# HW11 - momin(rrm497 Homework 11 cheng(58520 This print-out...

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momin (rrm497) – Homework 11 – cheng – (58520)1Thisprint-outshouldhave19questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.00110.0 points
in summation notation.00210.0 pointsIf thenthpartial sum ofn=1anis given bySn=4n+ 1n+ 4,what isanwhenn2?
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momin (rrm497) – Homework 11 – cheng – (58520)2Determine whether the seriesparenrightbiggnis convergent or divergent, and if convergent,find its sum.Determine whether the infinite seriessummationdisplayn=03 (cos)parenleftbigg131.convergent with sum492.convergent with sum923.convergent with sum944.divergent5.convergent with sum926.convergent with sum94correctExplanation:Sincecos= (1)n,the given series can be rewritten as an infinitegeometric seriessummationdisplayn=03parenleftbigg13parenrightbiggn=summationdisplayn=0a rnin whicha= 3,r=13.But the seriesn=0arnis(i) convergent with suma1rwhen|r|<1,and(ii) divergent when|r| ≥1.Consequently, the given series isconvergent with sum94.004summationdisplayn=13(n+ 1)2n(n+ 2)converges or diverges, and if converges, findits sum.1.converges with sum = 32.converges with sum =343.divergescorrect4.converges with sum =385.converges with sum =32Explanation:By the Divergence Test, an infinite seriesnandiverges whenlimn→∞annegationslash= 0.Now, for the given series,an=3(n+ 1)2n(n+ 2)=3n2+ 6n+ 3n2+ 2n.But then,limn→ ∞an= 3negationslash= 0.Consequently, the Divergence Test says thatthe given seriesdiverges.keywords: infinite series, Divergence Test, ra-tional function00510.0 pointsDetermine whether the infinite seriessummationdisplayn=1tan1parenleftBig2n225n+ 1parenrightBig
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momin (rrm497) – Homework 11 – cheng – (58520)3converges or diverges, and if it converges, findits sum.1.converges with sum =π22.divergescorrect3.converges with sum =25π4.converges with sum =18π5.converges with sum =15πExplanation:An infinite seriesnandiverges whenlimn→ ∞annegationslash= 0.For the given seriesan= tan1parenleftBig2n225n+ 1parenrightBig= tan1parenleftBig2n25 + 1/nparenrightBigButlimn→∞tan
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