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-ries∞summationdisplayn=1xn√n+ 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=xn√n+ 5,thenvextendsinglevextendsinglevextendsinglevextendsinglean+1anvextendsinglevextendsinglevextendsinglevextendsingle=vextendsinglevextendsinglevextendsinglevextendsinglexn+1√n+ 6√n+ 5xnvextendsinglevextendsinglevextendsinglevextendsingle=|x|parenleftbigg√n+ 5√n+ 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=11√n+ 5.Applying the Integral Test withf(x) =1√x+ 5we see thatfis continuous, positive, and de-creasing on [1,∞), but the improper integralI=integraldisplay∞1f(x)dxdiverges, so the infinite series (∗) divergesalso.On the other hand, atx=−1, the seriesbecomes(‡)∞summationdisplayn=1(−1)n√n+ 5.which is an alternating series∞summationdisplayn=1(−1)nan,an=f(n)withf(x) =1√x+ 5the same continuous, positive and decreasingfunction as before. Sincelimx→ ∞f(x) =limx→ ∞1√x+ 5= 0,however, the Alternating Series Test ensuresthat (‡) converges.Consequently, theinterval of convergence = [−1,1).00210.0pointsDetermine the radius of convergence,R, ofthe series∞summationdisplayn=1(−2)nn+ 3xn.