hw 5 - rhodes(ajr2283 HW05 Gilbert(56380 This print-out...

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rhodes (ajr2283) – HW05 – Gilbert – (56380)1Thisprint-outshouldhave21questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.00110.0 pointsDetermine whether the series3233+3435+36. . .is conditionally convergent, absolutely con-vergent or divergent.1.series is conditionally convergentcor-rect2.series is absolutely convergent3.series is divergentExplanation:In summation notation,3233+3435+36. . .=summationdisplayn=1(1)n1f(n),withf(x) =3x+ 1.Now the improper integralintegraldisplay1f(x)dx=integraldisplay13x+ 1dxis divergent, so by the Integral Test, the givenseries is not absolutely convergent.On theother hand,f(n) =3n+ 1>3n+ 1 + 1=f(n+ 1),whilelimn→∞3n+ 1= 0.Consequently, by the Alternating Series Test,the given seriesis conditionally convergent.1.summationdisplayn=1(1)363 +n2.summationdisplayn=1(1)n15 +n6 +n3.summationdisplayn=1(1)n15 +ncorrect4.summationdisplayn=1(1)2n63 +n5.summationdisplayn=136 +nExplanation:Sincesummationdisplayn=1(1)363 +n=summationdisplayn=163 +n,use of the Limit Comparison andp-seriesTests withp=12shows that this series isdivergent. Similarly, sincesummationdisplayn=1(1)2n63 +n=summationdisplayn=163 +n,the same argument shows that this series aswell assummationdisplayn=136 +nis divergent.On the other hand, by the Divergence Test,the seriessummationdisplayn=1(1)n15 +n6 +n
keywords: alternating series, Alternating se-ries test, conditionally convergent, absolutelyconvergent, divergent00210.0 pointsWhich one of the following series is conver-gent?
rhodes (ajr2283) – HW05 – Gilbert – (56380)2is divergent becauselimn→ ∞(1)n15 +n6 +nnegationslash= 0.This leaves only the seriessummationdisplayn=1(1)n15 +n.To see that this series is convergent, setbn=15 +n.Then(i)bn+1bn,(ii)limn→ ∞bn= 0.Consequently, by the Alternating Series Test,the seriessummationdisplayn=1(1)n15 +nis convergent.00310.0 pointsDetermine whether the seriessummationdisplayn=1(1)n1e1/n6nis absolutely convergent, conditionally con-vergent or divergent.1.absolutely convergent2.conditionally convergentcorrect3.divergentExplanation:Sincesummationdisplayn=1(1)n1e1/n6n=16summationdisplayn=1(1)ne1/nn,we have to decide if the seriessummationdisplayn=1(1)ne1/nnis absolutely convergent, conditionally con-vergent, or divergent.First we check for absolute convergence.Now, sincee1/n1 for alln1,e1/n6n16n>0.But by thep-series test withp= 1, the seriessummationdisplayn=116ndiverges, and so by the Comparison Test, theseriessummationdisplayn=1e1/n6ntoo diverges; in other words, the given seriesis not absolutely convergent.To check for conditional convergence, con-sider the seriessummationdisplayn=1(1)nf(n)wheref(x) =e1/x6x.Thenf(x)>0 on (0,). On the other hand,f(x) =16x3e1/xe1/x6x2=e1/xparenleftBig1 +x6x3parenrightBig.Thus

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