞an+1an= limn→∞n + 12n + 1=12< 1 .Consequently, the given series isabsolutely convergent .007 10.0 pointsDetermine which, if any, of the seriesA.∞Xn = 13n5n + 5n45nB.∞Xn = 15n + 1n!are divergent.1. both of them2. B only3. A only4. neither of them correctExplanation:To check for divergence we shall use eitherthe Ratio test or the Root test which meanscomputing one or other oflimn → ∞an+1an, limn → ∞|an|1/nfor each of the given series.A. The root test is the bett er one to applybecause of the nthpowers. For then|an|1/n=453n5n + 5−→1225< 1tapia (jat4858) – HW09 – clark – (52990) 5as n → ∞, so the given series converges.B. The ratio test is the better one to usebecausean+1an=5n + 65n + 1n!(n + 1)!.Thusan+1an=5n + 6(5n + 1)(n + 1)−→ 0 < 1as n → ∞, so the given series converges.008 10.0 pointsWhich of the following infinite series con-verges conditionally?1.∞Xn = 17(−n)n(n + 3)n2.∞Xn = 1−75n3.∞Xk = 1(−1)k58k ln(k) + 7correct4.∞Xn = 1cos(nπ)3n8n + 5n5.∞Xm = 187m − 5Explanation:The question requires us to find which ofthe five series has the property thatXnanconverges butXn|an|does not converge.(i) In the caseak= (−1)k58k ln(k) + 7the alternating series test applies since58k ln(k) + 7≥58(k + 1) ln(k + 1) + 7andlimk → ∞58k ln(k) + 7= 0 .Thus∞Xk = 1(−1)k58k ln(k) + 7converges.On the other hand, if we setbk=1k ln(k),thenlimk → ∞bk|ak|=8k