# HW14-solutions - thai (jvt293) HW14 he (53725) This...

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thai (jvt293) – HW14 – he – (53725)1Thisprint-outshouldhave25questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.00110.0pointsFind the interval of convergence of the se-riessummationdisplayn=1xnn+ 5.1.interval of cgce = [1,1]2.converges only atx= 03.interval of cgce = [1,1)correct4.interval of cgce = (1,1)5.interval of cgce = [5,5]6.interval of cgce = (5,5]Explanation:Whenan=xnn+ 5,thenvextendsinglevextendsinglevextendsinglevextendsinglean+1anvextendsinglevextendsinglevextendsinglevextendsingle=vextendsinglevextendsinglevextendsinglevextendsinglexn+1n+ 6n+ 5xnvextendsinglevextendsinglevextendsinglevextendsingle=|x|parenleftbiggn+ 5n+ 6parenrightbigg=|x|radicalbiggn+ 5n+ 6.Butlimn→ ∞n+ 5n+ 6= 1,solimn→ ∞vextendsinglevextendsinglevextendsinglevextendsinglean+1anvextendsinglevextendsinglevextendsinglevextendsingle=|x|.By the Ratio Test, therefore, the given series(i) converges when|x|<1,(ii) diverges when|x|>1.We have still to check what happens at theendpointsx=±1.Atx=1 the seriesbecomes()summationdisplayn=11n+ 5.Applying the Integral Test withf(x) =1x+ 5we see thatfis continuous, positive, and de-creasing on [1,), but the improper integralI=integraldisplay1f(x)dxdiverges, so the infinite series () divergesalso.On the other hand, atx=1, the seriesbecomes()summationdisplayn=1(1)nn+ 5.which is an alternating seriessummationdisplayn=1(1)nan,an=f(n)withf(x) =1x+ 5the same continuous, positive and decreasingfunction as before. Sincelimx→ ∞f(x) =limx→ ∞1x+ 5= 0,however, the Alternating Series Test ensuresthat () converges.Consequently, theinterval of convergence = [1,1).00210.0pointsDetermine the radius of convergence,R, ofthe seriessummationdisplayn=1(2)nn+ 3xn.
00310.0points
thai (jvt293) – HW14 – he – (53725)3But asn→ ∞, neither of4n,(1)n4n−→0holds, so both ofsummationdisplayn=14n,summationdisplayn=1(1)n4ndiverge. Consequently, theinterval of convergence = (1,1).

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