# Keywords 008 100 points determine whether the series

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keywords:00810.0 pointsDetermine whether the seriessummationdisplayn=3(1)n+26nlnnconverges conditionally, converges absolutely,or diverges.1.series converges conditionallycorrect2.series diverges3.series converges absolutelyExplanation:The given series can be rewritten assummationdisplayn=3(1)n+26nlnn= 6summationdisplayn=3(1)nf(n),wheref(x) =1xlnx= (xlnx)1.Now by the Chain and Product Rules,f(x) =1x(lnx)2parenleftBiglnx2x+xxparenrightBigWhenx >3, therefore,f(x) =1(lnx)2parenleftBiglnx+ 22xxparenrightBig<0,sof(x) is positive and decreasing for allx3.But the improper integralintegraldisplay3f(x)dx=integraldisplay31xlnxdxdoes not converge.By the Integral Test,therefore, the given series is not absolutelyconvergent. It could still converge condition-ally, however.To use the Alternating SeriesTest we have to check that(i)f(n)> f(n+ 1) forn3,(ii) limx→∞f(x) = 0.Sincefis decreasing for allx3, we see thatn3=f(n)> f(n+ 1).On the other hand,limx→ ∞1xlnx= 0,solimx→ ∞f(x) = 0.Consequently, by the Alternating series Test,the given series isconditionally convergent.00910.0 pointsDetermine whether the seriessummationdisplayn=02ln(n+ 4)cosparenleftBigparenrightBig
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tgo72 – Homework 12 – Gompf – (58370)7converges or diverges. (Similar such series arebasic to the mathematical description of allmusical sounds!)1.series diverges2.series convergescorrectExplanation:The basic idea is that cosπ2xis a peri-odic function taking values between1 and1. This will enable us to write the given seriesas an alternating series. Indeed,cos 0 = 1,cosπ2= 0,cosπ=1,cos3π2= 0,and so on for higher multiples ofπ2. Thus thegiven series can be written as2braceleftBig1ln 41ln 6+1ln 81ln 10+. . .bracerightBig.In summation notation, therefore, the seriesbecomes the alternating seriessummationdisplayn=0(1)n2ln(2n+ 4)= 2summationdisplayn=0(1)nf(n)wheref(x) =1ln(2x+ 4).Butf(n) =1ln(2n+ 4)>1ln(2n+ 6)=f(n+ 1),whilelimn→ ∞f(n) =limn→ ∞1ln(2n+ 4)= 0.Applying the Alternating Series Test, we nowsee that the given seriesconverges.Determine if the seriessummationdisplayn=2parenleftBign6parenrightBig6nconverges or diverges.1.series converges2.series divergescorrectExplanation:SinceparenleftBign6parenrightBig6n= (1)6nparenleftBign6parenrightBig6n,the Divergence Test will ensure that the givenseries diverges if we show thatlimn→ ∞parenleftBign6parenrightBig6nnegationslash= 0.But, by L’Hospital’s Rule,limx→ ∞6xlnparenleftBigx6parenrightBig=,so certainly,limn→ ∞parenleftBign6parenrightBig6n=.Consequently, the given seriesdiverges.01110.0 pointsDetermine all values ofpfor which the seriessummationdisplayn=2(1)n1(lnn)p7nis convergent, expressing your answer in in-terval notation.01010.0 points