# Vidal mav3385 hw14 schultz 53130 3 converges in the

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vidal (mav3385) – HW14 – schultz – (53130)3converges. In the first casean=parenleftbigg6n+ 74nparenrightbiggnparenleftbigg23parenrightbiggn=parenleftbigg6n+ 76nparenrightbiggn,in which caselimn→ ∞an=e7/6negationslash= 0;by the Divergence Test, therefore the seriessummationdisplayn=1andiverges. On the other hand,bn=parenleftbigg6n+ 74nparenrightbiggnparenleftbigg23parenrightbiggn= (1)nparenleftbigg6n+ 76nparenrightbiggn= (1)nan.But, as we have seen,limn→ ∞an=e7/6.Thusbnoscillates betweene7/6ande7/6asn→ ∞; in particular,limn→ ∞bndoes not exist. Again by the Divergence Test,therefore, the seriessummationdisplayn=1bndiverges.Consequently, the given power se-ries does not converge atx=±2/3 and soitsinterval of convergence =parenleftbigg23,23parenrightbigg.00510.0pointsDetermine the interval of convergence ofthe series1.parenleftbigg54,14parenrightbiggcorrect2.parenleftbigg14,14parenrightbigg3.(−∞,)4.series converges only atx=345.parenleftbigg14,54parenrightbiggExplanation:The given series has the formsummationdisplayk=0ck(xa)kwithck=k32k,a=34.But then,limk→ ∞ck+1ck=limk→ ∞2parenleftBigk+ 1kparenrightBig3= 2.By the Ratio Test, the series thus(i) converges when|xa|<12, and(ii) diverges when|xa|>12.Now at the pointsxa=12andxa=12the series reduces tosummationdisplayk=0k3,summationdisplayk=0(1)kk3respectively. But in both cases these series di-verge by the Divergence Test. Consequently,the interval of convergence of the given seriesisparenleftbigga12, a+12.
vidal (mav3385) – HW14 – schultz – (53130)400610.0pointsDetermine the interval of convergence ofthe infinite seriessummationdisplayk=0(1)k(k+ 1)!2k(8x+ 6)k.1.parenleftbigg1,12parenrightbigg2.series converges only atx=34correct3.(−∞,)4.parenleftbigg12,1parenrightbigg5.parenleftbigg12,12parenrightbiggExplanation:There are three possibilities for the intervalof convergence of the infinite seriessummationdisplayk=0ak(xa)k.First setL=limk→ ∞vextendsinglevextendsinglevextendsinglevextendsingleak+1akvextendsinglevextendsinglevextendsinglevextendsingle.Then(i) whenL>0, the interval of convergenceis given byparenleftbigga1L, a+1Lparenrightbigg;(ii) whenL= 0, the interval of convergenceis given by(−∞,);(iii) whenL=, the interval of conver-gence reduces to the point{a},i.e., the seriesconverges only atx=a.For the given series,summationdisplayk=0(1)k(k+ 1)!2k(8x+ 6)k=summationdisplayk=0(1)k(k+ 1)!8k2kparenleftbiggx+34parenrightbiggk,so the corresponding value ofakisak= (1)k8k(k+ 1)!2k.In this casevextendsinglevextendsinglevextendsinglevextendsingleak+1akvextendsinglevextendsinglevextendsinglevextendsingle=vextendsinglevextendsinglevextendsinglevextendsingle(k+ 2)!8k+12k+1·2k(k+ 1)!8kvextendsinglevextendsinglevextendsinglevextendsingle=8(k+ 2)2.

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